Knewton Alta Ch. 10

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H0:μd=0 Ha:μd>0 The researcher wants to test if the hypothesized mean of the differences for the paired data is greater than 0. Therefore, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd>0.

Suppose you computed r=−0.896 using n=28 data points. Using the critical values table below, determine if the value of r is significant or not.

r is significant because it is not between the positive and negative critical values. There are n−2=28−2=26 degrees of freedom. Looking at the table of critical values, the critical values corresponding to df=26 are −0.374 and 0.374. Since the value of r is less than −0.374, r is significant.

The following data set provides information on the lottery sales, proceeds, and prizes by year in Iowa. You decided to find the linear equation that corresponds to sales and year. Create a graph using the sales and year. Add the linear equation to the graph. What is the y-intercept of the linear equation? Round each value below to the nearest integer.

1$-1$−1 2$10$10 The y-intercept is −1E+10.

A linear equation, where x represents the number of years since 1990 and y represents after-tax corporate profits, contains the points (0,7) and (12,7). Plot the points in the graph below to determine the line.

The values for the number of years since 1990 correspond to x-values and the after-tax corporate profits correspond to y-values. (0,7) and (12,7) correspond to points (x,y) plotted in the graph. (0,7) is plotted along the y-axis and (12,7) is twelve units to the right and seven units up from the origin. The line is created by connecting these two points.

There is sufficient evidence at the 0.05 level of significance to conclude that the mean number of hours of sleep for pre-med students is less than the mean number of hours of sleep for pre-law students. Reject H0. The p-value of 0.002 is less than the level of significance, 0.05. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that the mean number of hours of sleep for pre-med students is less than the mean number of hours of sleep for pre-law students.

A linear equation, where x represents the number of years since 1990 and y represents after-tax corporate profits, contains the points (0,7) and (12,7). Given the line constructed in the graph above, what is the slope of the line?

Zero Remember that a graph has slope zero if it is a horizontal line. In other words, as the x values increase, the y values stay the same.

(a) The null and alternative hypotheses are: H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution. (b) χ20=13.167. (c) Determine the critical value using the following portion of the χ2-table.

$\text{Critical Value=}7.815$Critical Value=7.815 The degrees of freedom for a Goodness-of-Fit test is calculated as: df= (number of data categories) −1. So, df=4−1=3. We know df=3 and α=0.05, so, using the χ2-table, the critical value is χ20.05=7.815.

Dylan used the alternative hypothesis Ha:μ1−μ2<0 with α=0.01 and 75 degrees of freedom. What is/are the critical value(s) for this hypothesis test? Use the value(s) from the table. Use a comma and a space to separate answers as needed.

$\text{critical value(s)}=-2.377$critical value(s)=−2.377 Since this is a left-tailed test and α=0.01 with 75 degrees of freedom, the critical value occurs at −t0.01=−2.377.

ArtsHumanitiesSciencesRow TotalAthlete19.11414.41917.51851Non-Athlete17.92313.6916.51648Column Total37283499 In a chi-square test for independence, the expected frequency for each data cell (used in calculating the test statistic), can be found by using the formula:E=R⋅Cn...where R= the row total, C= the column total , and n= the total sample size. For example, the expected entry in the first row and first column (Athlete / Arts) isE=R⋅Cn=51⋅3799≈19.1

The following data set provides information on the lottery sales, proceeds, and prizes by year in Iowa. Create a graph using the sales and year. What approximate range of sales would you expect for the year 2017?

Between 375 and 400 million dollars The value for the last year is about 370 million dollars. You can expect growth in the next year to be between 375 and 400 million dollars.

A physician wants to determine if a supplement is effective in helping men lose weight. She takes a random sample of overweight men and records their weight before the trial. She then prescribes the supplement and instructs them to take it for four weeks while making no other lifestyle changes. After the four-week period, she records the weight of the men again. Identify the null and alternative hypotheses for a hypothesis test where d=weight after trial−weight before trial.

H0:μd=0 Ha:μd<0 The physician wants to test if the hypothesized mean of the differences for the paired data is less than 0. Therefore, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd<0.

The United Nations stores data about direct use of solar thermal heat around the world. The following data set gives the US information for direct use of solar thermal heat, consumption by households. Plot the data to identify the outliers, or examine the data. If the 3 outliers are removed, what is the slope of the new linear equation? Round to nearest thousandth.

$-85.495$−85.495 By taking out the 2012, 2013, and 2014 outliers, the slope changes from −2887.7 to −85.495. Outliers are points that have strong pulls on the line of best fit, thus greatly affecting predictions researchers want to make with the equation of that line.

In a sample of 35 randomly selected vehicles in City A, 7 use alternative energy sources. In a sample of 45 randomly selected vehicles in City B, 17 use alternative energy sources. Test the alternative hypothesis that the population proportion for City A is different from the population proportion for City B. The test statistic is z=−1.72. What is the corresponding p-value? Compute your answer using a value from the table below.

$0.086$0.086 Since the alternative hypothesis is that the population proportion for City A is different from the population proportion for City B, this is a two-tailed test. The p-value is twice the probability of an observed value of z=−1.72 or less if the null hypothesis is true because this hypothesis test is two-tailed. This probability is equal to twice the area under the standard normal curve to the left of z=−1.72. Using a normal distribution table, the area to the left of z=−1.72 is approximately 0.043. Twice this area is 2(0.043)=0.086. Therefore, the p-value is approximately 0.086.

In a survey of 100 randomly selected people in City A, 82 support increased government spending on roads and bridges. In a survey of 100 randomly selected people in City B, 72 support such spending. Test the alternative hypothesis that the population proportion of people in City A who support such spending is different from the population proportion of people in City B who do. The test statistic is z=1.68. What is the corresponding p-value? Compute your answer using a value from the table below.

$0.092$0.092 Since the alternative hypothesis is that the population proportion of people in City A who support such spending is different from the population proportion of people in City B who do, this is a two-tailed test. The p-value is twice the probability of an observed value of z=1.68 or greater if the null hypothesis is true because this hypothesis test is two-tailed. This probability is equal to twice the area under the standard normal curve to the right of z=1.68. Using a normal distribution table, the area to the left of z=1.68 is approximately 0.954. This means the area to the right of z=1.68 is approximately 1−0.954=0.046. Twice this area is 2(0.046)=0.092. Therefore, the p-value is approximately 0.092.

A researcher wants to show that the proportions of mosquitoes that carry a certain disease in Region 1 and Region 2 are different. In a sample of 1,392 mosquitoes trapped in Region 1, 1,173 test positive for the disease. In a sample of 1,457 mosquitoes trapped in Region 2, 1,196 test positive for the disease. Test the alternative hypothesis that the population proportion for Region 1 is different from the population proportion for Region 2. The test statistic is z=1.55. What is the corresponding p-value? Compute your answer using a value from the table below.

$0.122$0.122 Since the alternative hypothesis is that the population proportion for Region 1 is different from the population proportion for Region 2, this is a two-tailed test. The p-value is twice the probability of an observed value of z=1.55 or greater if the null hypothesis is true because this hypothesis test is two-tailed. This probability is equal to twice the area under the standard normal curve to the right of z=1.55. Using a normal distribution table, the area to the left of z=1.55 is approximately 0.939. This means the area to the right of z=1.55 is approximately 1−0.939=0.061. Twice this area is 2(0.061)=0.122. Therefore, the p-value is approximately 0.122.

The United Nations stores data about the kilowatt-hours of wind power produced around the world. The following data set gives information of the United States for 25 years, from 1990 to 2014, in units of million killowatt-hours. Country or Area Commodity - Transaction Year Unit Quantity United States Electricity - total wind production 2014 Kilowatt-hours, million 183892 United States Electricity - total wind production 2013 Kilowatt-hours, million 169713 United States Electricity - total wind production 2012 Kilowatt-hours, million 141922 United States Electricity - total wind production 2011 Kilowatt-hours, million 120854 United States Electricity - total wind production 2010 Kilowatt-hours, million 95148 United States Electricity - total wind production 2009 Kilowatt-hours, million 74226 United States Electricity - total wind production 2008 Kilowatt-hours, million 55696 United States Electricity - total wind production 2007 Kilowatt-hours, million 34603 United States Electricity - total wind production 2006 Kilowatt-hours, million 26676 United States Electricity - total wind production 2005 Kilowatt-hours, million 17881 HelpCopy to ClipboardDownload CSV Your company started producing wind power nine years ago, 2008, and has captured 10% of the market every year since then. Use the linear equation (y=20888x−4E+07) to predict your wind power production for the next year, 2018. Round your final answer to one decimal place.

$215198.4\text{ million kilowatt-hours}$215198.4 million kilowatt-hours Using the linear equation from the previous question, we have: y=20888x−4E+07 20888⋅2018=42151984 42151984−40000000=2151984 Now this is the predicted wind power for all of US and your company captures 10% of the market. We just need to muliply by 0.10.2151984⋅0.10=215198.4You predict your company's wind power production will be 215,198.4 million kilowatt-hours in year 2018.

A survey asked randomly selected people whether they thought it was rude for people to interact with their smart phones while in a restaurant. Among the 108 males surveyed, 62 responded "Yes." Of the 92 females surveyed, 57 responded "Yes." Compute the test statistic for a hypothesis test to compare the population proportions of males and females that think interacting with a smart phone in a restaurant is rude. Round your answer to two decimal places.

$\text{ z= }-0.65$ z= −0.65 To calculate the test statistic (z-value), use the formula z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2. pˆ1−pˆ2=62108−5792≈−0.0455 μpˆ1−pˆ2=p1−p2=0 σpˆ1−pˆ2=p¯¯¯q¯¯(1n1+1n2)−−−−−−−−−−−−−√ p¯¯¯=x1+x2n1+n2=62+57108+92=0.595 q¯¯=1−p¯¯¯=1−0.595=0.405 σpˆ1−pˆ2=0.595(0.405)(1108+192)−−−−−−−−−−−−−−−−−−−−−√≈0.0696 z-score=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2≈−0.0455−00.0696=−0.65

Dr. Klaus is conducting a study to determine if the mean body weights of males on vegetarian, vegan, pescatarian, and standard diets are statistically the same. A sample of five adult male vegetarians, four vegans, two pescatarians, and three on standard diets is selected, and the body weight of each individual is recorded in pounds. Assume that the conditions for a one-way ANOVA are met. Determine the critical value for this one-way ANOVA test using a 5% significance level based on the table of critical values of the F-distribution shown below.

$\text{Critical Value}=3.71$Critical Value=3.71 To find the critical value for a one-way ANOVA test using a level of significance of 5%, the degrees of freedom for the numerator and degrees of freedom for the denominator need to be determined. To determine the degrees of freedom for the numerator (dfnumerator), calculate the quantity k−1, where k is the number of groups. For this example, there are four groups corresponding to the four diets. dfnumerator=k−1=4−1=3 To determine the degrees of freedom for the denominator (dfdenominator), calculate the quantity N−k, where N is the sum of the sample sizes for each of the groups. For this example, there is a sample of five adult male vegetarians, four vegans, two pescatarians, and three on standard diets. N=5+4+2+3=14 Use N to find the degrees of freedom for the denominator. dfdenominator=N−k=14−4=10 To determine the critical value for this example, (a) Refer to a table of critical values of the F-distribution for α=0.05. (b) Use the column for degrees of freedom for the numerator of 3.(c) Use the row for degrees of freedom for the denominator of 10. At the intersection of the row and column, read the corresponding critical value. Degrees of Freedom for the Numerator Degrees of Freedom for the Denominator 1234561161.45199.50215.71224.58230.16233.99218.5119.0019.1619.2519.3019.33310.139.559.289.129.018.9447.716.946.596.396.266.1656.615.795.415.195.054.9565.995.144.764.534.394.2875.594.744.354.123.973.8785.324.464.073.843.693.5895.124.263.863.633.483.37104.964.103.713.483.333.22114.843.983.593.363.203.09124.753.893.493.263.113.00134.673.813.413.183.032.92144.603.743.343.112.962.85154.543.683.293.062.902.79 For this example, the critical value will be 3.71. Thus, any values of the F-test statistic that are greater than this critical value of 3.71 will fall in the rejection region, and the decision will be to reject the null hypothesis.Note that the one-way analysis of variance (ANOVA) hypothesis test is always a right-tailed test.

$\text{Critical value(s)}=1.796$Critical value(s)=1.796 Since this is a right-tailed test and α=0.05, the critical value has a left-tail area of 1−0.05=0.95 or a right-tail area of 0.05. To find the critical value, use a t-distribution table to find the t-value that corresponds to one of these areas. Notice that t=1.796 corresponds to a right-tail area of 0.05 and df=11, so the critical value is tα=t0.05=1.796.

(a) The null and alternative hypotheses are: H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution. (b) Compute the test statistic, rounded to three decimal places.

$\text{Test Statistic=}13.167$Test Statistic=13.167 To find the test statistic χ20, we take each observed value minus the expected value, square it, and divide by the expected value. Then we add up these values. So we find χ20=(12−12)212+(9−12)212+(5−12)212+(22−12)212≈13.167.

A doctor keeps track of the number of babies she delivers in each season. She expects that the distribution will be uniform (the same number of babies in each season). The data she collects is shown in the table below. Find the test statistic, χ20, for the chi-square goodness-of-fit test. Round the final answer to three decimal places.

$\text{chi-square test statistic=}10.294$chi-square test statistic=10.294 To find χ20, we take each observed value minus the expected value, square it, and divide by the expected value. Then we add up these values. So we find χ20=(41−34)234+(18−34)234+(37−34)234+(40−34)234≈10.294

$\text{chi-square test statistic=}10.833$chi-square test statistic=10.833 To find χ20, we take each observed value minus the expected value, square it, and divide by the expected value. Then we add up these values. So we find χ20=(8−12)212+(5−12)212+(16−12)212+(19−12)212≈10.833

Continuing with the previous question, a survey asks a group of college students about their majors. It also groups the students by whether they are athletes or not. These values are recorded in the contingency table below. What is the test statistic, χ20, for this test for homogeneity? Round the final answer to one decimal place.

$\text{chi-square test statistic=}2.9$chi-square test statistic=2.9 The formula for the test statistic is ∑(O−E)2E In words, for each entry in the table, we take the observed entry O and expected entry E, and compute (O−E)2E, and then we add these up for each entry in the table. So we findχ20 =(9−7.7)27.7+(8−11.4)211.4+(9−7)27+(12−13.3)213.3+(23−19.6)219.6+(10−12)212≈2.9

Manufacturers are testing a die to make sure that it is fair (has a uniform distribution). They roll the die 66 times and record the outcomes in the table below. Find the test statistic, χ20, for the chi-square goodness-of-fit test. Round the final answer to three decimal places. χ20=∑k(O−E)2E

$\text{chi-square test statistic=}6.364$chi-square test statistic=6.364 To find χ20, we take each observed value minus the expected value, square it, and divide by the expected value. Then we add up these values. So we find χ20=(13−11)211+(12−11)211+(11−11)211+(9−11)211+(16−11)211+(5−11)211≈6.364

Continuing with the previous question, a bill is brought to a group of politicians - democrats and republicans. Researchers polled the group on whether they would vote "in favor," "opposed," or "indifferent." These values are recorded in the contingency table below. What is the test statistic, χ20, for this test for homogeneity? Round the final answer to one decimal place

$\text{chi-square test statistic=}6.7$chi-square test statistic=6.7 The formula for the test statistic is ∑(O−E)2E In words, for each entry in the table, we take the observed entry O and expected entry E, and compute (O−E)2E, and then we add these up for each entry in the table. So we findχ20 =(25−19.1)219.1+(17−18.7)218.7+(6−10.2)210.2+(18−23.9)223.9+(25−23.3)223.3+(17−12.8)212.8≈6.7

You are trying to determine if a certain die is fair (has a uniform distribution). You roll the die 72 times and record the outcomes in the table below. Find the test statistic, χ20, for the chi-square goodness-of-fit test. If necessary, round the final answer to three decimal places. χ20=∑k(O−E)2E

$\text{chi-square test statistic=}9.000$chi-square test statistic=9.000 To find χ20, we take each observed value minus the expected value, square it, and divide by the expected value. Then we add up these values. So we find χ20=(13−12)212+(16−12)212+(13−12)212+(17−12)212+(5−12)212+(8−12)212=9.000

A consumer group claims that the proportion of vehicles in City A that use alternative energy sources is less than the proportion of vehicles in City B that use alternative energy sources. In a sample of 35 randomly selected vehicles in City A, 7 vehicles use alternative energy sources. In a sample of 45 randomly selected vehicles in City B, 11 vehicles use alternative energy sources. Let p1 be the population proportion for City A, and let p2 be the population proportion for City B. What are the critical value(s) of the z-test statistic for this hypothesis test, where α=0.01? Use a comma and a space to separate answers as needed.

$\text{critical value(s)}=-2.326$critical value(s)=−2.326 Since this is a left-tailed test and α=0.01, the critical value occurs at −zα=−z0.01. To find the critical value, use a normal distribution table to find the z-score that corresponds to α=0.01 for a left-tailed test. z0.10z0.05z0.025z0.01z0.0051.2821.6451.9602.3262.576 Therefore, −z0.01=−2.326 is the critical value for this hypothesis test.

Alex is a manager of a baseball team. He is wondering if a recently acquired reliever would be a better closer than the team's current closer. Craig, the current closer, has converted 305 of 349 save opportunities in his career. Kenley, the new reliever, has converted 242 of 266 save opportunities in his career. Use a two-proportion hypothesis test to test whether Kenley's true success rate is greater than Craig's. Assume that the conditions for inference are satisfied. Let Craig correspond to the first sample and Kenley correspond to the second sample. Find the p-value for the test, rounded to three decimal places.

$\text{p-value=}0.080$p-value=0.080 Use Excel to perform the hypothesis test. 1. For the purposes of this explanation, we'll enter labels in column A and values and calculations in column B. In cells A1 through A4, enter the labels x1, n1, x2, and n2. In the corresponding cells in column B, enter the corresponding values of 305, 349, 242, and 266. 2. Next, compute pˆ1=x1n1 and pˆ2=x2n2. In this case, compute these in cells B6 and B7, respectively, using =B1/B2 and =B3/B4. This gives, pˆ1≈0.87393 and pˆ2≈0.90977. 3. Now, calculate p¯¯¯=x1+x2n1+n2. Enter the calculation =(B1+B3)/(B2+B4) into cell B8. This gives, p¯¯¯≈0.88943. 4. To find the test statistic z, enter the calculation =(B6−B7)/SQRT(B8∗(1−B8)∗(1/B2+1/B4)) into cell B9. Thus, z≈−1.40. 5. Use the NORM.S.DIST function to find the p-value. Since the null hypothesis is H0:p1=p2 and the alternative hypothesis is Ha:p1<p2, this is a left-tailed test. Therefore, the p-value can be found using =NORM.S.DIST(B9,1). This gives a p-value of 0.080, rounding to three decimal places.

Do college athletes have similar majors as non-athletes? A survey asks a group of college students about their majors. It also groups the students by whether they are athletes or not. The researchers want to determine if the distribution of college majors is the same for athletes as for non-athletes. The data are recorded in the contingency table below, and a chi-square Homogeneity Test at the 1% significance level is performed. Arts Humanities Sciences Row Total Athlete 12 20 20 52 Non-Athlete 21 6 11 38 Column Total 33 26 31 90 (a) The null and alternative hypotheses are: H0: The distribution of the two populations are the same. Ha: The distribution of the two populations are not the same. (b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)

$\text{test statistic = }10.8$test statistic = 10.8 The test statistic, χ20 is the sum of the chi-square subtotals, using the formula: χ20=∑(O−E)2E=(12−19.1)219.1+(20−15)215+(20−17.9)217.9 +(21−13.9)213.9+(6−11)211+(11−13.1)213.1≈10.8

(a) The null and alternative hypotheses are: H0: The two variables are independent, so being an athlete does not affect college major. Ha: The two variables are dependent, so being an athlete does affect college major. (b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)

$\text{test statistic = }11.4$test statistic = 11.4 To compute the test statistic, we first determine the expected frequencies (rounded to one decimal place in bold in the table below). AthleteNon-AthleteColumn TotalArts10.7715.31926Humanities16.92524.11641Sciences12.4817.62230Row Total405797 The test statistic, χ20 is the sum of the chi-square subtotals, using the formula: χ20=∑(O−E)2E=(7−10.7)210.7+(25−16.9)216.9+(8−12.4)212.4 +(19−15.3)215.3+(16−24.1)224.1+(22−17.6)217.6≈11.4.

(a) The null and alternative hypotheses are: H0: The distribution of the two populations are the same. Ha: The distribution of the two populations are not the same. (b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)

$\text{test statistic = }3.8$test statistic = 3.8 The test statistic, χ20 is the sum of the chi-square subtotals, using the formula: χ20=∑(O−E)2E=(25−25.7)225.7+(17−13.1)213.1+(19−22.2)222.2 +(20−19.3)219.3+(6−9.9)29.9+(20−16.8)216.8≈3.8

$critical\ value\ =\ 9.210$critical value = 9.210 The significance level is given in the question. In this case we are asked to perform a test at the 1% significance level, so α=0.01. The degrees of freedom for a homogeneity test is calculated as: df=(r−1)(c−1), where r= the number of rows in the contingency table, and c= the number of columns. So, df=(2−1)(3−1)=2.We know df=2 and α=0.01, so, using the χ2-table, the critical value is χ20.01=9.210.

$critical\ value\ =\ 9.210$critical value = 9.210 The significance level is given in the question. In this case we are asked to perform a test at the 1% significance level, so α=0.01. The degrees of freedom for a test of independence is calculated as: df=(r−1)(c−1), where r= the number of rows in the contingency table, and c= the number of columns. So, df=(2−1)(3−1)=2.We know df=2 and α=0.01, so, using the χ2-table, the critical value is χ20.01=9.210.

A web developer wanted to test which types of advertisements were the most cost effective. He suspected that advertisements with animations generated more clicks than advertisements without animations but wanted to make sure before he started to recommend advertisements with animations to his clients. He created advertisements with animations for 30 different products and then created advertisements without animations for those same projects. Then he measured the click-through rate for those advertisements. The data is given below (given as percentages of total visitors to the site that clicked on the advertisement). Use Excel to test the paired data and determine whether the true mean difference in the click rates of advertisements without animations and the click rates of advertisements with animations is less than zero. Assign rates from advertisements without animations to population 1 and the others to population 2. Identify the t-test statistic, rounding to two decimal places, and the p-value, rounding to three decimal places.

$t=-2.70,\ \text{p-value}=0.006$t=−2.70, p-value=0.006 Step 1: Enter the data into Excel, with each sample entered as a single row or column. Step 2: Click the Data tab. Then select Data Analysis and t-Test: Paired Two Sample for Means. Step 3: In the dialog box, enter the appropriate ranges of data for Variable 1 Range and Variable 2 Range. Enter the Hypothesized Mean Difference, 0. Select Labels if the range contains labels. Click OK. Step 4: The t-test statistic is in the cell labeled t Stat, which contains a value, rounding to two decimal places, of −2.70. Step 5: The test uses a one-sided alternative hypothesis (μd<0), so the p-value is in the cell labeled P(T<=t) one-tail, which contains a value, rounding to three decimal places, of 0.006.

A political analyst believes that a senator's recent decision to support a bill resulted in a drop of approval ratings. To test this claim, he selects random cities in the state that voted the senator in and compares the approval ratings before the decision to the approval ratings after the decision. If we let d=approval rating after−approval rating before, based on the data below, what are the test statistic and degrees of freedom of an appropriate hypothesis test? Assume that the approval ratings are normally distributed. Round the test statistic to three decimal places.

$t=-3.935,\ df=7$ The mean of the differences is d¯¯¯=−2.625, the hypothesized mean is μd=0, the standard deviation of the differences is sd≈1.887, and the number of paired data is n=8. The standard deviation of the differences is calculated as: sd=n(∑d2)−(∑d)2n(n−1)−−−−−−−−−−−−−−−√=8(80.06)−(−21)28(8−1)−−−−−−−−−−−−−−−√≈1.887 The t-test statistic is calculated below. t≈−2.625−01.887/8-√≈−3.935 This test statistic has 8−1=7 degrees of freedom.

Jack wanted to prove to his identical twin brother Tim that he was the twin who performed statistically better in school. He randomly selected 30 pairs of tests that they had both taken in school and recorded their scores, which are given below. Use Excel to test if the true mean difference between Jack's test scores and Tim's test scores is significantly greater than zero. Let Jack's scores represent sample 1 and Tim's scores represent sample 2. Identify the t-test statistic, rounding to two decimal places, and the p-value, rounding to three decimal places.

$t=1.51,\ \text{p-value}=0.071$t=1.51, p-value=0.071 Step 1: Enter the data into Excel, with each sample entered as a single row or column. Step 2: Click the Data tab. Then select Data Analysis and t-Test: Paired Two Sample for Means. Step 3: In the dialog box, enter the appropriate ranges of data for Variable 1 Range and Variable 2 Range. Enter the Hypothesized Mean Difference, 0. Select Labels if the range contains labels. Click OK. Step 4: Identify the t-test statistic from the output. The t-test statistic is in the cell labeled t Stat, which contains a value, rounding to two decimal places, of 1.51. Step 5: The test uses a one-sided alternative hypothesis (μd>0), so the p-value is in the cell labeled P(T<=t) one-tail, which contains a value, rounding to three decimal places, of 0.071.

An automotive company used two different drivers to test the fuel economy of its latest crossover model. Each driver tested the same 20 vehicles from the factory. When the supervisor reviewed the data, she thought it looked like Driver A had different fuel economies than Driver B on average for the different vehicles and decided to test the data. To test the data, she first separated the data into pairs, with each car tested as a pair. The data is given below, with the fuel economy for each car given in miles per gallon. Assume that both populations are normally distributed. Use Excel to test if the true mean difference between Driver A and Driver B is significantly different from zero, where α=0.05, where each difference is calculated as Driver A−Driver B. Identify the t-test statistic, rounding to two decimal places, and the p-value, rounding to three decimal places.

$t=1.98,\ \text{p-value}=0.063$t=1.98, p-value=0.063 Step 1: Enter the data into Excel, with each sample entered as a single row or column. Step 2: Click the Data tab. Then select Data Analysis and t-Test: Paired Two Sample for Means. Step 3: In the dialog box, enter the appropriate ranges of data for Variable 1 Range and Variable 2 Range. Enter the Hypothesized Mean Difference, 0. Select Labels if the range contains labels. Click OK. Step 4: The t-test statistic is in the cell labeled t Stat, which contains a value, rounding to two decimal places, of 1.98. Step 5: The test uses a two-sided alternative hypothesis (μd≠0), so the p-value is in the cell labeled P(T<=t) two-tail, which contains a value, rounding to three decimal places, of 0.063.

The owners of an ice cream shop franchise, looking to reduce wait times for customers, launched a pilot program where customers could use a smartphone app to place orders. The accompanying data set shows the mean wait times (in seconds) for in-store and drive-through customers at 30 shops participating in the pilot program and at 30 shops not in the program. Assume that the population variance of mean wait times is equal for both groups and that the mean wait times are normally distributed for both groups. Let the mean wait times of the shops not in the program be the first sample, and let the mean wait times of the shops in the program be the second sample. At the 0.10 level of significance, is there evidence that the smartphone app reduces wait times? Perform the test using Excel. Find the test statistic, rounded to two decimal places, and the p-value, rounded to three decimal places.

$t=4.10,\ \text{p-value}=0.000$t=4.10, p-value=0.000 Use Excel to perform the hypothesis test. 1. Since the owners of the franchise want to know whether the mean wait time for shops not in the program is greater than for shops in the program, the null hypothesis is H0:μ1=μ2 and the alternative hypothesis is Ha:μ1>μ2. 2. Copy the data into a blank Excel worksheet. 3. Since it is assumed that the population variances are equal, select Data, and then Data Analysis and t-Test: Two-Sample Assuming Equal Variances. 4. In the dialog box, enter $A$1:$A$31 for Variable 1 Range and $B$1:$B$31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. Select Labels if the range contains the labels. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a one-sided alternative hypothesis (Ha:μ1>μ2), read the p-value from the cell labeled P(T<=t) two-tail. The test statistic is t≈4.10. The p-value is approximately 0.000. 6. The p-value of 0.000 is less than the level of significance, 0.10. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that the smartphone app reduces wait times

In a survey of 100 randomly selected taxi drivers in a city, 76 support increased government spending on roads and bridges. In a survey of 100 randomly selected bus drivers in the same city, 82 support such spending. Compute the test statistic for a hypothesis test to compare the population proportion of taxi drivers to the proportion of bus drivers in the city that support such spending. Assume that the conditions for a hypothesis test for the difference between the population proportions are met. Round your answer to two decimal places.

$z=-1.04$z=−1.04 To calculate the test statistic (z-value), use the formula z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2. pˆ1−pˆ2=0.76−0.82=−0.06 μpˆ1−pˆ2=p1−p2=0 σpˆ1−pˆ2=p¯¯¯q¯¯(1n1+1n2)−−−−−−−−−−−−−√ p¯¯¯=x1+x2n1+n2=76+82100+100=0.79 q¯¯=1−p¯¯¯=1−0.79=0.21 σpˆ1−pˆ2=0.79(0.21)(1100+1100)−−−−−−−−−−−−−−−−−−−−√≈0.0576 z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2≈−0.06−00.0576≈−1.04

In a survey of 100 randomly selected people in City A, 81 support increased government spending on education. In a survey of 100 randomly selected people in City B, 87 support such spending. Let p1 be the population proportion of people in City A who support such spending, and let p2 be the population proportion of people in City B who support such spending. To test the alternative hypothesis Ha:p1−p2<0, what is the critical value? Use the level of significance α=0.10 and round your answer to three decimal places.

$z=-1.282$z=−1.282 Notice that this is a left-tailed test. The z-value that corresponds to the level of significance is −1.282.

In a survey of 150 randomly selected coal miners in State A, 7 support government subsidies to encourage producing solar energy. In a survey of 150 randomly selected farmers in State B, 132 support such subsidies. Compute the test statistic for a hypothesis test to compare the population proportion of coal miners in State A that support such subsidies to the proportion of farmers in State B that support such subsidies. Assume that the conditions for a hypothesis test for the difference between the population proportions are met. Round your answer to two decimal places.

$z=-14.47$z=−14.47 To calculate the test statistic (z-value), use the formula z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2. pˆ1−pˆ2=7150−132150=−56 μpˆ1−pˆ2=p1−p2=0 σpˆ1−pˆ2=p¯¯¯q¯¯(1n1+1n2)−−−−−−−−−−−−−√ p¯¯¯=x1+x2n1+n2=7+132150+150=139300 q¯¯=1−p¯¯¯=1−139300=161300 σpˆ1−pˆ2=139300161300(1150+1150)−−−−−−−−−−−−−−−−−−√≈0.05758 z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2≈−56−00.05758≈−14.47

A survey asked randomly selected drivers if they would be willing to pay higher taxes on gasoline as long as all of the additional revenue went to improving roadways. Among the 41 males surveyed, 12 responded "Yes." Of the 59 females surveyed, 21 responded "Yes." Let p1 be the population proportion for males, and let p2 be the population proportion for females. To test the alternative hypothesis Ha:p1−p2<0, what is the critical value? Use the level of significance α=0.01 and round your answer to two decimal places.

$z=-2.33$z=−2.33 Notice that this is a left-tailed test. The z-value that corresponds to the level of significance is −2.33.

A large university is well known for both its business school and its mechanical engineering program. The dean of the career services office wants to know if there is a difference in starting job salary between recently graduated business majors and mechanical engineering majors. The following table shows the starting annual salaries for a random sample of 30 business majors and 30 mechanical engineering majors from the most recent graduating class. Assume that the population standard deviation of the business majors' starting salaries is $12,000 and the population standard deviation of the mechanical engineering majors' starting salaries is $7,000, and that the starting salaries for both majors are normally distributed. Let the business majors' salaries be the first sample, and let the mechanical engineering majors' salaries be the second sample. At the 0.05 level of significance, is there evidence of a difference in average annual starting salary? Perform the test using Excel.Find the test statistic, rounded to two decimal places, and the p-value, rounded to three decimal places.

$z=-2.83,\ \text{p-value}=0.005$z=−2.83, p-value=0.005 Use Excel to perform the hypothesis test. 1. Since the dean wants to know whether there is a difference in starting salary, the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1≠μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the squares of the known population standard deviations, or 144000000 and 49000000, respectively. Make sure Labels is selected. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a two-sided alternative hypothesis (μ1≠μ2), read the p-value from the cell labeled P(Z<=z) two-tail. The test statistic is z≈−2.83. The p-value is 0.005. 6. The p-value of 0.005 is less than the level of significance, 0.05. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that there is a difference in average annual starting salary between business majors and mechanical engineering majors.

Janet is a vegan chef looking to open a restaurant in either Portland, Oregon, or Seattle, Washington. She hires a company to conduct a survey to find out the proportions of people who are vegan in these two cities. Out of 569 people surveyed in Portland, 43 are vegan. Out of 884 people surveyed in Seattle, 49 are vegan. Perform a hypothesis test to determine if there is evidence that the proportion of vegans in Portland is greater than the proportion of vegans in Seattle. Assume that the conditions for the hypothesis test are satisfied. Let the Portland data correspond to sample 1 and the Seattle data correspond to sample 2. Identify the test statistic, rounding to two decimal places, and the p-value, rounding to three decimal places.

$z=1.54,\text{ p-value}=0.062$z=1.54, p-value=0.062 Use Excel to perform the hypothesis test. 1. For the purposes of this explanation, we'll enter labels in column A and values and calculations in column B. In cells A1 through A4, enter the labels x1, n1, x2, and n2. In the corresponding cells in column B, enter the corresponding values of 43, 569, 49, and 884. 2. Next, compute pˆ1=x1n1 and pˆ2=x2n2. In this case, compute these in cells B6 and B7, respectively, using =B1/B2 and =B3/B4. This gives pˆ1≈0.07557 and pˆ2≈0.05543. 3. Now, calculate p¯¯¯=x1+x2n1+n2. Enter the calculation =(B1+B3)/(B2+B4) into cell B8. This gives p¯¯¯≈0.06332. 4. To find the test statistic z, enter the calculation =(B6−B7)/SQRT(B8∗(1−B8)∗(1/B2+1/B4)) into cell B9. Thus, z≈1.54. 5. Use the NORM.S.DIST function to find the p-value. Since the null hypothesis is H0:p1=p2 and the alternative hypothesis is Ha:p1>p2, this is a right-tailed test. Therefore, the p-value can be found using =1−NORM.S.DIST(B9,1). This gives a p-value of 0.062, rounding to three decimal places.

To test a drug intended to increase memory, 80 subjects were randomly given a pill that either contained the drug or was a placebo. After ten minutes, the subjects were asked to look at a poster with pictures of 20 common objects for five minutes. After waiting ten minutes, the subjects were asked to list as many of the objects as they could recall. For the 44 subjects who took the drug, 75% were able to list at least half of the objects. For the 36 subjects who took the placebo, 50% were able to list at least half of the objects. Compute the test statistic for a hypothesis test to compare the population proportion of subjects that took the drug that could list at least half of the objects to the proportion of subjects that took the placebo that could list at least half of the objects. Assume that the conditions for a hypothesis test for the difference between the population proportions are met. Round your answer to two decimal places.

$z=2.31$z=2.31 To calculate the test statistic (z-value), use the formula z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2. pˆ1−pˆ2=0.75−0.50=0.25 μpˆ1−pˆ2=p1−p2=0 σpˆ1−pˆ2=p¯¯¯q¯¯(1n1+1n2)−−−−−−−−−−−−−√ p¯¯¯=x1+x2n1+n2=(0.75)(44)+(0.50)(36)44+36=0.6375 q¯¯=1−p¯¯¯=1−0.6375=0.3625 σpˆ1−pˆ2=0.6375(0.3625)(140+140)−−−−−−−−−−−−−−−−−−−−−−√≈0.1080 z=pˆ1−pˆ2−μpˆ1−pˆ2σpˆ1−pˆ2≈0.25−00.1080≈2.31

Elizabeth is a financial adviser for a car dealership with locations throughout the country. She wondered if sales of convertibles are relatively higher in warm weather cities than in cold weather cities. To test her theory, she collected random samples of cars sold throughout a one-year period for locations in San Diego and Boston. Out of 500 cars sold in San Diego, 59 were convertibles, and 37 out of 500 cars sold in Boston were convertibles. Is there evidence that the proportion of convertibles sold in San Diego is greater than the proportion in Boston? Assume that the conditions for the hypothesis test are satisfied. Let San Diego cars correspond to population 1 and Boston cars correspond to population 2. Identify the test statistic, rounding to two decimal places, and the p-value, rounding to three decimal places.

$z=2.36,\text{ p-value}=0.009$z=2.36, p-value=0.009 Use Excel to perform the hypothesis test. 1. For the purposes of this explanation, we'll enter labels in column A and values and calculations in column B. In cells A1 through A4, enter the labels x1, n1, x2, and n2. In the corresponding cells in column B, enter the corresponding values of 59, 500, 37, and 500. 2. Next, compute pˆ1=x1n1 and pˆ2=x2n2. In this case, compute these in cells B6 and B7, respectively, using =B1/B2 and =B3/B4. This gives pˆ1=0.118 and pˆ2=0.074. 3. Now, calculate p¯¯¯=x1+x2n1+n2. Enter the calculation =(B1+B3)/(B2+B4) into cell B8. This gives p¯¯¯=0.096. 4. To find the test statistic z, enter the calculation =(B6−B7)/SQRT(B8∗(1−B8)∗(1/B2+1/B4)) into cell B9. Thus, z≈2.36. 5. Use the NORM.S.DIST function to find the p-value. Since the null hypothesis is H0:p1=p2 and the alternative hypothesis is Ha:p1>p2, this is a right-tailed test. Therefore, the p-value can be found using =1−NORM.S.DIST(B9,1). This gives a p-value of 0.009, rounding to three decimal places.

The owners of a fast-food franchise, looking to reduce wait times for customers, launched a pilot program where customers could use a smartphone app to place orders. The accompanying data set shows the mean wait times (in seconds) for in-store and drive-through customers at 30 restaurants participating in the pilot program and at 30 restaurants not participating in the program. Assume that the population standard deviation of mean wait times is 30 seconds for both groups of restaurants and that the mean wait times are normally distributed for both groups of restaurants. Let the mean wait times at the restaurants not in the program be the first sample, and let the mean wait times at the stores in the program be the second sample. At the 0.05 level of significance, is there evidence that the smartphone app reduces wait times? Perform the test using Excel. Find the test statistic, rounded to two decimal places, and the p-value, rounded to three decimal places.

$z=6.92,\ \text{p-value}=0.000$z=6.92, p-value=0.000 Use Excel to perform the hypothesis test. 1. Since the owners of the franchise want to know whether the program reduces wait times, the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1>μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the square of the known population standard deviation, or 900. Make sure Labels is selected. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a one-sided alternative hypothesis (μ1>μ2), read the p-value from the cell labeled P(Z<=z) one-tail. The test statistic is z≈6.92. The p-value is 0.000. 6. The p-value of 0.000 is less than the level of significance, 0.05. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that the smartphone app reduces wait times for restaurants participating in the program.

A potato chip company makes potato chips in two flavors, Regular and Ranch. Riley is a quality control manager for the company who is trying to ensure that each bag contains about the same number of chips, regardless of flavor. He collects two random samples of 30 bags of chips from each flavor, and counts the number of chips in each bag. The results are shown in the following table. Assume that both flavors have a standard deviation of 9.5 chips per bag and that the number of chips per bag for both flavors is normally distributed. Let the Regular chips be the first sample, and let the Ranch chips be the second sample. At the 0.01 level of significance, is there evidence that both flavors have the same number of chips in each bag? Perform the test using Excel. Find the test statistic, rounded to two decimal places, and the p-value, rounded to three decimal places.

$z=8.72,\ \text{p-value}=0.000$z=8.72, p-value=0.000 Use Excel to perform the hypothesis test. 1. Since Riley is trying to ensure that each bag contains about the same number of chips , the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1≠μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the square of the known population standard deviation, or 90.25. Make sure Labels is selected. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a two-sided alternative hypothesis (μ1≠μ2), read the p-value from the cell labeled P(Z<=z) two-tail. The test statistic is z≈8.72. The p-value is 0.000. 6. The p-value of 0.000 is less than the level of significance, 0.01. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that Regular and Ranch chips have different amounts of chips per bag.

A physician wants to determine if a supplement is effective in helping men lose weight. She takes a random sample of overweight men and records their weights before the trial. She then prescribes the supplement and instructs them to take it for four weeks while making no other lifestyle changes. After the four-week period, she records the weight of the men again. Suppose that data were collected for a random sample of 6 men, where each difference is calculated by subtracting the weight before the trial from the weight after the trial. Assume that the populations are normally distributed. The physician uses the alternative hypothesis Ha:μd<0. Using a test statistic of t≈−2.795, which has 5 degrees of freedom, determine the range that contains the p-value.

0.01< p-value <0.05 Notice that this is a left-tailed test because the alternative hypothesis is Ha:μd<0. The t-test statistic is −2.795 with 5 degrees of freedom. The t-distribution table gives the values for a right-tailed test, so to find the p-value for a left-tailed test, use the absolute value of the test statistic. Using the table of areas in the right tail for the t-distribution, in the row for 5 degrees of freedom, 2.795 is greater than 2.015 and less than 3.365. So, the p-value is less than 0.05 but greater than 0.01, or 0.01< p-value <0.05.

A statistician believes that women received more bachelor's degrees than men did last year. To test this claim, she selects colleges and universities randomly and compares the number of bachelor's degrees awarded to men and women. Suppose that data were collected for a random sample of 7 colleges and universities, where each difference is calculated by subtracting the number of degrees earned by men from the number of degrees earned by women. Assume that the number of degrees is normally distributed. The statistician uses the alternative hypothesis Ha:μd>0. Using a test statistic of t≈2.968, which has 6 degrees of freedom, determine the range that contains the p-value.

0.01< p-value <0.05 Notice that this is a right-tailed test because the alternative hypothesis is Ha:μd>0. The t-test statistic is 2.968 with 6 degrees of freedom. The t-distribution table gives the values for a right-tailed test. Using the table of areas in the right tail for the t-distribution, in the row for 6 degrees of freedom, 2.968 is greater than 1.943 but less than 3.143. So, the p-value is between 0.01 and 0.05.

Your friend claims that his commute time to work is less than his commute time returning home on the same day. Suppose that data were collected for a random set of 11 days, where each difference is calculated by subtracting the commute time to work from the commute time returning home. Assume that the populations are normally distributed. Suppose the test statistic is computed to be t≈−1.788, with 10 degrees of freedom. What range contains the p-value?

0.05< p-value <0.10 Notice that this is a left-tailed test because the alternative hypothesis is Ha:μd<0. The t-test statistic is −1.788 with 10 degrees of freedom. The t-distribution table gives the values for a right-tailed test, so to find the p-value for a left-tailed test, use the absolute value of the test statistic. Using the table of areas in the right tail for the t-distribution, in the row for 10 degrees of freedom, 1.788 is greater than 1.372 and less than 1.812. So, the p-value is between 0.05 and 0.10.

A researcher wants to test to see if husbands are significantly older than their wives. To do this, he collects the ages of husbands and pairs them with the ages of their respective wives for a random set of married couples. Suppose that data were collected for a random sample of 12 couples, where each difference is calculated by subtracting the age of the wife from the age of the husband. Assume that the ages are normally distributed. The researcher uses the alternative hypothesis Ha:μd>0. Using a test statistic of t≈1.434, which has 11 degrees of freedom, determine the range that contains the p-value.

0.05< p-value <0.10 Notice that this is a right-tailed test because the alternative hypothesis is Ha:μd>0. The t-test statistic is 1.434 with 11 degrees of freedom. The t-distribution table gives the values for a right-tailed test. Using the table of areas in the right tail for the t-distribution, in the row for 11 degrees of freedom, 1.434 is greater than 1.363 and less than 1.796. So, the p-value is between 0.05 and 0.10.

A researcher wants to compare the heights of males between generations to see if they differ. To do this, he samples random pairs of males who are at least 18 years old and their fathers. He then splits them into a sample of fathers and a sample of sons. Suppose that data were collected for a random sample of 11 pairs, where each difference is calculated by subtracting the height of the son from the height of the father. Assume that the heights are normally distributed. The researcher uses the alternative hypothesis Ha:μd≠0. Using a test statistic of t≈1.971, which has 10 degrees of freedom, determine the range that contains the p-value.

0.05< p-value <0.10 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μd≠0. The t-test statistic is 1.971 with 10 degrees of freedom. The t-distribution table gives the values for a right-tailed test, so to find the p-value for a two-tailed test, find the range containing the test statistic and then multiply the areas of the range by 2. Using the table of areas in the right tail for the t-distribution, in the row for 10 degrees of freedom, 1.971 is less than 2.228 and greater than 1.812. So, the area in the right tail is greater than 0.025 and less than 0.05. Since this is a two-tailed test, multiply the area by 2. The p-value is between 0.05 and 0.10.

A school district has a standardized test that it uses to sort students into magnet schools, where the test is on a 200-point scale. The superintendent's office wants to determine whether male and female students have approximately equal scores on the test. Scores for a random sample of 30 male students and 30 female students are taken. Assume that the population standard deviation of scores is 18 points for both male and female students, and that the scores are normally distributed for both male and female students. Let the scores of the male students be the first sample, and let the scores of the female students be the second sample. The consulting group conducts a two-mean hypothesis test at the 0.10 level of significance, to test if there is evidence that male and female students have different scores, on average. (a) H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test. Male Score Female Score 136 128 138 154 133 137 132 110 87 106 115 128 99 135 108 132 153 110 HelpCopy to ClipboardDownload CSV

1$-0.46$−0.46 2$0.646$0.646 Use Excel to perform the hypothesis test. 1. Since the superintendent's office wants to know whether there is a difference in test scores between genders, the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1≠μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the square of the known population standard deviation, or 324. Make sure Labels is selected. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a two-sided alternative hypothesis (μ1≠μ2), read the p-value from the cell labeled P(Z<=z) two-tail. The test statistic is z≈−0.46. The p-value is 0.646.

A hardware store refills propane tanks. One of its employees, Brandon, received a warning for overfilling the tanks. In order to prove his innocence, he accused his fellow employee, Hunter, of filling the tanks even more than he does. Brandon randomly selected 20 tanks that he had filled and then paired each of those tanks with a randomly selected tank of the same size filled by Hunter. Brandon weighed all 20 pairs of tanks in order to determine how much each tank was filled. The weights of the tanks (in pounds) are given below. Assume that both populations are normally distributed. He tests the paired data, where α=0.01, in order to evaluate the claim that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is significantly less than zero. (a) H0:μd=0, Ha:μd<0, which is a left-tailed test.

1$-1.99$−1.99 2$0.030$0.030 Step 1: Enter the data into Excel, with each sample entered as a single row or column. Step 2: Click the Data tab. Then select Data Analysis and t-Test: Paired Two Sample for Means. Step 3: In the dialog box, enter the appropriate ranges of data for Variable 1 Range and Variable 2 Range. Enter the Hypothesized Mean Difference, 0. Select Labels if the range contains labels. Click OK. Step 4: The t-test statistic is in the cell labeled t Stat, which contains a value, rounding to two decimal places, of −1.99. Step 5: The test uses a one-sided alternative hypothesis (μd<0), so the p-value is in the cell labeled P(T<=t) one-tail, which contains a value, rounding to three decimal places, of 0.030.

The above table shows the average number of hours of sleep per night for a random sample of 30 pre-med students and 30 pre-law students. (b) Use Excel to test if the mean sleep for pre-med students is less than the mean sleep for pre-law students. Identify the test statistic, t, and p-value from the Excel output. Round your test statistic to two decimal places and your p-value to three decimal places.

1$-2.94$−2.94 2$0.002$0.002 Use Excel to perform the hypothesis test. 1. Since Christine in inquiring on whether the pre-med students get less sleep than the pre-law students, the null hypothesis is H0:μ1=μ2 and the alternative hypothesis is Ha:μ1<μ2. 2. Copy the data into a blank Excel worksheet. 3. Since it is assumed that the population variances are not equal, select Data, and then Data Analysis and t-Test: Two-Sample Assuming Unequal Variances. 4. In the dialog box, enter $A$1:$A$31 for Variable 1 Range and $B$1:$B$31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. Select Labels if the range contains the labels. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a one-sided alternative hypothesis (Ha:μ1<μ2), read the p-value from the cell labeled P(T<=t) one-tail. The test statistic is t≈−2.94. The p-value is approximately 0.002.

Claire selects a random sample of 30 students from each group and records the standardized test score for each student, as shown in the above table. (b) Use Excel to test if there is a difference in student loan debt between the two groups. Identify the test statistic, z, and p-value from the Excel output. Round your test statistic to two decimal places and your p-value to three decimal places.

1$0.44$0.44 2$0.662$0.662 Use Excel to perform the hypothesis test. 1. Since Claire is looking to determine whether there is a difference in test scores, the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1≠μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the squares of the known population standard deviations, or 36 and 121, respectively. Make sure Labels is selected. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a two-sided alternative hypothesis (μ1≠μ2), read the p-value from the cell labeled P(Z<=z) two-tail. The test statistic is z≈0.44. The p-value is 0.662. 6. The p-value of 0.662 is greater than the level of significance, 0.10. Thus, we fail to reject the null hypothesis. There is insufficient evidence to conclude that there is a difference in test scores between the teaching methods.

In a survey of randomly selected 21-year-old females and males, 259 of 500 females and 241 of 500 males are currently enrolled in college. Is there evidence that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds? Assume that the conditions for inference are satisfied. Use a 0.10 level of significance. Assign females to population 1 and males to population 2. (a) H0:p1=p2; Ha:p1>p2, which is a right-tailed test. (b) Use Excel to test if the proportions are different. Identify the test statistic, z, and p-value from the Excel output. Round the test statistic to two decimal places and the p-value to three decimal places.

1$1.14$1.14 2$0.127$0.127 Use Excel to perform the hypothesis test. 1. For the purposes of this explanation, we'll enter labels in column A and values and calculations in column B. In cells A1 through A4, enter the labels x1, n1, x2, and n2. In the corresponding cells in column B, enter the corresponding values of 259, 500, 241, and 500. 2. Next, compute pˆ1=x1n1 and pˆ2=x2n2. In this case, compute these in cells B6 and B7, respectively, using =B1/B2 and =B3/B4. This gives pˆ1=0.518 and pˆ2=0.482. 3. Now, calculate p¯¯¯=x1+x2n1+n2. Enter the calculation =(B1+B3)/(B2+B4) into cell B8. This gives p¯¯¯=0.5. 4. To find the test statistic z, enter the calculation =(B6−B7)/SQRT(B8∗(1−B8)∗(1/B2+1/B4)) into cell B9. Thus, z≈1.14. 5. Use the NORM.S.DIST function to find the p-value. Since the null hypothesis is H0:p1=p2 and the alternative hypothesis is Ha:p1>p2, this is a right-tailed test. Therefore, the p-value can be found using =1−NORM.S.DIST(B9,1). This gives a p-value of 0.127, rounding to three decimal places. 6. Since the p-value is less than α, we reject the null hypothesis. There is sufficient evidence to conclude that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds.

Find the linear regression line for the following table of values, where x represents the number of work site accidents and y represents the average percentage of employee turnover. You will need to use a calculator, spreadsheet, or statistical software. Round your final answers to two decimal places. x 1 2 3 4 5 6 y 7.97 7.85 11.3 10 12.58 15.41

1$1.43$1.43 2$5.84$5.84 If you use a TI-83 or TI-84 calculator, you press STAT, and then ENTER, which brings you to the edit menu where you can enter values. In the L1 list, you enter the values of x from the table above, 1,2,3,4,5,6. Then, in the L2 list, you enter the values of y from the table above, 7.97,7.85,11.3,10.0,12.58,15.41.Now, press STAT again, and arrow to the right, to CALC. Arrow down to the LinReg option and press ENTER. The resulting a and b are the slope m and y-intercept b of the linear regression line. You should find that m≈1.43 and b≈5.84. So the final answer is y=1.43x+5.84 Using spreadsheet software or other statistical software should give you the same result.

(b) Use Excel to test the paired data and evaluate the claim that the true mean difference is significantly greater than zero. Let the old club distances correspond to sample 1 and the new club distances correspond to sample 2. Identify the test statistic, t, and p-value from the Excel output. Round your test statistic to two decimal places and your p-value to three decimal places.

1$1.59$1.59 2$0.061$0.061 Step 1: Enter the data into Excel, with each sample entered as a single row or column. Step 2: Click the Data tab. Then select Data Analysis and t-Test: Paired Two Sample for Means. Step 3: In the dialog box, enter the appropriate ranges of data for Variable 1 Range and Variable 2 Range. Enter the Hypothesized Mean Difference, 0. Select Labels if the range contains labels. Click OK. Step 4: The t-test statistic is in the cell labeled t Stat, which contains a value, rounding to two decimal places, of 1.59. Step 5: The test uses a one-sided alternative hypothesis (μd>0), so the p-value is in the cell labeled P(T<=t) one-tail, which contains a value, rounding to three decimal places, of 0.061.

Going into Game 7 of a playoff hockey series, the goalie for the home team had faced 151 shots and stopped 142 of them. The goalie for the road team saved 126 of 145 shots. Use this information and the appropriate hypothesis test to determine whether the home goalie has a better true save percentage than the road goalie. Assume that the conditions for inference are satisfied. Use a 0.05 level of significance. Assign the home goalie data to sample 1 and the other goalie to sample 2. Identify all of the appropriate conclusions. (a) H0:p1=p2; Ha:p1>p2, which is a right-tailed test. (b) Use Excel to test if the proportions are different. Identify the test statistic, z, and p-value from the Excel output. Round your test statistic to two decimal places and your p-value to three decimal places.

1$2.10$2.10 2$0.018$0.018 Use Excel to perform the hypothesis test. 1. For the purposes of this explanation, we'll enter labels in column A and values and calculations in column B. In cells A1 through A4, enter the labels x1, n1, x2, and n2. In the corresponding cells in column B, enter the corresponding values of 142, 151, 126, and 145. 2. Next, compute pˆ1=x1n1 and pˆ2=x2n2. In this case, compute these in cells B6 and B7, respectively, using =B1/B2 and =B3/B4. This gives pˆ1≈0.94040 and pˆ2≈0.86897. 3. Now, calculate p¯¯¯=x1+x2n1+n2. Enter the calculation =(B1+B3)/(B2+B4) into cell B8. This gives p¯¯¯≈0.90541. 4. To find the test statistic z, enter the calculation =(B6−B7)/SQRT(B8∗(1−B8)∗(1/B2+1/B4)) into cell B9. Thus, z≈2.10. 5. Use the NORM.S.DIST function to find the p-value. Since the null hypothesis is H0:p1=p2 and the alternative hypothesis is Ha:p1>p2, this is a right-tailed test. Therefore, the p-value can be found using =1−NORM.S.DIST(B9,1). This gives a p-value of 0.018, rounding to three decimal places.

Julia is a real estate agent and is conducting a test to determine if the average price of a house is statistically the same in Colorado, New York, Massachusetts, and Virginia. Assume the populations are normally distributed and have equal variances. To collect data for this ANOVA test, Julia randomly selects houses in each state and records the estimated market value of each in dollars, as shown in the table below.

1$3$3 2$23$23 To determine the degrees of freedom for the numerator (dfnumerator) for a one-way ANOVA hypothesis test, calculate the quantity k−1, where k is the number of groups. There are four groups corresponding to the four states, so k=4. dfnumerator=k−1=4−1=3 To determine degrees of freedom for the denominator (dfdenominator), calculate the quantity N−k, where N is the sum of the sample sizes for each of the groups, and k is the number of groups.For this example, N is the sum of the sample sizes for each of the groups, and thus N=7+7+7+6=27. Thus, dfdenominator=N−k=27−4=23.

Dr. Klaus is conducting a study to determine if the mean body weights of males on vegetarian, vegan, pescatarian, and standard diets are statistically the same. Assume the populations are normally distributed and have equal variances. To collect data for this ANOVA test, Klaus randomly selects adult males on each type of diet and records their body weights in pounds, as shown in the table below.

1$3$3 2$30$30 To determine the degrees of freedom for the numerator (dfnumerator) for a one-way ANOVA hypothesis test, calculate the quantity k−1, where k is the number of groups. There are four groups corresponding to the four diets, so k=4. dfnumerator=k−1=4−1=3 To determine degrees of freedom for the denominator (dfdenominator), calculate the quantity N−k, where N is the sum of the sample sizes for each of the groups, and k is the number of groups.For this example, N is the sum of the sample sizes for each of the groups, and thus N=9+8+8+9=34. Thus, dfdenominator=N−k=34−4=30.

The table shows data collected on the relationship between the average yearly overtime and time spent on a project per day. The line of best fit for the data is yˆ=−0.66x+88.5. Yearly Overtime (Hours) 35455565 Minutes on a Project 66585246 (a) According to the line of best fit, what would be the predicted number of minutes spent on a project per day for an average yearly overtime of 46 hours?

1$58.14$58.14 Substitute 46 for x into the line of best fit to estimate the number of minutes spent on a project per day for an average yearly overtime of 46 hours: yˆ=−0.66(46)+88.5=58.14.

In which of the following situations should the chi-square Test for Goodness-of-Fit be used? Select the correct answer below: A researcher is trying to determine if salaries for male and female chefs have the same distribution. He surveys a random sample of chefs and records the distribution of salaries for each gender. He wants to determine if the distributions are the same. A casino wants to decide if a certain pair of dice is fair. They roll the dice 100 times and compare the distribution of outcomes with the expected distribution of outcomes if the dice were fair. A political polling company puts out a survey asking people two questions. First, it asks for the person's political party (whether the person identifies as a Republican, Democrat, or Independent). Second, it asks whether or not the person supports a proposed gun control law. The data are recorded in a contingency table. The company wants to determine if support for the bill depends on political party.

A casino wants to decide if a certain pair of dice is fair. They roll the dice 100 times and compare the distribution of outcomes with the expected distribution of outcomes if the dice were fair. The Goodness-of-Fit Test determines if data fit a particular distribution. It is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution.In this case, that applies to the casino trying to determine if the dice are fair.

The chi-square curve is skewed to the left. The chi-square curve is symmetrical. The total area under the χ2-curve is equal to the degrees of freedom, df. The following are all characteristics of the chi-square distribution. The chi-square curve is skewed to the right (it has a long tail to the right). The chi-square curve is nonsymmetrical, because it is skew to the right. The mean of the chi-square distribution is located to the right of the peak, because being skew to the right pulls the mean to the right. The total area under the χ2-curve is equal to 1, as is the case for any probability density function As the degrees of freedom increases, the chi-square curves look more and more like a normal curve. The chi-square curve approaches, but never touches, the positive horizontal axis, because the long right tail is technically never 0.

A referee wants to make sure the coin he uses for the opening coin toss is fair. He flips the coin 30 times and compares the number of heads and tails with the numbers he would expect to get if the coin were fair. The Goodness-of-Fit Test determines if data fit a particular distribution. It is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution.In this case, that applies to the referee comparing the number of heads and tails with the expected number of heads and tails.

A survey asks a group of college students about their majors. It also groups the students by whether they are athletes or not. These values are recorded in the contingency table below. Which of the following tables correctly shows the expected values for the chi-square homogeneity test? (The observed values are above the expected values.) Arts Humanities Sciences Row Total Athlete 9 8 9 26 Non-Athlete 12 23 10 45 Column Total 21 31 19 71

AthleteNon-AthleteColumn TotalArts7.7913.31221Humanities11.4819.62331Sciences79121019Row Total264571 In a chi-square test for homogeneity, as in the test of independence, the expected frequency for each data cell (used in calculating the test statistic), can be found by using the formula: E=R⋅Cn ...where R= the row total, C= the column total , and n= the total sample size. For example, the expected entry in the first row and first column (Athlete / Arts) isE=R⋅Cn=26⋅2171≈7.7

A real estate agent believes that the value of houses in the neighborhood she works in has increased from last year. To test this claim, she selects random houses in this neighborhood and compares their estimated market value in the current year to their estimated market value in the previous year. Suppose that data were collected for a random sample of 8 houses, where each difference is calculated by subtracting the market value of the previous year from the market value of the current year. Assume that the values are normally distributed. The agent uses the alternative hypothesis Ha:μd>0. Using a test statistic of t≈7.496, which has 7 degrees of freedom, determine the range that contains the p-value.

Correct answer: p-value <0.01 Notice that this is a right-tailed test because the alternative hypothesis is Ha:μd>0. The t-test statistic is 7.496 with 7 degrees of freedom. The t-distribution table gives the values for a right-tailed test. Using the table of areas in the right tail for the t-distribution, in the row for 7 degrees of freedom, 7.496 is greater than 2.998. So, the p-value is less than 0.01.

A physician wants to determine if a supplement is effective in helping men lose weight. She takes a random sample of overweight men and records their weights before the trial. She then prescribes the supplement and instructs them to take it for four weeks while making no other lifestyle changes. After the four-week period, she records the weights of the men again. Suppose that data were collected for a random sample of 6 men, where each difference is calculated by subtracting the weight before the trial from the weight after the trial. Assume that the weights are normally distributed. Identify the critical value of the t-test statistic for a left-tailed hypothesis test, where α=0.05.

Correct answers:$\text{Critical value}=-2.015$Critical value=−2.015 Since this is a left-tailed test and α=0.05, the critical value has a right-tail area of 1−0.05=0.95 or a left-tail area of 0.05. To find the critical value, use a t-distribution table to find the t-value that corresponds to one of these areas. Notice that t=−2.015 corresponds to a left-tail area of 0.05 and df=5, so the critical value is −tα=−t0.05=−2.015.

In a survey of randomly selected 21-year-old females and males, 259 of 500 females and 241 of 500 males are currently enrolled in college. Is there evidence that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds? Assume that the conditions for inference are satisfied. Use a 0.10 level of significance. Assign females to population 1 and males to population 2. (a) H0:p1=p2; Ha:p1>p2, which is a right-tailed test. (b) z≈1.14 , p-value is approximately 0.127 (c) Which of the following are appropriate conclusions for this hypothesis test?

Fail to reject H0. There is insufficient evidence at the 0.10 level of significance to conclude that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds. The p-value of 0.127 is greater than the level of significance, 0.10. Thus, we fail to reject the null hypothesis. There is insufficient evidence at the 0.10 level of significance to conclude that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds.

Claire is an education researcher looking to determine the relative effectiveness of two different teaching methods on standardized test scores. One group of high school seniors is taught using a traditional method, while a second group is taught using an experimental method. Claire selects a random sample of 30 students from each group and records the standardized test score for each student. Assume that the population standard deviation test score is 6 points for the traditional group, and 11 points for the experimental group, and that the test scores for both groups are normally distributed. Let the test scores of students taught using the traditional method be the first sample, and let the test scores of students taught using the experimental method be the second sample. The consulting group conducts a two-mean hypothesis test at the 0.10 level of significance, to test if there is evidence of a difference in test scores between the students taught using the two methods. (a) H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test. (b) z≈0.44 , p-value is approximately 0.662 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

Fail to reject H0. There is insufficient evidence at the 0.10 level of significance to conclude that there is a difference in test scores between the teaching methods. The p-value of 0.662 is greater than the level of significance, 0.10. Thus, we fail to reject the null hypothesis. There is insufficient evidence to conclude that there is a difference in test scores between the teaching methods.

A school district has a standardized test that it uses to sort students into magnet schools, where the test is on a 200-point scale. The superintendent's office wants to determine whether male and female students have approximately equal scores on the test. Scores for a random sample of 30 male students and 30 female students are taken. Assume that the population standard deviation of scores is 18 points for both male and female students, and that the scores are normally distributed for both male and female students. Let the scores of the male students be the first sample, and let the scores of the female students be the second sample. The consulting group conducts a two-mean hypothesis test at the 0.10 level of significance, to test if there is evidence that male and female students have different scores, on average. (a) H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test. (b) z≈−0.46 , p-value is approximately 0.646 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

Fail to reject H0. There is insufficient evidence at the 0.10 level of significance to conclude that there is a difference in the mean scores between male and female students. The p-value of 0.646 is greater than the level of significance, 0.10. Thus, we fail to reject the null hypothesis H0. There is insufficient evidence to conclude that there is a difference in scores between male and female students.

Agricultural scientists are testing a new pig feed to determine if it increases the pigs' weight gain. The scientists randomly selected a herd of 60 pigs and divided them into two groups, where 30 of the pigs were fed with the new feed and the other 30 pigs were fed with their regular feed. The following table shows the weight gain in pounds after 6 months for 30 pigs given the new feed and 30 pigs given the regular feed. Assume that the population variances of weight gained is the same for both groups and that the weight gains are normally distributed. Let the pigs given the new feed be the first sample, and let the pigs given the regular feed be the second sample. At the 0.05 level of significance, is there evidence that the new feed increases the amount of weight gained? Perform the test using Excel.

Fail to reject H0. There is insufficient evidence at the 0.05 level of significance to conclude that the mean weight gain of the pigs fed with the new feed is greater than the mean weight gain of the pigs fedwith the regular feed. Use Excel to perform the hypothesis test. 1. Since the scientists are testing whether the new feed increases the amount of weight gained, the null hypothesis is H0:μ1=μ2 and the alternative hypothesis is Ha:μ1>μ2. 2. Copy the data into a blank Excel worksheet. 3. Since it is assumed that the population variances are equal, select Data, and then Data Analysis and t-Test: Two-Sample Assuming Equal Variances. 4. In the dialog box, enter $A$1:$A$31 for Variable 1 Range and $B$1:$B$31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. Select Labels if the range contains the labels. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a one-sided alternative hypothesis (Ha:μ1>μ2), read the value from the cell labeled P(T<=t) one-tail. The test statistic is t≈1.31. The p-value is approximately 0.098. 6. The p-value of 0.098 is greater than the level of significance, 0.05. Thus, we fail to reject the null hypothesis. There is insufficient evidence at the 0.05 level of significance to conclude that the mean weight gain of the pigs fed with the new feed is greater than the mean weight gain of the pigs fed with the regular feed.

A researcher wants to test to see if husbands are significantly older than their wives. To do this, he collects the ages of husbands and pairs them with the ages of their respective wives for a random set of married couples. Suppose that data were collected for a random sample of 12 couples, where each difference is calculated by subtracting the age of the wife from the age of the husband. Assume that the ages are normally distributed. The test statistic is t≈1.434, α=0.05, the corresponding rejection region is t>1.796, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd>0. Which of the following statements are accurate for this hypothesis test in order to evaluate the claim that the true mean difference between the age of the husband and the age of the wife is greater than zero?

Fail to reject the null hypothesis that the true mean difference between the age of the husband and the age of his respective wife is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.05 level of significance to suggest that the true mean difference between the age of the husband and the age of his respective wife is greater than zero. Note that the t-test statistic is not within the rejection region. Therefore, fail to reject H0, and conclude that there is not enough evidence to suggest that the true mean difference between the age of the husband and the age of his respective wife is greater than zero. That is, there is not enough evidence to suggest that husbands are significantly older than their wives.

A physician wants to test two blood glucose meters to see if there is any difference in their measurements. To do this, he measures the glucose in blood samples taken from several randomly selected patients with both meters. Suppose that data were collected for a random sample of 10 patients, where each difference is calculated by subtracting the blood glucose rating from Meter A from the blood glucose reading from Meter B. Assume that the measurements are normally distributed. The test statistic is t≈1.015, α=0.01, the corresponding rejection regions are t<−3.250 and t>3.250, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd≠0. Which of the following statements are accurate for this hypothesis test in order to evaluate the claim that the true mean difference between the blood glucose reading from Meter B and the reading from Meter A is significantly not equal to zero?

Fail to reject the null hypothesis that the true mean difference between the blood glucose reading from Meter B and the reading from Meter A is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.01 level of significance to suggest that the true mean difference between the blood glucose reading from Meter B and the reading from Meter A is not equal to zero. Note that the t-test statistic is not within either of the rejection regions. Therefore, fail to reject H0, and conclude that there is not enough evidence to suggest that the true mean difference between the blood glucose reading from Meter B and the reading from Meter A is not equal to zero. That is, there is not enough evidence to suggest that the blood glucose measurements are different on the two meters.

A researcher wants to compare the heights of males between generations to see if they differ. To do this, he samples random pairs of males who are at least 18 years old and their fathers. He then splits them into a sample of fathers and a sample of sons. Suppose that data were collected for a random sample of 11 pairs, where each difference is calculated by subtracting the height of the son from the height of the father. Assume that the heights are normally distributed. The test statistic is t≈1.971, α=0.05, the corresponding rejection regions are t<−2.228 and t>2.228, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd≠0.

Fail to reject the null hypothesis that the true mean difference between the height of the father and the height of the son is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.05 level of significance to suggest that the true mean difference between the height of the father and the height of the son is not equal to zero. Note that the t-test statistic is not within either of the rejection regions. Therefore, fail to reject H0, and conclude that there is not enough evidence to suggest that the true mean difference between the height of the father and the height of the son is not equal to zero. That is, there is not enough evidence to suggest that the heights of males between generations differ.

A farmer wants to test if a new fertilizer will produce more massive crops. Suppose that data were collected for a random set of 6 crops, where each difference is calculated by subtracting the mass of the crop the previous year with the old fertilizer from the mass of the crop this year with the new fertilizer. Assume that the populations are normally distributed. The test statistic is t≈1.266, α=0.10, the corresponding rejection region is t>1.476, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd>0. Which of the following statements are accurate for this hypothesis test in order to evaluate the claim that the true mean difference between the mass of the crops with the new fertilizer and the mass of the crops with the old fertilizer is greater than zero?

Fail to reject the null hypothesis that the true mean difference between the mass of the crops with the new fertilizer and the mass of the crops with the old fertilizer is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.10 level of significance to suggest that the true mean difference between the mass of the crops with the new fertilizer and the mass of the crops with the old fertilizer is greater than zero. Note that the t-test statistic is not within the rejection region. Therefore, fail to reject H0, and conclude that there is not enough evidence to suggest that the true mean difference between the mass of the crops with the new fertilizer and the mass of the crops with the old fertilizer is greater than zero.

Fail to reject the null hypothesis that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is equal to zero. There is not sufficient evidence at the α=0.01 level of significance to suggest that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is significantly less than zero. Since the p-value is greater than α=0.01, fail to reject the null hypothesis. There is not enough evidence to suggest that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is significantly less than zero. Your answer: Reject the null hypothesis that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is equal to zero. This response is not correct. In this example, it was determined that the p-value was approximately 0.030. Since the p-value is greater than the level of significance, then the decision should be to fail to reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to suggest that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is significantly less than zero. This response is not correct. The decision is to fail to reject the null hypothesis. Failing to reject the null hypothesis means there is not enough evidence to suggest that the true mean difference in the weight of tanks filled by Brandon and the weight of tanks filled by Hunter is significantly less than zero.

A researcher wants to test to see if husbands are significantly older than their wives. To do this, he collects the ages of husbands and pairs them with the ages of their respective wives for a random set of married couples. Suppose that data were collected for a random sample of 12 couples, where each difference is calculated by subtracting the age of the wife from the age of the husband. Assume that the ages are normally distributed. Using a test statistic of t≈1.434, the significance level α=0.05, and the corresponding p-value between 0.05 and 0.10, draw a conclusion for the appropriate hypothesis test, where the null hypothesis is H0:μd=0 and the alternative hypothesis is Ha:μd>0.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that husbands are significantly older than their wives. Compare the p-value to α=0.05. Since the p-value is between 0.05 and 0.10, it is greater than α, so we fail to reject H0. Therefore, there is insufficient evidence at the α=0.05 level of significance to suggest that husbands are significantly older than their wives.

A doctor wants to research if there is any difference in the birth weights of a mother's first child and second child. Suppose that data were collected for a random sample of 10 mothers with 2 children, where each difference is calculated by subtracting the weight of the first child from the weight of the second child. Assume that the weights are normally distributed. The doctor uses the alternative hypothesis Ha:μd≠0. Suppose the test statistic t is calculated as −1.101, which has 9 degrees of freedom. If the p-value is greater than 0.10 and the significance level is α=0.10, what conclusion can be made about the birth weights of a mother's first 2 children?

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that the birth weights of a mother's first 2 children are different. Compare the p-value to α=0.10. Since it is given that the p-value is greater than α, fail to reject H0. Therefore, there is insufficient evidence at the α=0.10 level of significance to suggest that the birth weights of a mother's first 2 children are different.

A researcher wants to compare the heights of males between generations to see if they differ. To do this, he samples random pairs of males who are at least 18 years old and their fathers. He then splits them into a sample of fathers and a sample of sons. Suppose that data were collected for a random sample of 11 pairs, where each difference is calculated by subtracting the height of the son from the height of the father. Assume that the heights are normally distributed. Using a test statistic of t≈1.971, the significance level α=0.05, and the corresponding p-value between 0.05 and 0.10, draw a conclusion for the appropriate hypothesis test, where the null hypothesis is H0:μd=0 and the alternative hypothesis is Ha:μd≠0.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that the heights of males between generations are different. Compare the p-value to α=0.05. Since the p-value is between 0.05 and 0.10, it is greater than α, so we fail to reject H0. Therefore, there is insufficient evidence at the α=0.05 level of significance to suggest that the heights of males between generations are different.

A farmer wants to test if a new fertilizer will produce more massive crops. Suppose that data were collected for a randomly selected set of 6 crops, where each difference is calculated by subtracting the mass of the crop the previous year with the old fertilizer from the mass of the crop this year with the new fertilizer. Assume that the masses are normally distributed. Using a test statistic of t≈1.266, the significance level α=0.10, and the corresponding p-value greater than 0.10, draw a conclusion for the appropriate hypothesis test, where the null hypothesis is H0:μd=0 and the alternative hypothesis is Ha:μd>0.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that the new fertilizer will produce more massive crops. Compare the p-value to α=0.10. Since the p-value is greater than α, fail to reject H0. Therefore, there is insufficient evidence at the α=0.10 level of significance to suggest that the fertilizer produces more massive crops.

A physician wants to test two blood glucose meters to see if there are any differences in measurements. To do this, he uses both meters to measure the glucose in blood samples taken from several randomly selected patients. Suppose that data were collected for a random sample of 10 patients, where each difference is calculated by subtracting the blood glucose rating from Meter A from the blood glucose reading from Meter B. Assume that the measurements are normally distributed. Using a test statistic of t≈1.015, the significance level α=0.01, and the corresponding p-value greater than 0.10, draw a conclusion for the appropriate hypothesis test, where the null hypothesis is H0:μd=0 and the alternative hypothesis is Ha:μd≠0.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that there is a difference in measurements of the two blood glucose meters. Compare the p-value to α=0.01. Since the p-value is greater than 0.10, it is greater than α=0.01, so we fail to reject H0. Therefore, there is insufficient evidence at the α=0.01 level of significance to suggest that there is a difference in measurements of the two blood glucose meters.

An insurance agent is curious whether the accident rate for single drivers is different from the rate for married drivers. A random sample of 1,000 insurance policies included 572 single drivers and 428 married drivers. During the previous three years, 176 of the single drivers and 152 of the married drivers had been involved in at least one accident. Test the alternative hypothesis that the population proportion for single drivers is different from the population proportion for married drivers. The critical values for the level of significance α=0.05 are z0.025=−1.960 and z0.975=1.960. The test statistic is z=−1.58. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the accident rates are different for single drivers and married drivers. Notice that this is a two-tailed test. The rejection regions contain values of the test statistic less than the negative critical value and values of the test statistic greater than the positive critical value. Since the test statistic −1.58 is greater than the critical value −1.960 and less than the critical value 1.960, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the accident rates are different for single drivers and married drivers.

In a survey of 75 randomly selected people in Country A, 12 would like to travel abroad. In a survey of 60 randomly selected people in Country B, 12 would like to travel abroad. Test the alternative hypothesis that the population proportion for Country A is less than the population proportion for Country B. The critical value for the level of significance α=0.05 is z0.05=−1.645. The test statistic is z=−0.60. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion for Country A is less than the population proportion for Country B. Notice that this is a left-tailed test. The rejection region contains values of the test statistic less than the critical value. Since the test statistic of −0.60 is greater than the critical value of −1.645, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion for Country A is less than the population proportion for Country B.

A survey asked randomly selected drivers if they would be willing to pay higher taxes for gasoline as long as all of the additional revenue went to improving roadways. Among the 41 males surveyed, 12 responded "Yes." Of the 59 females surveyed, 21 responded "Yes." Test the alternative hypothesis that the population proportion for males is less than the population proportion for females. Use the level of significance α=0.10. The test statistic is z≈−0.66, and the p-value is approximately 0.255. Identify all of the appropriate conclusions to the hypothesis test below.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion for males is less than the population proportion for females The p-value is greater than the level of significance, so fail to reject the null hypothesis. Conclude that there is not enough evidence to support the claim that the population proportion for males is less than the population proportion for females.

A survey asked randomly selected people if they think it is rude for people to interact with their smartphones while in a restaurant. Among the 92 females surveyed, 57 responded "Yes." Of the 108 males surveyed, 62 responded "Yes." Test the alternative hypothesis that the population proportion for females is greater than the population proportion for males. The critical value for the level of significance α=0.01 is z0.99=2.326. The test statistic is z=0.65. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of females who think interacting with a smartphone while in a restaurant is rude is greater than the population proportion for males. Notice that this is a right-tailed test. The rejection region contains values of the test statistic that are greater than the critical value. Since the test statistic of 0.65 is less than the critical value of 2.326, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of females who think interacting with a smartphone while in a restaurant is rude is greater than the population proportion for males.

A survey asked randomly selected drivers if they would be willing to pay higher taxes for gasoline as long as all of the additional revenue went to improving roadways. Among the 41 males surveyed, 12 responded "Yes." Of the 59 females surveyed, 21 responded "Yes." Test the alternative hypothesis that the population proportion for males is less than the population proportion for females. The critical value for the level of significance α=0.10 is z0.10=−1.282. The test statistic is z=−0.66. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of males willing to pay higher taxes is less than the population proportion for females. Notice that this is a left-tailed test. The rejection region contains values of the test statistic that are less than the critical value. Since the test statistic of −0.66 is greater than the critical value of −1.282, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of males willing to pay higher taxes is less than the population proportion for females.

In a survey of 100 randomly selected people in City A, 82 support increased government spending on roads and bridges. In a survey of 100 randomly selected people in City B, 72 support such spending. Test the alternative hypothesis that the population proportion of people in City A that support such spending is different from the population proportion of people in City B that support such spending. The critical values for the level of significance α=0.05 are z0.025=−1.960 and z0.975=1.960. The test statistic is z=1.68. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of people in City A that support such spending is different from the population proportion of people in City B that support such spending. Notice that this is a two-tailed test. The rejection regions contains values of the test statistic that are less than the negative critical value and values of the test statistic that are greater than the positive critical value. Since the test statistic of 1.68 is greater than the critical value of −1.960 and less than the critical value of 1.960, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of people in City A that support such spending is different from the population proportion of people in City B that support such spending.

In a survey of 100 randomly selected people in City A, 73 support increased government spending on education. In a survey of 100 randomly selected people in City B, 81 support such spending. Test the alternative hypothesis that the population proportion of people in City A that support such spending is different from the population proportion of people in City B. Use the level of significance α=0.05. The test statistic is z≈−1.34, and the p-value is approximately 0.180. Identify all of the appropriate conclusions to the hypothesis test below.

In a survey of 100 randomly selected people in City A, 81 support increased government spending on education. In a survey of 100 randomly selected people in City B, 87 support such spending. Test the alternative hypothesis that the population proportion of people in City A that support such spending is less than the population proportion of people in City B that support such spending. The critical value for the level of significance α=0.01 is z0.01=−2.326. The test statistic is z=−1.16. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of people in City A that support such spending is less than the population proportion of people in City B that support such spending. Notice that this is a left-tailed test. The rejection region contains values of the test statistic that are less than the critical value. Since the test statistic of −1.16 is greater than the critical value of −2.326, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of people in City A that support such spending is less than the population proportion of people in City B that support such spending.

To test a drug intended to increase memory, 80 subjects were randomly given a pill that either contained the drug or was a placebo. After 10 minutes, the subjects were asked to look at a poster with pictures of 20 common objects for 5 minutes. After waiting 10 minutes, the subjects were asked to list as many of the objects as they could recall. For the 44 subjects who took the drug, 75% were able to list at least half of the objects. For the 36 subjects who took the placebo, 50% were able to list at least half of the objects. Test the alternative hypothesis that the population proportion of subjects who took the drug and were able to list at least half of the objects is different from the population proportion of subjects who took the placebo and were able do so. The critical values for the level of significance α=0.01 are z0.005=−2.576 and z0.995=2.576. The test statistic is z=2.31. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of subjects who were able to list at least half of the objects is different for the groups. Notice that this is a two-tailed test. The rejection regions contain values of the test statistic that are less than the negative critical value and values of the test statistic that are greater than the positive critical value. Since the test statistic of 2.31 is greater than the critical value of −2.576 and less than the critical value of 2.576, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportions of subjects who were able to list at least half of the objects is different for the groups.

In a survey of 100 randomly selected taxi drivers in a city, 76 support increased government spending on roads and bridges. In a survey of 100 randomly selected bus drivers in the same city, 82 support such spending. Test the alternative hypothesis that the population proportion of taxi drivers in the city that support such spending is less than the population proportion of bus drivers in the city that support such spending. The critical value for the level of significance α=0.01 is z0.01=−2.326. The test statistic is z=−1.04. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of taxi drivers in the city that support increased government spending on roads and bridges is less than the population proportion of bus drivers in the city that support increased government spending on roads and bridges. Notice that this is a left-tailed test. The rejection region contains values of the test statistic less than the critical value. Since the test statistic of −1.04 is greater than the critical value of −2.326, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of taxi drivers in the city that support increased government spending on roads and bridges is less than the population proportion of bus drivers in the city that support increased government spending on roads and bridges.

In a survey of 100 randomly selected taxi drivers in a city, 76 support increased government spending on roads and bridges. In a survey of 100 randomly selected bus drivers in the same city, 82 support such spending. Test the alternative hypothesis that the population proportion of taxi drivers in the city that support such spending is less than the population proportion of bus drivers in the city that do. Use the level of significance α=0.05. The test statistic is z≈−1.04 and the p-value is approximately 0.149. Identify all of the appropriate conclusions to the hypothesis test below.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of taxi drivers in the city that support such spending is less than the population proportion of bus drivers in the city that do. The p-value is greater than the level of significance, so fail to reject the null hypothesis. Conclude that there is not enough evidence to support the claim that the population proportion of taxi drivers in the city that support such spending is less than the population proportion of bus drivers in the city that do.

In a sample of 35 randomly selected vehicles in City A, 7 use alternative energy sources. In a sample of 45 randomly selected vehicles in City B, 11 use alternative energy sources. Test the alternative hypothesis that the population proportion for City A is less than the population proportion for City B. The critical value for the level of significance α=0.01 is z0.01=−2.326. The test statistic is z=−0.47. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportion of vehicles that use alternative energy sources is less in City A than in City B. Notice that this is a left-tailed test. The rejection region contains values of the test statistic that are less than the critical value. Since the test statistic of −0.47 is greater than the critical value of −2.326, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of vehicles that use alternative energy sources is less in City A than in City B.

In a survey of 75 randomly selected people in Country A, 12 would like to travel abroad. In a survey of 60 randomly selected people in Country B, 12 would like to travel abroad. Test the alternative hypothesis that the population proportion for Country A is different from the population proportion for Country B. The critical values for the level of significance α=0.02 are z0.01=−2.326 and z0.99=2.326. The test statistic is z=−0.60. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportions are different for the two countries. Notice that this is a two-tailed test. The rejection regions contain values of the test statistic that are less than the negative critical value and values of the test statistic that are greater than the positive critical value. Since the test statistic of −0.60 is greater than the critical value of −2.326 and less than the critical value of 2.326, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion for Country A is different from the population proportion for Country B.

A survey asked randomly selected drivers if they would be willing to pay higher taxes for gasoline as long as all of the additional revenue went to improving roadways. Among the 52 males surveyed, 12 responded "Yes." Of the 59 females surveyed, 21 responded "Yes." Test the alternative hypothesis that the population proportion for males is different from the population proportion for females. The critical values for the level of significance α=0.02 are z0.01=−2.326 and z0.99=2.326. The test statistic is z=−1.44. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportions of males and females willing to pay higher taxes are different. Notice that this is a two-tailed test. The rejection regions contain values of the test statistic less than the negative critical value and values of the test statistic greater than the positive critical value. Since the test statistic −1.44 is greater than the critical value −2.326 and less than the critical value 2.326, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of males and females willing to pay higher taxes are different.

A researcher wants to show that the proportions of mosquitoes that carry a certain disease in Region 1 and Region 2 are different. In a sample of 1392 mosquitoes trapped in Region 1, 1173 test positive for the disease. In a sample of 1457 mosquitoes trapped in Region 2, 1196 test positive for the disease. Test the alternative hypothesis that the population proportion for Region 1 is different from the population proportion for Region 2. The critical values for the level of significance α=0.05 are z0.025=−1.960 and z0.975=1.960. The test statistic is z=1.55. Identify all of the appropriate conclusions to the hypothesis test.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to support the claim that the population proportions of mosquitoes that carry the disease are different for Region 1 and Region 2. Notice that this is a two-tailed test. The rejection regions contain values of the test statistic less than the negative critical value and values of the test statistic greater than the positive critical value. Since the test statistic 1.55 is greater than the critical value −1.960 and less than the critical value 1.960, fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the population proportion of mosquitoes that carry the disease is different for Region 1 and Region 2.

A professor expects that the grades on a recent exam are normally distributed with a mean of 80 and a standard deviation of 10. She records the actual distribution of grades and wants to compare it to the normal distribution.Which of the following χ2 tests should be used in the situation above? Select the correct answer below: Test of Independence Goodness-of-Fit Test Test for Homogeneity

Goodness-of-Fit Test The Goodness-of-Fit Test determines if data fit a particular distribution. It is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution.In this case, the professor wants to compare the distribution of scores with the normal distribution, so the Goodness-of-Fit Test is most appropriate.

Do people in different political parties use similar modes of transportation? Researchers ask a group of people about how they get to work. They also group the people by their political party. They want to determine if the distribution of transportation is the same for democrats and republicans. The data are recorded in the contingency table below, and a chi-square Homogeneity Test at the 1% significance level is performed. Car Public Transit Walk Row Total Democrats 25 17 19 61 Republicans 20 6 20 46 Column Total 45 23 39 107 (a) Select the correct null and alternative hypotheses for this test.

H0: The distribution of the two populations are the same. Ha: The distribution of the two populations are not the same. For a chi-square Homogeneity test, the null and alternative hypotheses will always look like: H0: The two populations have the same distribution.Ha: The two populations do not have the same distribution.

A professor is trying to determine if her students guessed on a certain multiple choice question. She expects that if the students guessed, the distribution of answers would be uniform for that question. She compares the observed distribution of answers with the uniform distribution. The professor conducts a chi-square Goodness-of-Fit hypothesis test at the 5% significance level. (a) Select the correct null and alternative hypotheses for this test. Select all that apply: H0: The student answers have the uniform distribution. H0: The student answers do not have the uniform distribution. Ha: The student answers do not have the uniform distribution. Ha: The student answers have the uniform distribution.b

H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution. For the chi-square Goodness-of-Fit test, the null hypothesis assumes the expected distribution and looks at how likely the observed data is under that assumption. So the hypotheses will always look like: H0: The variable has the specified distribution. Ha: The variable does not have the specified distribution.

Does college major depend on whether you are an athlete or not? A survey asks a group of college students about their majors. It also groups the students by whether they are athletes or not. The data are recorded in the contingency table below, and a chi-square Test of Independence at the 1% significance level is performed.

H0: The two variables are independent, so being an athlete does not affect college major. Ha: The two variables are dependent, so being an athlete does affect college major. For a chi-square Independence test, the null and alternative hypotheses will always look like: H0: The two variables are independent.Ha: The two variables are dependent.

Let p1 be the percentage of car accidents involving teen drivers in Country A, and let p2 be the percentage of car accidents involving teen drivers in Country B. What is the null hypothesis for a test to determine whether Country A has a higher percentage of car accidents involving teen drivers than Country B?

H0: p1−p2=0 The null hypothesis (H0) states that there is no difference between the proportion of successes for the populations, so p1=p2, or p1−p2=0. H0:p1−p2=0

Suppose that for two species of bird, the proportion of eggs that hatch for Species 1 is p1 and the proportion of eggs that hatch for Species 2 is p2. State the null hypothesis for a test to determine whether the proportion of eggs that hatch for Species 1 is greater than the proportion for Species 2.

H0: p1−p2=0 The null hypothesis (H0) states that there is no difference between the proportion of successes for the populations, so p1=p2, or p1−p2=0. H0:p1−p2=0

Suppose the fraction of white sheep in Herd A is p1 and the fraction of white sheep in Herd B is p2. State the null hypothesis for a test to determine if Herd A has a lower proportion of white sheep.

H0: p1−p2=0 The null hypothesis (H0) states that there is no difference between the proportion of successes for the populations, so p1=p2, or p1−p2=0. H0:p1−p2=0

The proportion of mango trees on an island that produce only dwarf fruit is p1. The proportion of mango trees on a different island that produce only dwarf fruit is p2. What is the null hypothesis for a test to determine if the proportions are different between the two islands?

H0: p1−p2=0 The null hypothesis (H0) states that there is no difference between the proportion of successes for the populations, so p1=p2, or p1−p2=0. H0:p1−p2=0

Going into Game 7 of a playoff hockey series, the goalie for the home team had faced 151 shots and stopped 142 of them. The goalie for the road team saved 126 of 145 shots. Use this information and the appropriate hypothesis test to determine whether the home goalie has a better true save percentage than the road goalie. Assume that the conditions for inference are satisfied. Use a 0.05 level of significance. Assign the home goalie data to sample 1 and the other goalie to sample 2. Identify all of the appropriate conclusions. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:p1=p2; Ha:p1>p2, which is a right-tailed test. The null hypothesis should be an equality: H0:p1=p2. The study wants to know if the home goalie has a better true save percentage than the road goalie. This means that we want to test if the home goalie has a greater true save percentage than the road goalie, so the alternative hypothesis is H0:p1=p2; Ha:p1>p2, which is a right-tailed test.

In a survey of randomly selected 21-year-old females and males, 259 of 500 females and 241 of 500 males are currently enrolled in college. Is there evidence that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds? Assume that the conditions for inference are satisfied. Use a 0.10 level of significance. Assign females to population 1 and males to population 2. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:p1=p2; Ha:p1>p2, which is a right-tailed test. The null hypothesis should be an equality: H0:p1=p2. The study wants to know if the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds. This means that we want to test if the proportion of female 21-year-olds currently enrolled in college is more than the proportion of male 21-year-olds, so the alternative hypothesis isH0:p1=p2; Ha:p1>p2, which is a right-tailed test.

In a survey of randomly selected 21-year-old females and males, 261 of 500 females and 232 of 500 males are currently enrolled in college. Is there evidence that the proportion of female 21-year-olds currently enrolled in college is greater than the proportion of male 21-year-olds? Assume that the conditions for inference are satisfied. Use a 0.05 level of significance. Assign females to population 1 and males to population 2. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

Christine is the dean of students at a college. She is concerned about the average amount of sleep students in the pre-med program are getting. Christine makes an inquiry on whether students who are enrolled in the pre-med program are getting less sleep on average compared to pre-law students. The average number of hours of sleep per night for a random sample of 30 pre-med students and 30 pre-law students is recorded. Assume that the population variances of hours of sleep per night for pre-med students and for pre-law students are not equal and that the hours of sleep per night for both groups of students are normally distributed. Let the average number of hours of sleep per night for pre-med students be the first sample, and let the average number of hours of sleep per night for pre-law students be the second sample. She conducts a two-mean hypothesis test at the 0.05 level of significance, to test if there is evidence that pre-med students are getting less sleep than pre-law students. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ1=μ2; Ha:μ1<μ2, which is a left-tailed test. The null hypothesis should be an equality: H0:μ1=μ2. The study wants to know if pre-med students are getting less sleep than pre-law students. This means that we want to test if the mean sleep for pre-med students is less than the mean sleep for other students, so the alternative hypothesis is Ha:μ1<μ2, which is a left-tailed test.

A baseball pitcher, concerned about losing speed from his fastball, undertook a new training regimen during the offseason. His team's pitching coach measured the speed of 20 random fastballs (in miles per hour) thrown by the pitcher during spring training, and compared it with a sample of 20 random fastballs thrown during the pitcher's last five starts in the previous season. The results are shown in the following table. Assume that the pitcher's fastball speeds had a standard deviation of 2.9 miles per hour both before and after the training regimen and that the speeds for both time periods are normally distributed. Let the pitcher's fastball speeds in the previous season be the first sample, and let the pitcher's fastball speeds in spring training be the second sample. At the 0.05 level of significance, is there evidence that the pitcher is throwing fastballs at higher speeds? Find the test statistic, rounded to two decimal places, and the p-value, rounded to three decimal places.

H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test. The null hypothesis should be an equality: H0:μ1=μ2. Since the group claims that there is a difference in test scores between the students taught using the two methods., the alternative hypothesis is H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test.

A school district has a standardized test that it uses to sort students into magnet schools, where the test is on a 200-point scale. The superintendent's office wants to determine whether male and female students have approximately equal scores on the test. Scores for a random sample of 30 male students and 30 female students are taken. Assume that the population standard deviation of scores is 18 points for both male and female students, and that the scores are normally distributed for both male and female students. Let the scores of the male students be the first sample, and let the scores of the female students be the second sample. The consulting group conducts a two-mean hypothesis test at the 0.10 level of significance, to test if there is evidence that male and female students have different scores, on average. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test. The null hypothesis should be an equality: H0:μ1=μ2. Since the superintendent's office wants to know whether there is a difference in test scores between genders, the alternative hypothesis is H0:μ1=μ2; Ha:μ1≠μ2, which is a two-tailed test.

H0:μd=0, Ha:μd<0, which is a left-tailed test. Brandon wants to test if the hypothesized mean of the differences for the paired data is less than 0. Therefore, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd<0, which is a left-tailed test.

A professional golfer wondered if she should switch to a new line of golf clubs that are supposed to put a spin on the golf ball that reduces the variance in the ball's flight. She obtained a new set of the clubs and paired off each new club with one of her old clubs. Then she hit golf balls with the pairs of clubs in 30 situations that would be appropriate to the type of club and measured how far the golf balls landed from her target. She tests the paired data, where α=0.10, in order to evaluate the claim that the true mean difference in distance is significantly greater than zero (determine if the new clubs perform better than the old clubs). (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μd=0; Ha:μd>0, which is a right-tailed test. She wants to test if the hypothesized mean of the differences for the paired data is greater than 0. Therefore, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd>0, which is a right-tailed test.

Do college athletes have similar majors as non-athletes? A survey asks a group of college students about their majors. It also groups the students by whether they are athletes or not. The researchers want to determine if the distribution of college majors is the same for athletes as for non-athletes. The data are recorded in the contingency table below, and a chi-square Homogeneity Test at the 1% significance level is performed. Arts Humanities Sciences Row Total Athlete 12 20 20 52 Non-Athlete 21 6 11 38 Column Total 33 26 31 90 (a) Select the correct null and alternative hypotheses for this test.

Ha: The distribution of the two populations are not the same. H0: The distribution of the two populations are the same. For a chi-square Homogeneity test, the null and alternative hypotheses will always look like: H0: The two populations have the same distribution.Ha: The two populations do not have the same distribution.

A bill is brought to a group of politicians - democrats and republicans. Researchers recorded the expected values, and then polled the group on whether they would vote "in favor," "opposed," or "indifferent." These values are recorded in the contingency table below. Which of the following tables correctly shows the expected values for the chi-square independence test? In favor Indifferent Opposed Row Total Republicans 10 16 24 50 Democrats 21 5 21 47 Column Total 31 21 45 97

In favorIndifferentOpposedRow TotalRepublicans161010.81623.22450Democrats152110.2521.82147Column Total31214597 In a chi-square test for independence, the expected frequency for each data cell (used in calculating the test statistic), can be found by using the formula:E=R⋅Cn...where R= the row total, C= the column total , and n= the total sample size. For example, the expected entry in the first row and first column (Republicans / In favor) isE=R⋅Cn=50⋅3197≈16

The following data set provides infromation on acres harvested and planted by year and value in the US. From the data set, use the planting data for the United States for recent years. Using a calculator or statistical software, find the equation for the linear regression line for 2010 to 2015. What is the pattern of the data set?

Negative linear pattern As the years increased, the acres of asparagus planted decreased. The fit of the data is close to the line, without striking deviations. (R2=0.9407)

The scatter plot below shows data relating total income and the number of children a family has. Which of the following patterns does the scatter plot show?

Negative linear pattern with deviations We can see that as the x-values (number of children) increase, the y-values (income) decrease. So it is a negative relationship. But there are points that deviate from the negative linear pattern, so it is a weaker pattern, instead of a strong one.

You want to conduct a hypothesis test to determine if there is a difference in the average salaries of police officers, versus firefighters, versus paramedics. You determine that the salaries do not have equal variances in that firefighter salaries are more variable than salaries for paramedics. Assume the populations are normally distributed. Determine if the conditions needed for conducting a hypothesis test using ANOVA are met. The five conditions needed for ANOVA are summarized below: Each population from which a sample is taken is assumed to be normal. All samples are randomly selected and independent. The populations are assumed to have equal standard deviations (or variances). The factor is a categorical variable. The response is a numerical variable. Select the correct answer below: Yes, the conditions the conditions needed for conducting a hypothesis test using ANOVA are met. No, the conditions needed for conducting a hypothesis test using ANOVA are not met.

No, the conditions needed for conducting a hypothesis test using ANOVA are not met. In this example, the conditions needed for conducting a hypothesis test using ANOVA are not met. Each population from which a sample is taken is assumed to be normal. Condition #1 is met since we assume the salaries follows a normal distribution. All samples are randomly selected and independent. Condition #2 is met since the sample of salaries for each of the occupations are being selected randomly and the samples are independent. The populations are assumed to have equal standard deviations (or variances). Condition #3 is not met since it was determined that the salaries do not have equal variances in that firefighter salaries are more variable than salaries for paramedics. The factor is a categorical variable. Condition #4 is met since the factor in this example is the type of occupation, which is a categorical variable. The response is a numerical variable. Condition #5 is met since the response here is the salaries which is a numerical variable.

A librarian would like to determine whether a new book appeals more to adults or to children. The librarian asks 10 randomly selected adults and their children whether or not they will be interested in reading the book. The librarian then sorts the responses into two samples: one sample for the adults' answers and one sample for their children's answers. Are these samples independent? If not, how can these samples be made independent?

No, the samples are dependent. To make the samples independent, the librarian should take a random sample of adults and then separately take a random sample of children who are not related to those adults. An adult's response can affect the response of the child, and the opposite is also true. Thus, the samples are dependent. To fix this, the librarian can separately sample a random sample of adults and a random sample of children to make sure that the two samples are not related.

The following data set provides information on the lottery sales, proceeds, and prizes by year in Iowa. When only comparing proceeds with prizes, which category is having the largest increase in amount over the entire time period?

Prizes The graph and the data set show the largest increase in prizes, from 39 million dollars to 222 million dollars, a difference of 183 million dollars. The proceeds went from 28 million dollars to 88 million dollars, a difference of only 60 million dollars.

Going into Game 7 of a playoff hockey series, the goalie for the home team had faced 151 shots and stopped 142 of them. The goalie for the road team saved 126 of 145 shots. Use this information and the appropriate hypothesis test to determine whether the home goalie has a better true save percentage than the road goalie. Assume that the conditions for inference are satisfied. Use a 0.05 level of significance. Assign the home goalie data to sample 1 and the other goalie to sample 2. Identify all of the appropriate conclusions. (a) H0:p1=p2; Ha:p1>p2, which is a right-tailed test. (b) z≈2.10 , p-value is approximately 0.018 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply

Reject H0. There is sufficient evidence at the 0.05 level of significance to conclude that the home goalie has a better true save percentage than the road goalie. The p-value of 0.018 is less than the level of significance, 0.05. Thus, we reject the null hypothesis. There is sufficient evidence at the 0.05 level of significance to conclude that the home goalie has a better true save percentage than the road goalie.

Agricultural scientists are testing a new pig feed to determine whether it increases the pigs' weights. The scientists divided a herd of 60 pigs into two random groups, where 30 of the pigs are fed with the new feed, while the other 30 pigs were fed with their regular feed. The following table shows the weight gain after 6 months for both groups of pigs. Assume that the population standard deviation of weight gained is 10 pounds for the pigs fed with the new feed and 8 pounds for the pigs fed with the regular feed. Also assume that the weight gains are normally distributed. Let the weight gained by the pigs fed with the new feed be the first sample, and let the weight gained by the pigs fed with the regular feed be the second sample. At the 0.01 level of significance, is there evidence that the new feed increases the amount of weight gained? Perform the test using Excel.

Reject H0. There is sufficient evidence at the 0.01 level of significance to conclude that the mean weight gained by the pigs fed with the new feed is greater than the mean weight gained by the pigs fed with the existing feed. Use Excel to perform the hypothesis test. 1. Since the scientists want to know whether the new feed increases the amount of weight gained, the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1>μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the square of the known population standard deviation, or 100 and 64, respectively. Make sure Labels is selected. Click OK. 5. Read the test statistic and P-value from the output. Since the test uses a one-sided alternative hypothesis (μ1>μ2), read the p-value from the cell labeled P(Z<=z) one-tail. The test statistic is z≈5.15. The p-value is 0.000. 6. The p-value of 0.000 is less than the level of significance, 0.01. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that the new feed increases the pigs' weights compared to the regular feed.

A manufacturer buys a particular electrical component used in its products from two suppliers, Supplier A and Supplier B. Raquel is a manager for the manufacturer and is looking to determine whether there is a difference in the reliability of the components from each supplier. She collected a random sample of 30 components from each supplier and determined the time to failure for each component, in days. Raquel records the failure time for each component, as shown in the accompanying table. Assume that the population standard deviation of time to failure for each supplier's components is 52 days and that the distribution of time to failure for each supplier's components is normally distributed. Let the time to failure for Supplier A's components be the first sample, and let the time to failure for Supplier B's components be the second sample. At the 0.05 level of significance, is there evidence of a difference in the mean time to failure between each supplier's components? Perform the test using Excel.

Reject H0. There is sufficient evidence at the 0.05 level of significance to conclude that there is a difference in the mean time to failure between the components from Supplier A and the components from Supplier B. Use Excel to perform the hypothesis test. 1. Since Raquel is looking to determine whether there is a difference in mean time to failure, the null hypothesis is μ1=μ2 and the alternative hypothesis is μ1≠μ2. 2. Copy the data into a blank Excel worksheet. 3. Select Data, and then Data Analysis and z-Test: Two Sample for Means. 4. In the dialog box, enter A1:A31 for Variable 1 Range and B1:B31 for Variable 2 Range. Enter 0 for the Hypothesized Mean Difference. For Variable 1 Variance and Variable 2 Variance, enter the square of the known population standard deviation, or 2704. Make sure Labels is selected. Click OK. 5. Read the test statistic and p-value from the output. Since the test uses a two-sided alternative hypothesis (μ1≠μ2), read the p-value from the cell labeled P(Z<=z) two-tail. The test statistic is z≈12.55. The p-value is 0.000. 6. The p-value of 0.000 is less than the level of significance, 0.05. Thus, we reject the null hypothesis. There is sufficient evidence to conclude that there is a difference in mean time to failure between each supplier's components.

Your friend believes that he has found a route to work that would make your commute faster than what it currently is under similar conditions. Suppose that data were collected for a random set of 7 days, where each difference is calculated by subtracting the time taken on the current route from the time taken on the new route. Assume that the populations are normally distributed. The test statistic is t≈−3.201, α=0.05, the corresponding rejection region is t<−1.943, the null hypothesis is H0:μd=0, and the alternative hypothesis is Ha:μd<0. Which of the following statements are accurate for this hypothesis test in order to evaluate the claim that the true mean difference between the commute time using the new route and the commute time using the current route is significantly less than zero?

Reject the null hypothesis that the true mean difference between the commute time using the new route and the commute time using the current route is equal to zero. Based on the results of the hypothesis test, there is enough evidence at the α=0.05 level of significance to suggest that the true mean difference between the commute time using the new route and the commute time using the current route is less than zero. Since the test statistic t≈−3.201 is less than the critical value −tα=−t0.05≈−1.943, the t-test statistic is in the rejection region. Therefore, reject H0 and conclude that there is enough evidence to suggest that the true mean difference between the commute time using the new route and the commute time using the current route is less than zero.

A professional golfer wondered if she should switch to a new line of golf clubs that are supposed to put a spin on the golf ball that reduces the variance in the ball's flight. She obtained a new set of the clubs and paired off each new club with one of her old clubs. Then she hit golf balls with the pairs of clubs in 30 situations that would be appropriate to the type of club and measured how far the golf balls landed from her target. She tests the paired data, where α=0.10, in order to evaluate the claim that the true mean difference in distance is significantly greater than zero (determine if the new clubs perform better than the old clubs). (a) H0:μd=0; Ha:μd>0, which is a right-tailed test. (b) t≈1.59 , p-value is approximately 0.061 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

Reject the null hypothesis that the true mean difference in distance is equal to zero. There is sufficient evidence at the α=0.10 level of significance to suggest that the true mean difference in distance is significantly greater than zero (determine if the new clubs perform better than the old clubs). Since the p-value is less than α=0.10, reject the null hypothesis. There is enough evidence to conclude that the true mean difference in distance is significantly greater than zero (determine if the new clubs perform better than the old clubs).

An analyst wants to determine if there is any difference in the amount of time teenagers play video games between two years. To do this, he takes a random sample of teenagers and gathers the average time they spent playing video games the previous year and compares it to the average time they spent playing video games this year. Suppose that data were collected for a random sample of 15 teenagers, where each difference is calculated by subtracting the time spent playing video games this year in hours per day from the time spent playing video games last year in hours per day. Assume that the times are normally distributed. Using a test statistic of t≈2.385, the significance level α=0.10, and the corresponding p-value between 0.01 and 0.05, draw a conclusion for the appropriate hypothesis test, where the null hypothesis is H0:μd=0 and the alternative hypothesis is Ha:μd≠0.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to suggest that there is a difference in the amount of time teenagers spend playing video games. Compare the p-value to α=0.10. Since the p-value is between 0.01 and 0.05, it is less than α, so we reject H0. Therefore, there is sufficient evidence at the α=0.10 level of significance to suggest that there is a difference in the amount of time teenagers play video games between the two years.

A researcher wants to show that the proportion of mosquitoes that carry a certain disease is greater in Region X than in Region Y. In a sample of 1,392 mosquitoes trapped in Region X, 1,173 test positive for the disease. In a sample of 1,457 mosquitoes trapped in Region Y, 1,196 test positive for the disease. Test the alternative hypothesis that the population proportion for Region X is greater than the population proportion for Region Y. Use the level of significance α=0.10. The test statistic is z≈1.55, and the p-value is approximately 0.061. Identify all of the appropriate conclusions to the hypothesis test below.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportion for Region X is greater than the population proportion for Region Y. The p-value is less than the level of significance, so reject the null hypothesis. Conclude that there is enough evidence to support the claim that the population proportion for Region X is greater than the population proportion for Region Y.

A researcher wants to show that the proportion of mosquitoes that carry a certain disease is greater in Region 1 than in Region 2. In a sample of 1392 mosquitoes trapped in Region 1, 1173 test positive for the disease. In a sample of 1457 mosquitoes trapped in Region 2, 1196 test positive for the disease. Test the alternative hypothesis that the population proportion for Region 1 is greater than the population proportion for Region 2. The critical value for the level of significance α=0.10 is z0.90=1.282. The test statistic is z=1.55. Identify all of the appropriate conclusions to the hypothesis test.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportion of mosquitoes that carry the disease is greater for Region 1 than for Region 2. Notice that this is a right-tailed test. The rejection region contains values of the test statistic that are greater than the critical value. Since the test statistic of 1.55 is greater than the critical value of 1.282, reject the null hypothesis. Therefore, there is enough evidence to conclude that the population proportion of mosquitoes that carry the disease is greater for Region 1 than for Region 2.

To test a drug intended to increase memory, 80 subjects were randomly given a pill that either contained the drug or was a placebo. After 10 minutes, the subjects were asked to look at a poster with pictures of 20 common objects for 5 minutes. After waiting 20 minutes, the subjects were asked to list as many of the objects as they could recall. For the 44 subjects who took the drug, 75% were able to list at least half of the objects. For the 36 subjects who took the placebo, 50% were able to list at least half of the objects. Test the alternative hypothesis that the population proportion of subjects who took the drug and were able to list at least half of the objects is greater than the population proportion of subjects who took the placebo and were able do so. The critical value for the level of significance α=0.05 is z0.95=1.645. The test statistic is z=2.31. Identify all of the appropriate conclusions to the hypothesis test.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportion of subjects who took the drug and were able to list at least half of the objects is greater than the population proportion of subjects who took the placebo and were able do so. Notice that this is a right-tailed test. The rejection region contains values of the test statistic that are greater than the critical value. Since the test statistic of 2.31 is greater than the critical value of 1.645, reject the null hypothesis. Therefore, there is enough evidence to conclude that the population proportion of subjects who took the drug and were able to list at least half of the objects is greater than the population proportion of subjects who took the placebo and were able do so.

In a survey of 100 randomly selected taxi drivers in a city, 72 support increased government spending on roads and bridges. In a survey of 100 randomly selected bus drivers in the same city, 89 support such spending. Test the alternative hypothesis that the population proportion of taxi drivers in the city that support such spending is different from the population proportion of bus drivers in the city that support such spending. The critical values for the level of significance α=0.01 are z0.005=−2.576 and z0.995=2.576. The test statistic is z=−3.03. Identify all of the appropriate conclusions to the hypothesis test.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportion of taxi drivers in the city that support increased government spending on roads and bridges is not the same as the population proportion of bus drivers in the city that support such spending. Notice that this is a two-tailed test. The rejection region contains values of the test statistic that are less than the negative critical value and values of the test statistic that are greater than the positive critical value. Since the test statistic of −3.03 is less than the critical value of −2.576, reject the null hypothesis. Therefore, there is enough evidence to conclude that the population proportion of taxi drivers in the city that support increased government spending on roads and bridges is less than the population proportion of bus drivers in the city that support such spending.

In a survey of 100 randomly selected taxi drivers in a city, 72 support increased government spending on roads and bridges. In a survey of 100 randomly selected bus drivers in the same city, 89 support such spending. Test the alternative hypothesis that the population proportion of taxi drivers in the city that support such spending is different from the population proportion of bus drivers in the city. Use the level of significance α=0.01. The test statistic is z≈−3.03, and the p-value is approximately 0.002. Identify all of the appropriate conclusions to the hypothesis test below.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportion of taxi drivers in the city that support such spending is different from the population proportion of bus drivers in the city. The p-value is less than the level of significance, so reject the null hypothesis. Conclude that there is enough evidence to support the claim that the population proportion of taxi drivers in the city that support such spending is different from the population proportion of bus drivers in the city.

In a sample of 300 randomly selected seniors at Upton High School, 145 own at least one car. At Danville High School, 120 of 400 randomly selected seniors own at least one car. For the α=0.01 level of significance, the critical values for a hypothesis test to determine whether the proportion of seniors who own at least one car is different between the two schools are z0.005=−2.576 and z0.995=2.576. The test statistic is z=4.95. Identify all of the appropriate conclusions to the hypothesis test.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportions of seniors who own at least one car are different between the two schools. Notice that this is a two-tailed test. The rejection regions contain values of the test statistic less than the negative critical value and values of the test statistic greater than the positive critical value. Since the test statistic 4.95 is greater than the critical value 2.576, reject the null hypothesis. Therefore, conclude that there is enough evidence to support the claim that the population proportions of seniors who own at least one car are different between the two schools.

In a sample of 35 randomly selected vehicles in City A, 7 use alternative energy sources. In a sample of 45 randomly selected vehicles in City B, 17 use alternative energy sources. Test the alternative hypothesis that the population proportion for City A is different from the population proportion for City B. The critical values for the level of significance α=0.10 are z0.05=−1.645 and z0.95=1.645. The test statistic is z=−1.72. Identify all of the appropriate conclusions to the hypothesis test.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to support the claim that the population proportions of vehicles that use alternative energy sources are different in City A and City B. Notice that this is a two-tailed test. The rejection region contains values of the test statistic less than the negative critical value and values of the test statistic greater than the positive critical value. Since the test statistic −1.72 is less than the critical value −1.645, reject the null hypothesis. Therefore, there is enough evidence to conclude that the population proportion of vehicles that use alternative energy sources are different in City A and City B.

A bill is brought to a group of politicians - democrats and republicans. Researchers polled the group on whether they would vote "in favor," "opposed," or "indifferent." These values are recorded in the contingency table below. Which of the following tables correctly shows the expected values for the chi-square homogeneity test? (The observed values are above the expected values.)

RepublicansDemocratsColumn TotalIn favor19.12523.91843Indifferent18.71723.32542Opposed10.2612.81723Row Total4860108 In a chi-square test for homogeneity, as in the test of independence, the expected frequency for each data cell (used in calculating the test statistic), can be found by using the formula: E=R⋅Cn ...where R= the row total, C= the column total , and n= the total sample size. For example, the expected entry in the first row and first column (Republicans / In favor) isE=R⋅Cn=48⋅43108≈19.1

The following data set provides information on the lottery sales, proceeds, and prizes by year in Iowa. When comparing sales, proceeds, and prizes, which category is having the largest increase in amount in the last year?

Sales The graph and the data set show that the last year increase was greater for sales. They increased from 325 million dollars to 367 million dollars, a difference of 42 million dollars. Prizes increased from 197 million dollars to 222 million dollars, a difference of 25 million dollars. Proceeds increased from 75 million dollars to 88 million dollars, a difference of 13 million dollars.

In a survey of 100 randomly selected taxi drivers in a city, 72 support increased government spending on roads and bridges. In a survey of 100 randomly selected bus drivers in the same city, 89 support such spending. Test the alternative hypothesis that the population proportion of taxi drivers in the city that support such spending is different from the population proportion of bus drivers in the city. The test statistic is z=−3.03. What is the corresponding p-value? Compute your answer using a value from the table below.

Since the alternative hypothesis is that the population proportion of taxi drivers in the city that support such spending is different from the population proportion of bus drivers in the city, this is a two-tailed test. The p-value is twice the probability of an observed value of z=−3.03 or less if the null hypothesis is true because this hypothesis test is two-tailed. This probability is equal to twice the area under the standard normal curve to the left of z=−3.03. Using a normal distribution table, the area to the left of z=−3.03 is approximately 0.001. Twice this area is 2(0.001)=0.002. Therefore, the p-value is approximately 0.002.

A university admissions officer is trying to determine if men and women have similar acceptance rates across the different schools at the university (engineering, arts and sciences, business). She finds the distribution of acceptances for each gender across the three schools and wants to compare the distributions.Which of the following χ2 tests should be used in the situation above?

Test for Homogeneity The Test for Homogeneity is used to determine if two populations with unknown distributions have the same distribution as each other.In this case, the college is trying to determine if acceptances of men and women have the same distribution across the three schools, so the Test for Homogeneity is appropriate.

Which of the following is not a characteristic of the chi-square distribution? Select all correct answers. Select all that apply: As the degrees of freedom increases, the chi-square curves look more and more like a normal curve. The chi-square curve is skewed to the left. The chi-square curve is symmetrical. The total area under the χ2-curve is equal to the degrees of freedom, df.

Which of the following is a characteristic of the chi-square distribution? Select all correct answers. Select all that apply: The chi-square curve is skewed to the right. The chi-square curve is symmetrical. The chi-square curve is nonsymmetrical. The total area under the χ2-curve is equal to the degrees of freedom, df.

The chi-square curve is skewed to the right. The chi-square curve is nonsymmetrical. The following are all characteristics of the chi-square distribution. The chi-square curve is skewed to the right (it has a long tail to the right). The chi-square curve is nonsymmetrical, because it is skew to the right. The mean of the chi-square distribution is located to the right of the peak, because being skew to the right pulls the mean to the right. The total area under the χ2-curve is equal to 1, as is the case for any probability density function As the degrees of freedom increases, the chi-square curves look more and more like a normal curve. The chi-square curve approaches, but never touches, the positive horizontal axis, because the long right tail is technically never 0.

Which of the following is not a characteristic of the chi-square distribution? Select all correct answers. Select all that apply: The chi-square curve is symmetrical. The chi-square curve is skewed to the left. The total area under the χ2-curve is equal to the degrees of freedom, df. The total area under the χ2-curve is equal to 1.

The chi-square curve is symmetrical. The chi-square curve is skewed to the left. The total area under the χ2-curve is equal to the degrees of freedom, df. The following are all characteristics of the chi-square distribution. The chi-square curve is skewed to the right (it has a long tail to the right). The chi-square curve is nonsymmetrical, because it is skew to the right. The mean of the chi-square distribution is located to the right of the peak, because being skew to the right pulls the mean to the right. The total area under the χ2-curve is equal to 1, as is the case for any probability density function As the degrees of freedom increases, the chi-square curves look more and more like a normal curve. The chi-square curve approaches, but never touches, the positive horizontal axis, because the long right tail is technically never 0.

Which of the following is not a characteristic of the chi-square distribution? Select all correct answers. Select all that apply: The total area under the χ2-curve is equal to 1. The chi-square curve is symmetrical. The chi-square curve is nonsymmetrical. As the degrees of freedom increases, the chi-square curves look more and more like a normal curve.

The chi-square curve is symmetrical. The following are all characteristics of the chi-square distribution. The chi-square curve is skewed to the right (it has a long tail to the right). The chi-square curve is nonsymmetrical, because it is skew to the right. The mean of the chi-square distribution is located to the right of the peak, because being skew to the right pulls the mean to the right. The total area under the χ2-curve is equal to 1, as is the case for any probability density function As the degrees of freedom increases, the chi-square curves look more and more like a normal curve. The chi-square curve approaches, but never touches, the positive horizontal axis, because the long right tail is technically never 0.

(a) According to the line of best fit, the predicted number of minutes spent on a project per day for an average yearly overtime of 46 hours is 58.14. (b) Is it reasonable to use this line of best fit to make the above prediction?

The estimate, a predicted time of 58.14 minutes, is reasonable. The data in the table only includes overtime between 35 and 65 hours, so the line of best fit only gives reasonable predictions for values of x between 35 and 65. Since 46 is between these values, the estimate is reasonable.

A researcher wants to test to see if husbands are significantly older than their wives. To do this, he collects the ages of husbands and pairs them with the ages of their respective wives for a random set of married couples. Find the test statistic and degrees of freedom for an appropriate hypothesis test using the data set below. Let the difference d for each couple be computed as d=husband−wife. Assume that the ages are normally distributed. Round the test statistic to three decimal places.

The mean of the differences is 1.25, the hypothesized mean is μd=0, the standard deviation of the differences is sd≈3.019, and the number of paired data is n=12. The standard deviation of the differences is calculated as: sd=n(∑d2)−(∑d)2n(n−1)−−−−−−−−−−−−−−−√=12(119)−(15)212(12−1)−−−−−−−−−−−−−√≈3.019 The t-test statistic is calculated below. t≈1.25−03.019/12−−√≈1.434 This test statistic has 12−1=11 degrees of freedom.

For a sample of 120 randomly selected adult couples, 62% of the men and 79% of the women prefer travel by train over travel by bus. Check that the conditions are met for a hypothesis test to compare the population proportion of men that prefer travel by train over travel by bus to the proportion of women.

The samples are not independent. The conditions for a hypothesis test for the difference between two population proportions are shown below. 1) The samples are random. 2) The samples are independent. 3) The values n1p¯¯¯, n1q¯¯, n2p¯¯¯, and n2q¯¯ are at least 5, where n1 is the size of sample 1, n2 is the size of sample 2, p¯¯¯=x1+x2n1+n2 is the overall proportion of successes in the sample, where x1 is the number of successes in sample 1 and x2 is the number of successes in sample 2, and q¯¯=1−p¯¯¯. In this case, the men and women are couples, so the samples are not independent.

To test a drug intended to increase memory, a random sample of 80 subjects were asked to look at a poster with pictures of 20 common objects for five minutes. After waiting ten minutes, the subjects were then asked to list as many of the objects as they could recall. The subjects were then given the drug and asked to look at a different poster with pictures of 20 common objects for five minutes. After waiting ten minutes, the subjects were then asked to list as many of the objects as they could recall. Before taking the drug, 40% were able to list at least half of the objects. After taking the drug, 65% were able to list at least half of the objects. Check that the conditions are met for a hypothesis test to compare the population proportion of subjects that could list at least half of the objects before the drug to the proportion that could after taking the drug.

The samples are not independent. The conditions for a hypothesis test for the difference between two population proportions are shown below. 1) The samples are random. 2) The samples are independent. 3) The values n1p¯¯¯, n1q¯¯, n2p¯¯¯, and n2q¯¯ are at least 5, where n1 is the size of sample 1, n2 is the size of sample 2, p¯¯¯=x1+x2n1+n2 is the overall proportion of successes in the sample, where x1 is the number of successes in sample 1 and x2 is the number of successes in sample 2, and q¯¯=1−p¯¯¯. In this case, the samples are not independent because the same subjects are in both samples.

In a survey of 50 randomly selected taxi drivers in City A, 27 support increased government spending on public transportation. In a survey of 50 randomly selected bus drivers in City B, 32 support such spending. Check that the conditions are met for a hypothesis test to compare the population proportion of people in City A that support such spending to the proportion of people in City B.

The samples are not random samples of the populations. The conditions for a hypothesis test for the difference between two population proportions are shown below. 1) The samples are random. 2) The samples are independent. 3) The values n1p¯¯¯, n1q¯¯, n2p¯¯¯, and n2q¯¯ are at least 5, where n1 is the size of sample 1, n2 is the size of sample 2, p¯¯¯=x1+x2n1+n2 is the overall proportion of successes in the sample, where x1 is the number of successes in sample 1 and x2 is the number of successes in sample 2, and q¯¯=1−p¯¯¯. In this case, the samples are not random samples of the populations of the cities.

Suppose 10% of a sample of 75 randomly selected diners at a steakhouse in City A prefer eating beef over chicken and 8% of a sample of 75 randomly selected diners at a seafood restaurant in City B prefer eating beef over chicken. Check that the conditions are met for a hypothesis test to compare the population proportion of people that prefer eating beef over chicken in City A to the proportion in City B.

The samples are not random. The conditions for a hypothesis test for the difference between two population proportions are shown below. 1) The samples are random. 2) The samples are independent. 3) The values n1p¯¯¯, n1q¯¯, n2p¯¯¯, and n2q¯¯ are at least 5, where n1 is the size of sample 1, n2 is the size of sample 2, p¯¯¯=x1+x2n1+n2 is the overall proportion of successes in the sample, where x1 is the number of successes in sample 1 and x2 is the number of successes in sample 2, and q¯¯=1−p¯¯¯. For these samples, n1p¯¯¯=6.75≥5, n2p¯¯¯=6.75≥5, n1q¯¯=68.25≥5, and n2q¯¯=68.25≥5. Thus, this condition is met. In this case, the samples are not random samples of the populations of the cities in general.

A manager keeps track of the amount of time she works on project planning each week and the number of deadlines she is able to complete on-time. The data are shown in the table below. Plot the points to create a scatter plot that accurately records the data.

The values for hours project planning correspond to x-values and the values for deadlines completed on-time correspond to y-values. Each row of the table of data corresponds to a point (x,y) plotted in the scatter plot. For example, the first row, 1,2 corresponds to the point (1,2). Doing this for every row in the table, we find the scatter plot should have points (1,2), (2,2), (3,4), (4,7), and (5,9).

This is a left-tailed test because the alternative hypothesis is Ha:p1−p2<0. Since the consumer group claims that the proportion of vehicles in City A that use alternative energy sources is less than the proportion of vehicles in City B that use alternative energy sources, it is looking for evidence that supports p1 being less than p2. Therefore, the alternative hypothesis is Ha:p1−p2<0, which means that this hypothesis test is a left-tailed test.

A physician wants to determine if a supplement is effective in helping men lose weight. She takes a random sample of overweight men and records their weights before the trial. She then prescribes the supplement and instructs them to take it for four weeks while making no other lifestyle changes. After the four-week period, she records the weights of the men again. Suppose that data were collected for a random sample of 6 men, where each difference is calculated by subtracting the weight before the trial from the weight after the trial. Assume that the weights are normally distributed. What type of test is this hypothesis test?

This is a left-tailed test because the alternative hypothesis is Ha:μd<0. The physician wants to determine if the mean of the differences in weight after the trial minus weight before the trial is less than 0. This corresponds to an alternative hypothesis of Ha:μd<0, which means that the test is a left-tailed test.

This is a right-tailed test because the alternative hypothesis is Ha:μd>0. The researcher wants to test whether the mean difference of the age of the husband minus the age of the wife is greater than 0. This corresponds to an alternative hypothesis of Ha:μd>0, which means that the test is a right-tailed test.

Do people in different political parties use similar modes of transportation? Researchers ask a group of people about how they get to work. They also group the people by their political party. They want to determine if the distribution of transportation is the same for democrats and republicans. A chi-square Homogeneity Test at the 1% significance level is performed. (a) The null and alternative hypotheses are: H0: The distribution of the two populations are the same. Ha: The distribution of the two populations are not the same. (b) χ20=3.8 (c) χ20.01=9.210 (d) What conclusions can be made? Select all that apply.

We should not reject H0. At the 1% significance level, there is not sufficient evidence to conclude that the distributions of the two populations are not the same. The homogeneity test is a right-tailed hypothesis test (as are all chi-square tests). This means that if the test statistic, χ20, is greater than the critical value, χ20.01, then we should reject H0.Here χ20 is less than the critical value (3.800<9.210), so, we do not reject H0. Interpretation: At the 1% significance level, there is not sufficient evidence to conclude that the distributions of the two populations are not the same.

(d) What conclusions can be made? Select all that apply.

We should reject H0. At the 1% significance level, there is sufficient evidence to conclude that being an athlete affects college major. The test of independence is a right-tailed hypothesis test (as are all chi-square tests). This means that if the test statistic, χ20, is greater than the critical value, χ20.01, then we should reject H0.Here χ20 is greater than the critical value (11.4>9.210), so, we reject H0. Interpretation: At the 1% significance level, there is sufficient evidence to conclude that being an athlete affects college major.

A professor is trying to determine if her students guessed on a certain multiple choice question. She expects that if the students guessed, the distribution of answers would be uniform for that question. She compares the observed distribution of answers with the uniform distribution. The professor conducts a chi-square Goodness-of-Fit hypothesis test at the 5% significance level. (a) The null and alternative hypotheses are: H0: The student answers have the uniform distribution. Ha: The student answers do not have the uniform distribution. (b) χ20=13.167. (c) χ20.05=7.815. (d) What conclusions can be made? Select all that apply.

We should reject H0. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing. The Goodness-of-fit hypothesis test is right-tailed. This means that if the test statistic, χ20, is greater than the critical value, χ20.05, then we should reject H0. Here χ20 is greater than the critical value (13.167>7.815), so, we reject H0. Interpretation: At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing (the outcomes do not occur with equal frequencies).

A farmer would like to determine if crops in the western fields are dying at a higher rate than crops in the eastern fields. The farmer takes a random sample of 12 crops in each set of fields. Should these samples be considered independent? Why or why not?

Yes, the samples are independent because the farmer is only looking into crops on his or her farm. With this being the case, whether a crop dies in a western field does not have an effect on a crop dying in the eastern field. The farmer is comparing fields in the western portion of their farm to fields in the eastern portion. Because there is no overlap and the farmer is not interested in how other farms are faring, these samples are independent. A dying crop in a western field does not affect a dying crop in an eastern field.

In which of the scenarios below would it be appropriate to use a One-way Analysis of Variance (ANOVA) method to determine whether or not there is a statistical difference among the groups? Select all that apply. Select all that apply: You want to conduct a hypothesis test to determine if there is a difference in average commute time going to work versus going home from work. You want to conduct a hypothesis test to determine if there is a difference in the average salaries of police officers, versus firefighters, versus paramedics. You want to conduct a hypothesis test to determine if there is a difference in the average amount of time spent doing physical exercise for men versus women. You want to conduct a hypothesis test to determine if the average time spent asleep daily is different for elementary school students, versus high school students, versus college students, versus adults not in school.

You want to conduct a hypothesis test to determine if there is a difference in the average salaries of police officers, versus firefighters, versus paramedics. You want to conduct a hypothesis test to determine if the average time spent asleep daily is different for elementary school students, versus high school students, versus college students, versus adults not in school. One-way Analysis of Variance (ANOVA) is a hypothesis testing method that compares three or more means from different populations. The intent is to determine if there is a statistical difference among the three or more means. Thus, for any applications that compare the averages (means) from three or more groups, ANOVA is the appropriate analysis method.

p-value <0.01 Notice that this is a left-tailed test because the alternative hypothesis is Ha:μd<0. The t-test statistic is −3.201 with 6 degrees of freedom. The t-distribution table gives the values for a right-tailed test, so to find the p-value for a left-tailed test, use the absolute value of the test statistic. Using the table of areas in the right tail for the t-distribution, in the row for 6 degrees of freedom, 3.201 is greater than 3.143. So, the p-value is less than 0.01.

A sports analyst wants to determine if the number of free throws basketball players make in their second year is more than the number of free throws they make in their rookie year. To do this, he selects several random players and compares the average number of free throws they made per game in their second year to the average number of free throws they made per game in their rookie year. Suppose that data were collected for a random sample of 11 players, where each difference is calculated by subtracting the average number of free throws made per game in the player's rookie year from the average number of free throws made per game in the player's second year. Assume that the number of throws is normally distributed. The analyst uses the alternative hypothesis Ha:μd>0. Using a test statistic of t≈4.842, which has 10 degrees of freedom, determine the range that contains the p-value.

p-value <0.01 Notice that this is a right-tailed test because the alternative hypothesis is Ha:μd>0. The t-test statistic is 4.842 with 10 degrees of freedom. The t-distribution table gives the values for a right-tailed test. Using the table of areas in the right tail for the t-distribution, in the row for 10 degrees of freedom, 4.842 is greater than 2.764. So, the p-value is less than 0.01.

The correlation coefficient for the difference between acres planted in asparagus and acres harvested in the United States is 0.452. The sample size is the years 2010 to 2015. Using the table below, determine if the value of r is significant or not. Select the correct answer below: r is significant because it is between the positive and negative critical values. r is not significant because it is between the positive and negative critical values. r is significant because it is not between the positive and negative critical values. r is not significant because it is not between the positive and negative critical values.

r is not significant because it is between the positive and negative critical values. The degrees of freedom is equal to the sample size minus 2. 6−2=4 Look up 4 in the table. The r value of 0.452 is not significant because it is between the values −0.811 and 0.811.

Continuing with the previous question, here is the contingency table. What is the test statistic, χ20, for this test of independence?

χ20=5.9 The formula for the test statistic is∑(O−E)2EIn words, for each entry in the table, we take the observed entry O and expected entry E, and compute (O−E)2E, and then we add these up for each entry in the table. So we findχ20 =(14−19.1)219.1+(19−14.4)214.4+(18−17.5)217.5+(23−17.9)217.9+(9−13.6)213.6+(16−16.5)216.5≈5.9

Continuing with the previous question, here is the contingency table. What is the test statistic, χ20, for this test of independence? In favor Indifferent Opposed Row Total Republicans 10 16 24 50 Democrats 21 5 21 47 Column Total 31 21 45 97

χ20=9.9 The formula for the test statistic is∑(O−E)2EIn words, for each entry in the table, we take the observed entry O and expected entry E, and compute (O−E)2E, and then we add these up for each entry in the table. So we findχ20 =(10−16)216+(16−10.8)210.8+(24−23.2)223.2+(21−15)215+(5−10.2)210.2+(21−21.8)221.8≈9.9

In a large high school, two calculus classes were conducted at the same time. At the beginning of the year, the 47 students taking calculus were randomly assigned to one of the classes. At the end of the course, all of the students took a standardized achievement exam. To determine if one of the calculus teachers was more effective than the other, the principal plans to conduct a two-mean hypothesis test to determine if there is a difference in the population mean scores on the exam between the two classes. Assuming the population variances are equal and using the pooled estimate of the standard deviation, how many degrees of freedom will the test have?

$df=45$df=45 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1 is the size of sample 1 and n2 is the size of sample 2. Note that in this case, n1+n2=47. 47−2=45

What is/are the critical value(s) of the z-test statistic for this hypothesis test, where α=0.05? Use the appropriate value(s) from the table. Use a comma and a space to separate answers as needed.

$\text{critical value(s)}=-1.645$critical value(s)=−1.645 Since this is a left-tailed test and α=0.05, the critical value occurs at −zα=−z0.05. To find the critical value, use a normal distribution table to find the z-score that corresponds to α=0.05 for a left-tailed test. z0.10z0.05z0.025z0.01z0.0051.2821.6451.9602.3262.576 Therefore, −z0.05=−1.645 is the critical value for this hypothesis test.

Lara uses the alternative hypothesis Ha:μ1−μ2>0 with α=0.10 and 108 degrees of freedom. What is/are the critical value(s) for this hypothesis test? Use the value(s) from the table. Use a comma and a space to separate answers as needed.

$\text{critical value(s)}=1.289$critical value(s)=1.289 Since this is a one-tailed test and α=0.10 with 108 degrees of freedom, the critical value occurs at and t0.10=1.289.

Brennan tests the alternative hypothesis Ha:μ1−μ2≠0 with α=0.05 and 68 degrees of freedom. What is/are the critical value(s) for this hypothesis test? Use the value(s) from the table. Use a comma and a space to separate answers as needed.

$\text{lower critical value}=-1.995\ ,\text{upper critical value}=\ 1.995$lower critical value=−1.995 ,upper critical value= 1.995 Since this is a two-tailed test and α=0.05 with 68 degrees of freedom, the critical values occur at −t0.025=−1.995 and t0.025=1.995.

Suppose you want to buy a car and are trying to decide between two models. A friend suggests that perhaps the annual maintenance costs are different for the two models, so you obtain random samples of the annual maintenance costs for 50 of each model of car. Assuming the population variances are equal and using the pooled estimate of the standard deviation, a two-mean hypothesis test would have how many degrees of freedom?

$df=98$df=98 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=50 is the size of sample 1 and n2=50 is the size of sample 2. 50+50−2=98

Marketers at PaperClips, a regional office supply store, are researching whether the population mean amount spent on back-to-school shopping for students in grades K-12 has changed from 2016 to 2018. Based on market research, the marketers assume the population standard deviation is $101.52 for 2016 and $96.47 for 2018. Households across the region with at least one student in grades K-12 were randomly selected in 2016 and again in 2018. The results are shown in the table below. Let μ1 be the population mean amount spent on back-to-school shopping per household in 2016 and μ2 be the population mean amount spent on back-to-school shopping per household in 2018. If the test statistic is z=2.39, what is the p-value for this hypothesis test?

$\text{p-value}=0.016$p-value=0.016 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0. Since this a two-tailed test and the z-test statistic is positive, find twice the area to the right of z=2.39. Using a standard normal table, the area to the left of z=2.39 is 0.992. So, the area to the right of z=2.39 is 1−0.992=0.008. Therefore, the p-value is 2⋅0.008=0.016.

A large drug store chain outsources the manufacturing of its pain relief tablets to two different independent manufacturers, Manufacturer A and Manufacturer B. The quality control department of the chain would like to ensure that the population mean amount of the active ingredient in the tablets is still the same between the two manufacturers. The department assumes the population standard deviation for the amount of the active ingredient in a tablet based on the specifications from the manufacturers. The quality control department selects a random sample of tablets from each manufacturer across dozens of lots. The tablets are then analyzed using chromatography to determine the amount of the active ingredient in each tablet. The results of the analysis and the population standard deviations that are assumed for this hypothesis test are provided in the table below. Let μ1 be the population mean amount of the active ingredient, in milligrams, in tablets produced by Manufacturer A and μ2 be the population mean amount of the active ingredient, in milligrams, in tablets produced by Manufacturer B. If the test statistic is z=−1.65, what is the p-value for this hypothesis test?

$\text{p-value}=0.098$p-value=0.098 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0. Since this a two-tailed test and the z-test statistic is negative, find twice the area to the left of z=−1.65. Using a standard normal table, the area to the left of z=−1.65 is 0.049. Therefore, the p-value of this hypothesis test is 2⋅0.049=0.098.

A team of archeologists is reexamining a site where two additional sets of artifacts have been discovered. The sets of artifacts are in two separate areas of the archeological site, where each set is adjacent to artifacts that were discovered in the past. Since both sets of artifacts were discovered at about the same time years after the first examination, the members of the team would like to find out whether the population mean age of the artifacts in Set A is the different than the population mean age of the artifacts in Set B. Using information from the first examination of the site, the members of the team assume that the population standard deviation of the age of artifacts found in Set A is 135 years and the population standard deviation of the age of the artifacts found in Set B is 119 years. The team takes a random sample of the artifacts and finds the ages of each artifact using radiocarbon dating. The results are provided in the table below. Let μ1 be the population mean age, in years, of the artifacts in Set A and μ2 be the population mean age, in years, of the artifacts in Set B. If the test statistic is z=1.56, what is the p-value for this hypothesis test?

$\text{p-value}=0.118$p-value=0.118 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0. Since this a two-tailed test and the z-test statistic is positive, find twice the area to the right of z=1.56. Using a standard normal table, the area to the left of z=1.56 is 0.941. So, the area to the right of z=1.56 is 1−0.941=0.059. Therefore, the p-value for this hypothesis test is 2⋅0.059=0.118.

Andy Thompson is a dean at a local college. He is trying to determine whether the population mean GPA of undergraduate students at his college who have at least one job during the semester is different than the population mean GPA of undergraduate students at his college who do not have a job during the semester. Andy conducts research on the topic and assumes that the population standard deviation is 0.59 for undergraduates who have at least one job during the semester and 0.68 for undergraduates who don't have a job during the semester. The results of the survey are displayed in the table below. Let μ1 be the population mean GPA of students who have at least one job during the semester and μ2 be the population mean GPA of students who do not have a job during the semester. If the test statistic is z=0.71, what is the p-value for this hypothesis test?

$\text{p-value}=0.478$p-value=0.478 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0. Since this a two-tailed test and the z-test statistic is positive, find twice the area to the right of z=0.71. Using a standard normal table, the area to the left of z=0.71 is 0.761. So, the area to the right of z=0.71 is 1−0.761=0.239. Therefore, the p-value for this hypothesis test is 2⋅0.239=0.478.

Arianna Estefan manages Salmon Falls Park, which has two flagship roller coasters, the Flyer and the Destroyer. When Arianna walks around the park during hours of operation, she notices that there are fewer people waiting in line for the Flyer than for the Destroyer. Since both roller coasters have similar capacities and similar run times, both roller coasters should have about the same number of riders. Based on the anecdotal evidence, Arianna claims that the population mean number of riders per hour on the Flyer is less than the population mean number of riders per hour on the Destroyer. Arianna reviews data collected in the past and assumes that the population standard deviation is $_48.29$_ for the Flyer and $_51.86$_ for the Destroyer. Arianna randomly selects hours during which the attendant of each ride counts the number of riders. The sampling occurs over the course of several weeks. The results are provided in the table shown below. Assuming the conditions needed for the hypothesis test have been met, what is the $_z$_ test statistic for this hypothesis test, rounded to two decimal places?

$\text{z=}-1.54$

A large drug store chain outsources the manufacturing of its pain relief tablets to two different independent manufacturers, Manufacturer A and Manufacturer B. The quality control department of the chain would like to ensure that the population mean amount of the active ingredient in the tablets is still the same between the two manufacturers. The department assumes the population standard deviation for the amount of the active ingredient in a tablet based on the specifications from the manufacturers. The department selects a random sample of tablets from both manufacturers across dozens of lots. The tablets are then analyzed using chromatography to determine the amount of the active ingredient in each tablet. The population standard deviations as well as the results of the test are provided in the table below. Assuming the conditions needed for the hypothesis test have been met, what is the z test statistic for this hypothesis test, rounded to two decimal places?

$\text{z=}-2.93$z=−2.93 To find the test statistic z, substitute 196.08 for x¯¯¯1, 251 for n1, 0.57 for σ1, 196.24 for x¯¯¯2, 228 for n2, and 0.62 for σ2 into the formula and evaluate. z=x¯¯¯1−x¯¯¯2σ21n1+σ22n2−−−−−−−−√=196.08−196.240.572251+0.622228−−−−−−−−−−−−√≈−2.93

The composition of a quarter dollar coin changed from silver and copper to a copper-nickel alloy in 1964. Hunter Rowland collects coins as a hobby and has access to large collections of quarters produced before and after 1964. He would like to show that the population mean mass of the quarters produced after 1964 is less than the population mean mass of the quarters produced before 1964. Hunter researches several coin collector websites and assumes the population standard deviation for each type of quarter from the information he gathered. Hunter randomly selects each type of quarter from the large collection and finds the mass of each one using a precise scale. The results of the samples and the population standard deviations that are assumed for this hypothesis test are shown in the table below. Assuming the conditions needed for the hypothesis test have been met, what is the $_z$_ test statistic for this hypothesis test, rounded to two decimal places?

$\text{z=}-30.26$z=−30.26 To find the test statistic z, substitute 5.648 for x¯¯¯1, 37 for n1, 0.070 for σ1, 6.191 for x¯¯¯2, 39 for n2, and 0.086 for σ2 into the formula and evaluate. z=x¯¯¯1−x¯¯¯2σ21n1+σ22n2−−−−−−−−√=5.648−6.1910.070237+0.086239−−−−−−−−−−−−−√≈−30.26

Andy Thompson is a dean at a local college. He is trying to find out whether the population mean GPA of undergraduate students at his college who have at least one job during the semester is different than the population mean GPA of undergraduate students at his college who do not have a job during the semester. Andy conducts research on the topic and assumes that the population standard deviation is 0.59 for undergraduates who have at least one job during the semester and 0.68 for undergraduates who don't have a job during the semester. Andy conducts a survey of the undergraduate student body and asks randomly selected students about their job status during the semester and their GPA. The results of the survey are displayed in the table below. Assuming the conditions needed for the hypothesis test have been met, what is the z test statistic for this hypothesis test, rounded to two decimal places?

$\text{z=}0.71$z=0.71 To find the test statistic z, substitute 2.79 for x¯¯¯1, 56 for n1, 0.59 for σ1, 2.71 for x¯¯¯2, 73 for n2, and 0.68 for σ2 into the formula and evaluate. z=x¯¯¯1−x¯¯¯2σ21n1+σ22n2−−−−−−−−√=2.79−2.710.59256+0.68273−−−−−−−−−−−−√≈0.71

A website designer wants to determine which of two home page format leads visitors to click on more links. She writes two scripts; one to randomly display one of the home pages and another to record the page displayed and the number of links clicked. For one of the pages, the 103 visitors in the sample clicked an average of 7.2 links with a standard deviation of 5.3. For the other page, the 147 visitors in the sample clicked an average of 4.3 links with a standard deviation of 1.9. Assuming the population variances are not equal, a hypothesis test for the difference between the two population means would have how many degrees of freedom?

$df=102$df=102 When the population variances are assumed unequal, the degrees of freedom for a two-mean hypothesis test are the lesser of n1−1 and n2−1, where n1=103 is the sample size for one of the pages and n2=147 is the sample size for the other page. Note that n1=103 is the lesser value. 103−1=102

A website designer wants to determine which of two home page formats leads visitors to click on more links. She writes two scripts: one to randomly display one of the home pages and another to record the page displayed and the number of links clicked. For one of the pages, the 132 visitors in the sample clicked an average of 5.7 links with a standard deviation of 2.3. For the other page, the 118 visitors in the sample clicked an average of 4.3 links with a standard deviation of 2.1. Assuming the population variances are equal and using the pooled estimate of the standard deviation, a hypothesis test for the difference between the two means would have how many degrees of freedom?

$df=248$df=248 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=132 is the sample size for one of the pages and n2=118 is the sample size for the other page. 132+118−2=248

In a large high school, two calculus classes were conducted at the same time. At the beginning of the year, the 52 students taking calculus were randomly assigned to one of the classes, with half in each. At the end of the course, all of the students took a standardized achievement exam. To determine if one of the calculus teachers was more effective than the other, the principal plans to conduct a two-mean hypothesis test to determine if there is a difference in the population mean scores on the exam between the two classes. Assuming the population variances are unequal, how many degrees of freedom will the hypothesis test have?

$df=25$df=25 When the population variances are assumed unequal, the degrees of freedom for a two-mean hypothesis test are the lesser of n1−1 and n2−1, where n1=522=26 is the size of one class and n2=26 is the size of the other class. Note that the samples are the same size. 26−1=25

In order to check for gender bias in a standardized national achievement test, researchers compared the means of random samples of scores for male and female students. The sample mean of the 463 scores for males was 382 with a sample standard deviation of 122. The sample mean of the 437 scores for females was 391 with a sample standard deviation of 64. Assuming the population variances are unequal, a hypothesis test for the difference between the two population means would have how many degrees of freedom?

$df=436$df=436 When the population variances are assumed unequal, the degrees of freedom for a two-mean hypothesis test are the lesser of n1−1 and n2−1, where n1=463 is the sample size of the male students and n2=437 is the sample size of the female students. Note that n2=437 is the lesser value. 437−1=436

Advertisements for a laundry detergent claim that the colors in clothing washed using this brand remain brighter, longer. To test this claim, a consumer rights group purchased 100 identical, brightly colored plaid shirts. Each shirt was randomly assigned to be washed in the advertised detergent or a different brand. After 50 washes, each shirt was compared to a new identical, brightly colored plaid shirt. The brightness of each shirt was assigned a score from 0 to 5 by three different judges. The sample mean brightness score for the 46 shirts washed with the advertised detergent was slightly higher than sample the mean score of the 54 shirts washed with other detergents. Assuming the population variances are unequal, a hypothesis test for the difference between the two population means would have how many degrees of freedom?

$df=45$df=45 When the population variances are assumed unequal, the degrees of freedom for a two-mean hypothesis test are the lesser of n1−1 and n2−1, where n1=46 is the sample size of the shirts washed with the advertised detergent and n2=54 is the sample size of the shirts washed with the other detergents. Note that n1=46 is the lesser value. 46−1=45

Engineers tested the shear strength of a new brand of rivets to ensure that they were at least as strong as the brand already approved for use building a new bridge. The engineers randomly selected 250 approved-brand rivets and 250 new-brand rivets and used a hydraulic press to apply shear forces on the rivets. The force necessary for failure was recorded. Determine the degrees of freedom for a two-mean hypothesis test for population variances assumed equal (pooled estimate of the standard deviation) to test whether the new rivets are at least as strong.

$df=498$df=498 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=250 is the sample size of the approved-brand rivets and n2=250 is the sample size of the new-brand rivets. 250+250−2=498

In a study, a random sample of 30 male college students ran 1 mile. Another random sample of 35 male college students walked the same mile. On average, those who ran burned 112 calories while those who walked burned 89 calories. Determine the degrees of freedom for a hypothesis test for the difference between the two means for population variances assumed equal (pooled estimate of the standard deviation) to test whether running burns more calories than walking.

$df=63$df=63 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=30 is the sample size of the students who ran and n2=35 is the sample size of the students who walked. 30+35−2=63

To determine whether there is a difference in the population mean price of a single family home in two nearby towns, a real estate agent obtained random samples of the sale prices of 42 single-family homes that recently sold in town A and 31 single-family homes that recently sold in town B. The sample mean selling price was $121,783 in town A and $118,997 in town B. The sample standard deviations were $35,123 for town A and $39,178 for town B. How many degrees of freedom would a two-mean hypothesis test have? Assume the population variances are equal.

$df=71$df=71 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=42 is the sample size for town A and n2=31 is the sample size for town B. 42+31−2=71

A bank has two branches. To determine whether the population mean wait time at the drive-through was different for the branches, the director of the bank had the manager at each branch use security camera footage to randomly select 40 customers at each branch who used the drive-through and record the wait time for each customer. The sample mean wait time at one branch was 73 seconds with a sample variance of 69 square seconds. The sample mean wait time at the other branch was 88 seconds with a sample variance of 67 square seconds. How many degrees of freedom would a two-mean hypothesis test have? Assume the population variances are equal.

$df=78$df=78 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=40 is the sample size of one branch and n2=40 is the sample size of the other branch. 40+40−2=78

The partners at an investment firm want to determine which of two financial planners produced a greater population mean rate of return for their clients. The partners reviewed the rates of return last quarter for random samples of the clients. For one of the planners, the sample mean rate of return for the 40 clients selected was 3.1% with a sample standard deviation of 2.1%. For the other planner, the sample mean rate of return for the 40 clients selected was 3.3% with a sample standard deviation of 2.3%. Determine the number of degrees of freedom for a two-mean hypothesis test for population variances assumed equal (pooled estimate of the standard deviation) to test if one planner produced a higher mean rate of return.

$df=78$df=78 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=40 is the sample size of one of the financial planners and n2=40 is the sample size of the other financial planner. 40+40−2=78

The quality manager at a breakfast cereal producer claimed that one of the production lines was over-filling the boxes. A random sample of 40 boxes produced by the line in question had a mean weight of 20.062 ounces with a standard deviation of 0.035. A sample of 40 boxes produced by a different line had a mean weight of 20.047 ounces with a standard deviation of 0.033. Determine the degrees of freedom for a two-mean hypothesis test for population variances assumed equal (pooled estimate of the standard deviation).

$df=78$df=78 When the population variances are assumed equal, the number of degrees of freedom for a two-mean hypothesis test is df=n1+n2−2, where n1=40 is the sample size of the boxes produced by the line in question and n2=40 is the sample size of the boxes produced by the other line. n1+n2−2=40+40−2=78

A professor of entomology claims that for a certain species of dragonfly, the ones that live by a nearby river have a greater population mean wingspan than those that live near a lake that is 10 kilometers away. To test this claim, the entomologist captures, measures, and releases 45 dragonflies at each site. Assuming the population variances are equal and using the pooled estimate of the standard deviation, a hypothesis test for the difference between the two population means would have how many degrees of freedom?

$df=88$df=88 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=45 is the sample size of one site and n2=45 is the sample size of the other site. 45+45−2=88

A tutoring company claims that attending their preparatory classes would improve students' performance on a widely used standardized test. One year, a check box was added to the test for students to indicate whether they attended preparatory classes. Researchers selected a random sample of 1,000 tests. The sample mean score for the 237 students who attended preparatory classes was 592 with a sample standard deviation of 152. The sample mean of the 763 students who did not attend preparatory classes was 551 with a sample standard deviation of 164. Determine the degrees of freedom for a two-mean hypothesis test for population variances assumed equal (pooled estimate of the standard deviation) to test whether the population mean score of students who attended preparatory classes is greater than the population mean score of students who did not attend the classes.

$df=998$df=998 When the population variances are assumed equal, the degrees of freedom for a two-mean hypothesis test are df=n1+n2−2, where n1=237 is the sample size of students who attended preparatory classes and n2=763 is the sample size of students who did not attend preparatory classes. 237+763−2=998

A farmer wants to test if a new fertilizer will produce more massive crops. In order to do this, he randomly gathers the masses of several types of crops from the last growing season and the current growing season and compares them to see if the current growing season's crops are more massive. Find the test statistic and degrees of freedom for an appropriate hypothesis test using the data set below. Let the difference d for each crop be computed as d=current year−previous year. Assume that the masses are normally distributed. Round the test statistic to three decimal places.

$t=1.266,\ df=5$ The mean of the differences is 1.2, the hypothesized mean is μd=0, the standard deviation of the differences is sd≈2.321, and the number of paired data is n=6. The standard deviation of the differences is calculated as: sd=n(∑d2)−(∑d)2n(n−1)−−−−−−−−−−−−−−−√=6(35.58)−(7.2)26(6−1)−−−−−−−−−−−−−−√≈2.321 The t-test statistic is calculated below. t≈1.2−02.321/6-√≈1.266 This test statistic has 6−1=5 degrees of freedom.

A statistician believes that women received more bachelor's degrees than men last year. To test this claim, she selects colleges and universities randomly and compares the number of bachelor's degrees awarded to men and women. Find the test statistic and degrees of freedom for an appropriate hypothesis test using the data set below. Let the difference d for each university be computed as d=women−men. Assume that the numbers of degrees earned are normally distributed. Round the test statistic to three decimal places.

$t=2.968,\ df=6$ The mean of the differences is d¯¯¯=0.46, the hypothesized mean is μd=0, the standard deviation of the differences is sd≈0.410, and the number of paired data is n=7. The standard deviation of the differences is calculated as: sd=n(∑d2)−(∑d)2n(n−1)−−−−−−−−−−−−−−−√=7(2.492)−(3.22)27(7−1)−−−−−−−−−−−−−−−√≈0.410 The t-test statistic is calculated below. t≈0.46−00.410/7-√≈2.968 This test statistic has 7−1=6 degrees of freedom.

Jordan Scarff works for a state's department of transportation. Recently, an inquiry has been made in Jordan's department about the number of vehicles that cross a certain bridge within the state. The engineers of the bridge decided to install tolls on the southbound lanes only. It appears that there are fewer vehicles traveling on the southbound lanes per day than on the northbound lanes. Jordan looks into the inquiry and tests the claim that the mean number of vehicles per day traveling in the southbound lanes is less than the mean number of vehicles per day traveling in the northbound lanes. Due to limited time, Jordan selects a random sample of days for the northbound lanes and another random sample of days for the southbound lanes, where the number of vehicles traveled each day was recorded. The sample statistics are shown in the table below. Let μ1 be the population mean number of vehicles traveling in the southbound lanes per day and μ2 be the population mean number of vehicles traveling in the northbound lanes per day. Jordan uses the alternative hypothesis Ha:μ1−μ2<0, assuming that the population standard deviations of the two groups are not equal and using 60 degrees of freedom. If the t-test statistic is t≈−2.15, what is the p-value for this hypothesis test?

0.01< p-value <0.05 Notice that this is a left-tailed test because the alternative hypothesis is Ha:μ1−μ2<0. The test statistic is t≈−2.15 with 60 degrees of freedom.The t-distribution table gives the values for a right-tailed test, so to find the area in the left tail, use the absolute value of the test statistic. Using the table of areas in the right tail for the t-distribution, in the row for 60 degrees of freedom, 2.15 is greater than 1.671 and less than 2.390. So, the p-value is between 0.01 and 0.05.

Casey Fitch is writing a report for a statistics class. He is investigating whether the mean price for a gallon of regular gasoline at stations of Brand A are currently higher than the mean price for a gallon of regular gasoline at stations of Brand B in the region. Casey randomly selects 37 service stations of Brand A and 35 service stations of Brand B and records the price for a gallon of regular gasoline. The sample mean price for Brand A is $2.836 with a sample standard deviation of $0.083, while Brand B has a sample mean price of $2.798 with a standard deviation of $0.049. Let μ1 be the population mean price for a gallon of regular gasoline at Brand A and μ2 be the population mean price for a gallon of regular gasoline at Brand B. Casey uses the alternative hypothesis Ha:μ1−μ2>0. He assumes that the population standard deviations of the two groups are not equal, so uses 34 degrees of freedom. If the t-test statistic is t≈2.38, what is the p-value for this hypothesis test?

0.01< p-value <0.05 Notice that this is a right-tailed test because the alternative hypothesis is Ha:μ1−μ2>0. The test statistic is t≈2.38 with 34 degrees of freedom. Using the table of areas in the right tail for the t-distribution, in the row for 34 degrees of freedom, 2.38 is greater than 1.691 and less than 2.441. So, the p-value is between 0.01 and 0.05.

A pet food manufacturer has a facility with two lines that blend, can, and cook cat food. Nolan Munroe is the production manager of the facility and curious whether there is a difference in the mean number of cans produced each day by the two lines, A and B. Nolan randomly selects 300 daily production reports, 141 of which were for production line A and 159 of which were for production line B. The sample statistics are shown in the table below. Let μ1 be the population mean number of cans produced per day on production line A, and let μ2 be the population mean number of cans produced per day on production line B. Nolan assumes the population standard deviations are not equal and tests the alternative hypothesis Ha:μ1−μ2≠0. If the test statistic is t≈−1.93 and the number of degrees of freedom is 140, what is the p-value for this hypothesis test?

0.05< p-value <0.10 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0, so the p-value is twice the area in one tail. The test statistic is t≈−1.93 with 140 degrees of freedom. The t-distribution table gives the values for a right-tailed test, so to find twice area in the left tail, use the absolute value of the test statistic. Using the table of areas in the right tail for the t-distribution, in the row for 140 degrees of freedom, 1.93 is greater than 1.656 and less than 1.977. So, the p-value is between 2(0.025)=0.05 and 2(0.05)=0.10.

Kayla Greene is a team lead for an environmental group for a certain region. She is investigating whether the population mean monthly number of kilowatt hours (kWh) used per residential customer in the region has changed from 2006 to 2017. She is concerned that changes such as more efficient lighting and the increased use of electronics and air conditioners are affecting the mean monthly number of kilowatt hours consumed per residential customer. Kayla investigates the data and assumes the population standard deviation for 2006 and 2017 using the data that were provided to her by local utility companies. Using data that was collected by her company, Kayla selects a random sample of residential customers who were active for all of 2006 and a separate sample of residential customers who were active for all of 2017. The population standard deviations and the results from the samples are provided in the accompanying table. Explain whether a hypothesis test for the difference between two means of independent samples is appropriate, and if so, determine the null and alternative hypotheses for this hypothesis test. Let μ1 be the population mean monthly number of kilowatt hours consumed per residential customer in 2006 and μ2 be the population mean monthly number of kilowatt hours consumed per residential customer in 2017.

All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:μ1−μ2=0Ha:μ1−μ2≠0. All of the conditions that are required for the hypothesis test have been met. The population standard deviations are known, the samples are randomly selected and are independent, and the sample sizes are sufficiently large. Since Kayla wants to find out whether the mean monthly number of kilowatt hours (kWh) used per residential customer has changed from 2006 to 2017, μ1 is not equal to μ2 and the difference, μ1−μ2, would be not equal to 0. The null and alternative hypotheses are shown below. {H0:μ1−μ2=0Ha:μ1−μ2≠0

Quinn Napoli is looking to purchase a vehicle and is deciding between two models, the Glacier and the Ravine. One of Quinn's friends suggests that perhaps the mean annual maintenance costs are different for the two models, so Quinn obtains random samples of the annual maintenance costs from 50 owners of each model of car. The sample statistics are in the table shown below. Let μ1 be the population mean annual maintenance cost for the Glacier, and let μ2 be the population mean annual maintenance cost for the Ravine. Quinn tests the alternative hypothesis Ha:μ1−μ2≠0 and assumes that the population standard deviations of the two models are not equal. If the p-value of the hypothesis test is greater than 0.10 and the significance level is α=0.10, what conclusion could be made about the population mean annual maintenance costs of the two models?

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that the population mean annual maintenance costs of the Glacier is different than the population mean annual maintenance costs of the Ravine. Compare the p-value that is greater than 0.10 to α=0.10. Since the p-value is greater than α, fail to reject H0. Therefore, there is insufficient evidence at the α=0.10 level of significance to suggest that the population mean annual maintenance costs of the Glacier is different than the population mean annual maintenance costs of the Ravine.

Lydia Hayward is an external research consultant hired by FirstCard, a major credit card company, to find evidence that the population mean total credit card balance for adults ages 25 to 34 with a college degree in the region is greater than for adults ages 25 to 34 without a college degree. Due to research already conducted by FirstCard, Lydia assumes that the population standard deviation is $2,047.24 for adults ages 25 to 34 with a college degree and $1,749.65 for those without a college degree. She conducts a survey at a local shopping center and selects every tenth adult ages 25 to 34 with a college degree and every tenth adult ages 25 to 34 without a college degree to ask for the total credit card balance. The results of the samples are shown in the table below. Explain whether a hypothesis test for the difference between two means of independent samples is appropriate, and if so, determine the null and alternative hypotheses for this hypothesis test. Let μ1 be the population mean total credit card balance for adults ages 25 to 34 with a college degree and μ2 be the population mean total credit card balance for adults ages 25 to 34 without a college degree.

Although the population standard deviations are known and the sample sizes are sufficiently large, the samples are not randomly selected and are not independent of each other. Lydia did not select a random sample for both groups because she selected her samples from patrons at a local shopping center. Since not all adults ages 25 to 34 in the region have an equal opportunity to be selected, the sample does not satisfy the randomization condition.

Dalton Castene is a public school teacher in Burlington. After a conversation with a colleague in Sudbury, Dalton realizes that his salary is much lower than his colleague's, even though they have similar backgrounds and work experiences. Dalton theorizes that the population mean annual salary for a public school teacher in Burlington is less than the population mean annual salary for a public school teacher in Sudbury. He conducts research on the inquiry and assumes that the population standard deviation is $2,409 for Burlington and $3,192 for Sudbury. Dalton randomly selects the salaries of public school teachers in Burlington and in Sudbury. The results of Dalton's findings are shown in the table below. Explain whether a hypothesis test for the difference between two means of independent samples is appropriate, and if so, find the null and alternative hypotheses, where μ1 is the population mean annual salary for a public school teacher in Burlington and μ2 is the population mean annual salary for a public school teacher in Sudbury.

Although the population standard deviations are known and the samples are randomly selected and independent, the sample sizes are not sufficiently large. Although the population standard deviations are known and the samples are randomly selected and are independent, the sample sizes for Burlington and Sudbury are 26 and 22, respectively. Since both sample sizes are less than 30, the sample sizes are not sufficiently large. Therefore, not all of the conditions to conduct the hypothesis test have been met and a hypothesis test for the difference between two means of independent samples is not appropriate.

Brandon Blake is a real estate agent working in several towns in a local area, mainly Littleton and Whitefield. Brandon has worked in the real estate business for several years and believes that the population mean selling price of a single-family home in Littleton during 2017 was greater than the population mean selling price of a single-family home in Whitefield during 2017. Brandon randomly selects a sample of selling prices for single-family homes for each of the two towns during 2017. The results of the samples are shown in the table below. What are the null and alternative hypotheses for this hypothesis test, where μ1 is the population mean selling price of a single-family home in Littleton during 2017 and μ2 is the population mean selling price of a single-family home in Whitefield during 2017?

Although the samples are randomly selected, independent, and sufficiently large, the population standard deviation is not known for both towns. Although the samples are randomly selected, independent, and sufficiently large, the population standard deviation of the selling price is not known for both towns. Therefore, not all of the conditions to conduct the hypothesis test have been met and a hypothesis test for the difference between two means of independent samples is not appropriate.

A farm has a large cornfield with two artificial ponds along the east and west borders. Brennan Castillo is an environmental science student who is investigating whether the mean concentrations of a particular pesticide are the same in the two ponds. Brennan collects water samples from random locations and depths in each pond. The sample mean concentration in the 35 samples Brennan collected from the pond to the east of the field is 8.52 parts per billion (ppb) with a standard deviation of 2.71 ppb. The 35 samples Brennan collected from the pond to the west of the field have a sample mean concentration of 9.84 ppb and a standard deviation of 2.93 ppb. Let μ1 be the population mean concentration of the pesticide in the pond to the east of the field, and let μ2 be the population mean concentration of the pesticide in the pond to the west of the field. Brennan tests the alternative hypothesis Ha:μ1−μ2≠0 and assumes that the population standard deviations of the two ponds are equal. If the p-value of the hypothesis test is greater than 0.05 and less than 0.10 and the significance level is α=0.01, what conclusion could be made about the population mean pesticide concentration of the two artificial ponds? Identify all of the appropriate conclusions to the hypothesis test below.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that the population mean pesticide concentration of the pond to the east of the field is different than the population mean pesticide concentration of the pond to the west of the field. Compare the p-value that is greater than 0.05 and less than 0.10 to α=0.01. Since the p-value is greater than α, fail to reject H0. Therefore, there is insufficient evidence at the α=0.01 level of significance to suggest that the population mean pesticide concentration of the pond to the east of the field is different than the population mean pesticide concentration of the pond to the west of the field.

Caleb Satterfield is a student in an automotive technology class. He is writing a report about trends in passenger vehicles, including the age. Caleb would like to claim in the report that the population mean age, in years, of a passenger vehicle owned by a city resident is greater in 2017 than it was in 2012. Caleb goes to the city's vehicle registration office to gather the information. Due to restrictions on the information that can be released, the city clerk provided Caleb with the population standard deviations and a random sample of the ages of passenger vehicles owned by city residents. The sample means and sizes and the population standard deviations Caleb assumes for this hypothesis test are shown in the table below. Let μ1 be the population mean age of passenger vehicles owned by city residents in 2017, and μ2 be the population mean age of passenger vehicles owned by city residents in 2012. The p-value rounded to three decimal places is 0.117, the significance level is α=0.01, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2>0.

Fail to reject the null hypothesis that the true difference between the population mean age of a passenger vehicle owned by a city resident in 2017 and the population mean age of a passenger vehicle owned by a city resident in 2012 is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.01 level of significance to suggest that the true difference between the population mean age of a passenger vehicle owned by a city resident in 2017 and the population mean age of a passenger vehicle owned by a city resident in 2012 is greater than zero. Compare the p-value, 0.117, to α=0.01. Since the p-value is greater than α, fail to reject the null hypothesis H0. Therefore, there is not enough evidence to suggest that the true difference between the population mean age of a passenger vehicle owned by a city resident in 2017 and the population mean age of a passenger vehicle owned by a city resident in 2012 is is greater than zero.

A cell phone manufacturer claims that the population mean battery life of its flagship smartphone model, the Black Bear, is greater than the population mean battery of the largest competitor, the Grizzly. A consumer advocacy publication tests this claim by purchasing a random sample of Black Bear smartphones and a random sample of Grizzly smartphones. Members of the publication charged each smartphone to full capacity and then had the smartphones play the same videos until the batteries were completely depleted. The sample statistics are in the table below. Based on data from the manufacturers, the publication assumes that the population standard deviation for the Black Bear is 0.71 hour and the population standard deviation for the Grizzly is 0.63 hour. Let μ1 be the population mean battery life for the Black Bear and μ2 be the population mean battery life for the Grizzly. The p-value rounded to three decimal places is 0.084, the significance level is α=0.01, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2>0.

Fail to reject the null hypothesis that the true difference between the population mean battery life of the Black Bear and the population mean battery life of the Grizzly is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.01 level of significance to suggest that the true difference between the population mean battery life of the Black Bear and the population mean battery life of the Grizzly is greater than zero. Compare the p-value, 0.084, to α=0.01. Since the p-value is greater than α, fail to reject the null hypothesis H0. Therefore, there is not enough evidence at the α=0.01 level of significance to suggest that the true difference between the population mean battery life of the Black Bear and the population mean battery life of the Grizzly is greater than zero.

Austin Clemens is writing a report for his high school environmental science class about the city's climate. To impress his teacher, Austin would like to show evidence that the population mean daily low temperature during the five-year period from 1998-2002 is less than the population mean daily low temperature during the five-year period from 2013-2017. He researches the claim using a website that records all of the weather data observed at a local airport. After conducting the research, Austin assumes that the population standard deviation is 10.48∘F for 1998-2002 and 11.29∘F for 2013-2017. Due to the large amount of data in each five-year period, Austin randomly selects the daily low temperatures for each group. The sample statistics are shown in the table below. Let μ1 be the population mean daily low temperature during the five-year period from 1998-2002 and μ2 be the population mean daily low temperature during the five-year period from 2013-2017. The p-value rounded to three decimal places is 0.192, the significance level is α=0.10, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2<0.

Fail to reject the null hypothesis that the true difference between the population mean daily low temperature during the five-year period from 1998-2002 and the population mean daily low temperature during the five-year period from 2013-2017 is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.10 level of significance to suggest that the true difference between the population mean daily low temperature during the five-year period from 1998-2002 and the population mean daily low temperature during the five-year period from 2013-2017 is less than zero. Compare the p-value, 0.192, to α=0.10. Since the p-value is greater than α, fail to reject H0. Therefore, there is not enough evidence at the α=0.10 level of significance to suggest that the true difference between the population mean daily low temperature during the five-year period from 1998-2002 and the population mean daily low temperature during the five-year period from 2013-2017 is less than zero.

Kayla Greene is a team lead for an environmental group for a certain region. She is investigating whether the population mean monthly number of kilowatt hours (kWh) used per residential customer in the region has changed from 2006 to 2017. She is concerned that changes such as more efficient lighting and the increased use of electronics and air conditioners are affecting how much electricity residential customers now consume. Kayla assumes the population standard deviation for 2006 and 2017 using the data that were provided to her by local utility companies. Using data that were collected by her organization, Kayla selects a random sample of residential customers who were active for all of 2006 and a separate sample of residential customers who were active for all of 2017. The results from the samples and the population standard deviations that are assumed for this hypothesis test are provided in the accompanying table. Let μ1 be the population mean monthly number of kilowatt hours consumed per residential customer in 2006 and μ2 be the population mean monthly number of kilowatt hours consumed per residential customer in 2017. The p-value rounded to three decimal places is 0.250, the significance level is α=0.01, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2≠0.

Fail to reject the null hypothesis that the true difference between the population mean monthly number of kilowatt hours consumed per residential customer in 2006 and the population mean monthly number of kilowatt hours consumed per residential customer in 2017 is equal to zero. Based on the results of the hypothesis test, there is not enough evidence at the α=0.01 level of significance to suggest that the true difference between the population mean monthly number of kilowatt hours consumed per residential customer in 2006 and the population mean monthly number of kilowatt hours consumed per residential customer in 2017 is not equal to zero. Compare the p-value, 0.250, to α=0.01. Since the p-value is greater than α, fail to reject H0. Therefore, there is not enough evidence to suggest that the true difference between the population mean monthly number of kilowatt hours consumed per residential customer in 2006 and the population mean monthly number of kilowatt hours consumed per residential customer in 2017 is not equal to zero.

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project. Dylan randomly selects 32 student athletes and 45 students who do not play any sports to walk along a board that was 16 feet long and 2 inches wide and raised 6 inches off the ground. Dylan records the number of times each participant touched the ground. The sample of athletes had a mean of 3.7 touches with a standard deviation of 1.1. The sample of non-athletes had a mean of 4.1 touches with a standard deviation of 1.3. Let μ1 be the population mean number of touches for student athletes, and let μ2 be the population mean number of students who do not play any sports. Dylan is testing the alternative hypothesis Ha:μ1−μ2<0 and assumes that the population standard deviation of the two groups of students are equal. If the p-value is greater than 0.05 and less than 0.10 and the significance level is α=0.01, what conclusion could be made about the balance of student athletes and the balance of students who do not play sports? Identify all of the appropriate conclusions to the hypothesis test below.

Fail to reject the null hypothesis. The conclusion of the hypothesis test is that there is insufficient evidence to suggest that the student athletes have a different population mean number of touches than students who do not play sports. Compare the p-value that is greater than 0.05 and less than 0.10 to α=0.01. Since the p-value is greater than α, fail to reject H0. Therefore, there is insufficient evidence at the α=0.01 level of significance to suggest that the student athletes have a different population mean number of touches than students who do not play sports.

Caleb Satterfield is a student in an automotive technology class. He is writing a report about trends in passenger vehicles, including their age. Caleb would like to claim in the report that the population mean age, in years, of a passenger vehicle owned by a city resident is greater in 2017 than it was in 2012. Caleb goes to the city's vehicle registration office to gather the information. Due to restrictions on the information that can be released, the city clerk provides Caleb with the population standard deviations and a random sample of the ages of passenger vehicles owned by city residents, where the same resident was not selected in both samples. The results of the samples and the population standard deviations Caleb assumes for this hypothesis test are shown in the table below. Let α=0.01, μ1 be the population mean age of passenger vehicles owned by city residents in 2017, and μ2 be the population mean age of passenger vehicles owned by city residents in 2012. If the test statistic is z≈1.19 and the rejection region is greater than z0.01≈2.326, what conclusion could be made about the population mean age of passenger vehicles owned by city residents in 2012 and 2017? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean age of passenger vehicles owned by city residents in 2017 is greater than the population mean age of passenger vehicles owned by city residents in 2012. Since z≈1.19 is less than z0.01≈2.326, the z-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean age of passenger vehicles owned by city residents in 2017 is greater than the population mean age of passenger vehicles owned by city residents in 2012.

An automobile manufacturer claims that the population mean braking distance of its premiere vehicle, the Hawk, is less than the population mean braking distance of its main competitor, the Wildcat. Ryan Pottier is a writer for a national automotive magazine and would like to verify the claim made by the manufacturer for an article. He contacts the manufacturer of each vehicle and uses the information to assume that the population standard deviation of the braking distance from 60 miles per hour to 0 is 4.59ft for the Hawk and 4.38ft for the Wildcat. Ryan randomly selects brand new vehicles of each model and conducts a brake test on each car, where each vehicle is stopped from 60 miles per hour in a controlled environment. The results of the test are provided in the table below. Let α=0.01, μ1 be the population braking distance in feet of the Hawk, and μ2 be the population mean braking distance in feet of the Wildcat. If the test statistic is z≈−2.24 and the rejection region is less than −z0.01≈−2.326, what conclusion could be made about the population mean braking distance of the Hawk and the Wildcat? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean braking distance of the Hawk is less than the population mean braking distance of the Wildcat. Since z≈−2.24 is greater than −z0.01≈−2.326, the z-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean braking distance of the Hawk is less than the population mean braking distance of the Wildcat.

The amount of time a plane deviates from its scheduled arrival time is a measure of performance for airports and airlines, where a positive value means the plane arrived late and a negative value means the plane arrived early. Noah Madbury is an aeronautical engineering student who is writing a report on the amount of time planes deviate from their scheduled arrival time. He would like to show in the report whether the population mean deviation of planes from their scheduled arrival times are different at two local airports, Skyhaven and MacArthur. Noah collects a random sample of arrivals at both airports and records the deviations from their scheduled arrival times in minutes. The results of the sample are shown in the table provided. Let μ1 be the population mean deviation, in minutes, of arrivals at Skyhaven and μ2 be the population mean deviation, in minutes, of arrivals at MacArthur. Noah is testing the alternative hypothesis Ha:μ1−μ2≠0, with α=0.01. He assumes that the population standard deviations of the two groups are not equal, so he uses 85 degrees of freedom. If the t-test statistic is t≈1.59 and the rejection regions are less than −t0.005=−2.635 and greater than t0.005=2.635, what conclusion could be made about the population mean deviation of arrivals at Skyhaven and MacArthur airports? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean deviation of arrivals at Skyhaven is different than the population mean deviation of arrivals at MacArthur. Since t≈1.59 is between −t0.005=−2.635 and t0.005=2.635, the t-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean deviation of arrivals at Skyhaven is different than the population mean deviation of arrivals at MacArthur.

Identify the null and alternative hypotheses in the following scenario: To determine if running shoes 1 have the same durability as running shoes 2, the mean durability, in miles, of the two competing running shoes is compared. Fourteen runners are randomly assigned to test each type of shoe. Both populations have normal distributions with known standard deviations.

H0:μ1=μ2; Ha:μ1≠μ2 In order for running shoes 1 to have the same durability as running shoes 2, the mean durability of running shoes 1 must be equal to the mean durability of running shoes 2: μ1=μ2. The null hypothesis must have some form of equality, so Ha is μ1≠μ2 and H0 is μ1=μ2.

A farm has a large corn field with two artificial ponds along the east and west borders. Brennan Castillo is an environmental science student who is investigating whether the population mean concentrations of a particular pesticide in the two ponds are different from each other. Brennan collects water samples from random locations and depths in each pond. The 35 samples Brennan collected from the pond to the east of the field has a mean concentration of 8.52 parts per billion (ppb) with a standard deviation of 2.71 ppb. The 35 samples Brennan collected from the pond to the west of the field have a mean concentration of 9.84 ppb and a standard deviation of 2.93 ppb. Let μ1 be the population mean concentration of the pesticide in the pond to the east of the field and μ2 be the population mean concentration of the pesticide in the pond to the west of the field. Brennan assumes that the population standard deviations of the two groups are equal and tests the alternative hypothesis Ha:μ1−μ2≠0 using α=0.01 and 68 degrees of freedom. If the t-test statistic is t≈−1.96 and the rejection regions are less than −t0.005≈−2.650 and greater than t0.005≈2.650, what conclusion could be made about the population mean pesticide concentration of the two ponds? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean pesticide concentration in the pond to the east of the field is different than the population mean pesticide concentration in the pond to the west of the field. Since t≈−1.96 is between −t0.005≈−2.650 and t0.005≈2.650, the t-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.01 level of significance to conclude that the population mean pesticide concentration in the pond to the east of the field is different than the population mean pesticide concentration in the pond to the west of the field.

A pet food manufacturer has a facility with two lines that blend, can, and cook cat food. Nolan Munroe is the production manager of the facility and curious whether there is a difference in the population mean number of cans produced each day by the two lines, A and B. Nolan randomly selects 300 daily production reports, 141 of which were for production line A and 159 of which were for production line B. The sample statistics are shown in the table below. Let μ1 be the population mean number of cans produced per day on production line A and μ2 be the population mean number of cans produced per day on production line B. Nolan tests the alternative hypothesis Ha:μ1−μ2≠0 with α=0.05, assuming that the population standard deviations of the two groups are not equal and using 140 degrees of freedom. If the t-test statistic is t≈−1.93 and the rejection regions are less than −t0.025=−1.977 and greater than t0.025=1.977, what conclusion could be made about the population mean number of cans produced per day between the two production lines? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.05 level of significance to conclude that the population mean number of cans produced per day by production line A is different than the population mean number of cans produced per day by production line B. Since t≈−1.93 is between −t0.025=−1.977 and t0.025=1.977, the t-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.05 level of significance to conclude that the population mean number of cans produced per day by production line A is different than the population mean number of cans produced per day by production line B.

Quinn Napoli is looking to purchase a vehicle and is deciding between two models, the Glacier and the Ravine. One of Quinn's friends suggests to her that perhaps the annual maintenance costs are different for the two models, so she obtains random samples of the annual maintenance costs for 50 owners of each model of car. The sample results are provided in the table shown below. Let μ1 be the population mean annual maintenance cost for the Glacier and μ2 be the population mean annual maintenance cost for the Ravine. Quinn tests the alternative hypothesis Ha:μ1−μ2≠0, with α=0.10, assuming that the population standard deviations of the two groups are not equal and using 49 degrees of freedom. If the t-test statistic is t≈−1.55 and the rejection regions are less than −t0.05=−1.677 and greater than t0.05=1.677, what conclusion could be made about the population mean annual maintenance cost for the two models? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.10 level of significance to conclude that the population mean annual maintenance cost of the Glacier is different than the population mean annual maintenance cost of the Ravine. Since t≈−1.55 is between −t0.05≈−1.677 and t0.05≈1.67, the t-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.10 level of significance to conclude that the population mean annual maintenance cost of the Glacier is different than the population mean annual maintenance cost of the Ravine.

Ella Mayfield is researching the cost of preschool for a regional parenting magazine by comparing the population mean hourly costs between two large metropolitan areas in the region, Waterboro and Sanford. Based on anecdotal evidence, she is testing the claim that the population mean hourly cost of preschool in Sanford is less than the population mean hourly cost of preschool in Waterboro. She selects a random sample of preschools from both areas and records the rates. The sample statistics are shown in the table below. Let μ1 be the population mean hourly cost of preschool in Sanford and μ2 be the population mean hourly cost of preschool in Waterboro. Ella tests the alternative hypothesis Ha:μ1−μ2<0 with α=0.10, assuming that the population standard deviations of the two groups are not equal and using 40 degrees of freedom. If the t-test statistic is t≈−0.99 and the rejection region is less than −t0.10=−1.303, what conclusion could be made about the population mean hourly costs of preschool in Sanford and Waterboro? Identify all of the appropriate conclusions.

Fail to reject the null hypothesis. There is insufficient evidence at the α=0.10 level of significance to conclude that the population mean hourly cost of preschool in Sanford is less than the population mean hourly cost of preschool in Waterboro. Since t≈−0.99 is greater than −t0.10=−1.303, the t-test statistic is not in the rejection region, so fail to reject the null hypothesis. There is insufficient evidence at the α=0.10 level of significance to conclude that the population mean hourly cost of preschool in Sanford is less than the population mean hourly cost of preschool in Waterboro.

What are the null and alternative hypotheses in the following scenario?The mean loan rate of two competing car dealerships is to be compared. Twenty-six loans are randomly sampled from each dealership. Both populations have normal distributions with known standard deviation.

H0:μ1=μ2; Ha:μ1≠μ2 In order for loans from dealership 1 to be different from loans from dealership 2, the mean loan rate of loans from dealership 1 must be greater or less than the mean loan rate of loans from dealership 2. In other words, the mean loan rate must be unequal: μ1≠μ2. The null hypothesis must have some form of equality, so Ha is μ1≠μ2 and H0 is μ1=μ2.

In the search to determine if car 1 is slower to accelerate than car 2, the mean time it takes to accelerate to 30 miles per hour is recorded (Note: a car is slower to accelerate if it takes more time to accelerate). Twenty trials of the acceleration time for each car are recorded, and both populations have normal distributions with known standard deviations. What are the hypotheses used in this test?

H0:μ1≤μ2; Ha:μ1>μ2 If car 1 is slower to accelerate than car 2, then the time for car 1 to accelerate must be greater than the time for car 2: μ1>μ2. The null hypothesis must have some form of equality, so Ha is μ1>μ2 and H0 is μ1≤μ2.

Identify the null and alternative hypotheses in the following scenario.To determine if computer 1 starts up slower than computer 2, the mean startup time of the two competing computers is compared (the longer the startup time, the slower the computer). Twenty-two startup times are randomly sampled from each computer. Both populations have normal distributions with known standard deviations.

H0:μ1≤μ2; Ha:μ1>μ2 In order for computer 1 to start up slower than computer 2, the mean startup time of computer 1 must be greater than the mean startup time of computer 2: μ1>μ2. The null hypothesis must have some form of equality, so Ha is μ1>μ2 and H0 is μ1≤μ2.

The mean durability, in miles, of two competing running shoes is to be compared. Fifteen runners are randomly assigned to test each type of shoe. Both populations have normal distributions with known standard deviations. In the search to determine if running shoes 1 are more durable running shoes 2, state the null and alternative hypotheses.

H0:μ1≤μ2; Ha:μ1>μ2 In order for running shoes 1 to be more durable than running shoes 2, the mean durability of running shoes 1 must be more than the mean durability of running shoes 2: μ1>μ2. The null hypothesis must have some form of equality, so Ha is μ1>μ2 and H0 is μ1≤μ2.

The mean lasting time of two competing car air fresheners is to be compared. Eighteen cars are randomly assigned to test each air freshener. Both populations have normal distributions with unknown standard deviations. In the search to determine if air freshener 1 is less effective than air freshener 2, state the null and alternative hypotheses.

H0:μ1≥μ2; Ha:μ1<μ2 In order for air freshener 1 to be less effective than air freshener 2, the mean lasting time of air freshener 1 must be less than the mean lasting time of air freshener 2: μ1<μ2. The null hypothesis must have some form of equality, so Ha is μ1<μ2 and H0 is μ1≥μ2.

The mean lasting time of two competing car air fresheners is to be compared. Sixteen cars are randomly assigned to test each air freshener. Both populations have normal distributions with known standard deviations. In the search to determine if air freshener 1 is less effective than air freshener 2, what are the null and alternative hypotheses?

Identify the null and alternative hypotheses in the following scenario.To determine if computer 1 starts up faster than computer 2, the mean startup time of the two competing computers is compared (the shorter the startup time, the faster the computer). Twenty-five startup times are randomly sampled from each computer. Both populations have normal distributions with known standard deviations.

H0:μ1≥μ2; Ha:μ1<μ2 In order for computer 1 to have a faster startup speed than computer 2, the mean startup time of computer 1 must be less than the mean startup time of computer 2: μ1<μ2. The null hypothesis must have some form of equality, so Ha is μ1<μ2 and H0 is μ1≥μ2.

We are doing a two-mean pooled t-test. We have two samples with sizes n1=21 and n2=13. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=10.9 and x¯1=29 For sample 2: s2=11.5 and x¯2=26 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=11.129, then t=29−26(11.129) 121+113√≈0.76df=32 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(21−1)(10.9)2+(13−1)(11.5)221+13−2−−−−−−−−−−−−−−−−−−−−−−−−−−√≈11.129Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=29−26(11.129)121+113−−−−−−√≈3(11.129)0.125−−−−√≈0.76The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=21+13−2=32

Some researchers are doing a two-mean pooled t-test. They have two samples with sizes n1=15 and n2=19. The population standard deviations are unknown but assumed to be equal, so the sample standard deviations are found and used to calculate a pooled standard deviation, sp. For sample 1: s1=11.8 and x¯1=31 For sample 2: s2=12.5 and x¯2=38 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=12.199, then t=31−38(12.199) 115+119√≈−1.66 df=32 For a pooled two-mean t-test, we calculate sp using the formula sp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(15−1)(11.8)2+(19−1)(12.5)215+19−2−−−−−−−−−−−−−−−−−−−−−−−−−−√≈12.199 Now, we can plug this into the formula for the <b>test statistic</b>:t=x¯1−x¯2sp1n1+1n2−−−−−−√=31−38(12.199)115+119−−−−−−√≈−7(12.199)0.119−−−−√≈−1.66 The degrees of freedom for a two-mean pooled t-test is found using the formula: df=n1+n2−2=15+19−2=32

We are doing a two-mean pooled t-test. We have two samples with sizes n1=20 and n2=12. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=13.5 and x¯1=43 For sample 2: s2=12.8 and x¯2=36 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=13.248, then t=43−36(13.248) 120+112√≈1.45df=30 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(20−1)(13.5)2+(12−1)(12.8)220+12−2−−−−−−−−−−−−−−−−−−−−−−−−−−√≈13.248Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=43−36(13.248)120+112−−−−−−√≈7(13.248)0.133−−−−√≈1.45The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=20+12−2=30

Some researchers are doing a two-mean pooled t-test. They have two samples with sizes n1=18 and n2=25. The population standard deviations are unknown but assumed to be equal, so the sample standard deviations are found and used to calculate a pooled standard deviation, sp. For sample 1: s1=13.9 and x¯1=50 For sample 2: s2=13.1 and x¯2=55 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=13.437, then t=50−55(13.437) 118+125√≈−1.2df=41 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(18−1)(13.9)2+(25−1)(13.1)218+25−2−−−−−−−−−−−−−−−−−−−−−−−−−−√≈13.437Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=50−55(13.437)118+125−−−−−−√≈−5(13.437)0.096−−−−√≈−1.2The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=18+25−2=41

We are doing a two-mean pooled t-test. We have two samples with sizes n1=21 and n2=19. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=13.3 and x¯1=50 For sample 2: s2=13.8 and x¯2=40 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=13.539, then t=50−40(13.539) 121+119√≈2.33df=38 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(21−1)(13.3)2+(19−1)(13.8)221+19−2−−−−−−−−−−−−−−−−−−−−−−−−−−√≈13.539Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=50−40(13.539)121+119−−−−−−√≈10(13.539)0.1−−−√≈2.34The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=21+19−2=38

Some researchers are doing a two-mean pooled t-test. They have two samples with sizes n1=25 and n2=13. The population standard deviations are unknown but assumed to be equal, so the sample standard deviations are found and used to calculate a pooled standard deviation, sp. For sample 1: s1=13.2 and x¯1=28 For sample 2: s2=14.2 and x¯2=37 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=13.542, then t=28−37(13.542) 125+113√≈−1.94df=36 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(25−1)(13.2)2+(13−1)(14.2)225+13−2−−−−−−−−−−−−−−−−−−−−−−−−−−√≈13.542Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=28−37(13.542)125+113−−−−−−√≈−9(13.542)0.117−−−−√≈−1.94The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=25+13−2=36

We are doing a two-mean pooled t-test. We have two samples with sizes n1=20 and n2=18. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=14 and x¯1=39 For sample 2: s2=13.2 and x¯2=36 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=13.628, then t=39−36(13.628) 120+118√≈0.68df=36 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(20−1)(14)2+(18−1)(13.2)220+18−2−−−−−−−−−−−−−−−−−−−−−−−−−√≈13.628Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=39−36(13.628)120+118−−−−−−√≈3(13.628)0.106−−−−√≈0.68The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=20+18−2=36

We are doing a two-mean pooled t-test. We have two samples with sizes n1=16 and n2=21. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=4.1 and x¯1=48 For sample 2: s2=5.1 and x¯2=39 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=4.698, then t=48−39(4.698) 116+121√≈5.77df=35 For a pooled two-mean t-test, we calculate sp using the formula sp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(16−1)(4.1)2+(21−1)(5.1)216+21−2−−−−−−−−−−−−−−−−−−−−−−−−√≈4.698 Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=48−39(4.698)116+121−−−−−−√≈9(4.698)0.11−−−−√≈5.77The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=16+21−2=35

We are doing a two-mean pooled t-test. We have two samples with sizes n1=19 and n2=14. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=6.4 and x¯1=43 For sample 2: s2=5.5 and x¯2=33 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=6.039, then t=43−33(6.039) 119+114√≈4.7df=31 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(19−1)(6.4)2+(14−1)(5.5)219+14−2−−−−−−−−−−−−−−−−−−−−−−−−√≈6.039Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=43−33(6.039)119+114−−−−−−√≈10(6.039)0.124−−−−√≈4.7The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=19+14−2=31

We are doing a two-mean pooled t-test. We have two samples with sizes n1=17 and n2=13. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=7.1 and x¯1=35 For sample 2: s2=7.6 and x¯2=26 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=7.318, then t=35−26(7.318) 117+113√≈3.34df=28 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(17−1)(7.1)2+(13−1)(7.6)217+13−2−−−−−−−−−−−−−−−−−−−−−−−−√≈7.318Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=35−26(7.318)117+113−−−−−−√≈9(7.318)0.136−−−−√≈3.33The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=17+13−2=28

We are doing a two-mean pooled t-test. We have two samples with sizes n1=13 and n2=24. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=9.4 and x¯1=47 For sample 2: s2=8.7 and x¯2=44 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=8.946, then t=47−44(8.946) 113+124√≈0.97df=35 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(13−1)(9.4)2+(24−1)(8.7)213+24−2−−−−−−−−−−−−−−−−−−−−−−−−√≈8.946Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=47−44(8.946)113+124−−−−−−√≈3(8.946)0.119−−−−√≈0.97The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=13+24−2=35

We are doing a two-mean pooled t-test. We have two samples with sizes n1=12 and n2=16. The population standard deviations are unknown but assumed to be equal, so we find the sample standard deviations and use them to calculate a pooled standard deviation, sp. For sample 1: s1=10.5 and x¯1=25 For sample 2: s2=9.5 and x¯2=29 What are the test statistic (t) and the degrees of freedom to perform this test?

If sp=9.935, then t=25−29(9.935) 112+116√≈−1.05df=26 For a pooled two-mean t-test, we calculate sp using the formulasp=(n1−1)s21+(n2−1)s22n1+n2−2−−−−−−−−−−−−−−−−−−−√=(12−1)(10.5)2+(16−1)(9.5)212+16−2−−−−−−−−−−−−−−−−−−−−−−−−−√≈9.935Now, we can plug this into the formula for the test statistic:t=x¯1−x¯2sp1n1+1n2−−−−−−√=25−29(9.935)112+116−−−−−−√≈−4(9.935)0.146−−−−√≈−1.05The degrees of freedom for a two-mean pooled t-test is found using the formula:df=n1+n2−2=12+16−2=26

A television station executive would like to determine whether there is more viewership during the 6 P.M. time slot or the 7 P.M. time slot. The executive takes a random sample of 25 days and collects the viewership numbers for the 6 P.M. time slot into one sample and the viewership numbers for the 7 P.M. time slot into a second sample. The executive claims that these samples are independent. Do you agree? Explain.

No, because the data points in the first sample can be linked to data points in the second sample. Two samples are independent if one sample has no effect on the other sample. In this case, the data for both samples are from the same day. It is possible that, on any given day, many of the people that like to watch the program at 6 P.M. also like to watch the program at 7 P.M. This means that a good viewership number at 6 P.M. could make it more likely that there is a good viewership number at 7 P.M. Therefore, these samples are dependent.

A magazine would like to determine how long it takes people to solve the crossword puzzles included inside their magazine. The magazine creates 2 samples to conduct the study. Are the samples dependent or independent?

Not enough information is provided to determine independence. It is unknown how the samples were constructed. To determine whether the samples are dependent or independent, it is necessary to determine if one of the samples can be affected by the other. Are the same people solving crossword puzzles across multiple issues? Or are different groups of people solving one puzzle? Until these types of procedural questions are known, independence cannot be determined.

An executive believes that a new energy drink his company developed will increase an individual's stamina. In order to test this, he selects random individuals and times how long they can run without stopping. He then instructs the individuals to drink the energy drink for two weeks. After the two weeks, he times how long they can run again without stopping. If we were to perform a hypothesis test on the difference between running time after drinking the energy drink and running time before drinking the energy drink, what would be the null and alternative hypotheses?

Null: The population average difference in running time is equal to zero. Alternative: The population average difference in running time is greater than zero. The executive is testing whether the hypothesized population mean of the differences for the paired data is less than 0. Therefore, the null hypothesis is "the population average difference is equal to zero" and the alternative hypothesis is "the population average difference is greater than zero."

A therapist believes that the music a child listens to may have an impact on the amount of words on a list they can memorize. In order to test this, she selects a random sample of children for an experiment. She has the children look over a list of one hundred words for five minutes while listening to classic rock music. After the five minutes, she tests the children to see how many words they could remember in the order of the list. She then does the same experiment, this time with a different list of words, while listening to classical music. If we let d=classical−classic rock, identify the null and alternative hypotheses for an appropriate hypothesis test.

Null: The population average difference in words recited is equal to zero. Alternative: The population average difference in words recited is not equal to zero. The therapist wants to test if the hypothesized mean of the differences for the paired data is different from 0. Therefore, the null hypothesis is "the population average difference in words recited is equal to zero" and the alternative hypothesis is "the population average difference in words recited is not equal to zero."

Derby Leicester is a city planner preparing for a meeting with the mayor. He would like to show that the population mean age of the houses on Lincoln Street is less than the population mean age of the houses on Maple Street so that more resources are allotted to repair Maple Street. Derby uses information from a previous study and assumes that the population standard deviation for the ages of the houses on Lincoln Street is 7.72 years and 8.39 years for the houses on Maple Street. Due to limited time, Derby randomly selects the houses on Lincoln Street and the houses on Maple Street from the city's property records, and then records the age of each house in years. The results of the samples are shown in the table below. Let μ1 be the population mean age, in years, of the houses on Lincoln Street and μ2 be the population mean age, in years, of the houses on Maple Street. The p-value is less than 0.001, the significance level is α=0.05, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2<0.

Reject the null hypothesis that the true difference between the population mean age of the houses on Lincoln Street and the population mean age of the houses on Maple Street is equal to zero. Based on the results of the hypothesis test, there is enough evidence at the α=0.05 level of significance to support the claim that the true difference between the population mean age of the houses on Lincoln Street and the population mean age of the houses on Maple Street is less than zero. Compare the p-value that is less than 0.001 to α=0.05. Since the p-value is less than α, reject H0. Therefore, there is enough evidence at the α=0.05 level of significance to support the claim that the true difference between the population mean age of the houses on Lincoln Street and the population mean age of the houses on Maple Street is less than zero

Leah Peschel is the bottling department manager for a bottling company that produces various soft drinks and juices. The company uses two different machines from different manufacturers to fill the bottles of its popular cola. Leah periodically verifies that the population mean amount of cola in the bottles filled by Machine 1 is the same as the population mean amount in the bottles filled by Machine 2. The manufacturers calibrated the machines at the time of installation and provided that information to the bottling company. Leah uses the manufacturer's specification to assume that the population standard deviation for Machine 1 is 0.021 ounce and the population standard deviation for Machine 2 is 0.019 ounce. She randomly selects samples of bottles filled by Machine 1 and Machine 2. The amount of cola in each bottle is recorded for both samples, and the results are shown in the table. Let μ1 be the population mean amount of cola in bottles filled by Machine 1 and μ2 be the population mean amount of cola in bottles filled by Machine 2. The p-value is less than 0.001, α=0.05, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2≠0.

Reject the null hypothesis that the true difference between the population mean amount of cola in bottles filled by Machine 1 and the population mean amount of cola in bottles filled by Machine 2 is equal to zero. Based on the results of the hypothesis test, there is enough evidence at the α=0.05 level of significance to support the claim that the true difference between the population mean amount of cola in bottles filled by Machine 1 and the population mean amount of cola in bottles filled by Machine 2 is not equal to zero. Compare the p-value, less than 0.001, to α=0.05. Since the p-value is less than α, reject H0. Therefore, there is enough evidence at the α=0.05 level of significance to support the claim that the true difference between the population mean amount of cola in bottles filled by Machine 1 and the population mean amount of cola in bottles filled by Machine 2 is not equal to zero.

Jeremy Remland is a civil engineering student who lives in a large city that has a vast public transportation network. He is writing a report for one of his classes that evaluates the city's public bus and subway services. Jeremy would like to make the claim in the report that the population mean commute time for city residents who take only the bus to commute to work is different than the population mean commute time for city residents who take only the subway to commute to work. Jeremy uses the results of similar studies that have already been completed to estimate that the population standard deviation is 8.07 minutes for commuters who take only the bus and 7.14 minutes for commuters who take only the subway. Jeremy conducts a survey using a random sample of city residents who take only the bus and who take only the subway to commute. The results of the survey are given in the table provided. Let μ1 be the population mean commute time, in minutes, for city residents who take only the bus and μ2 be the population mean commute time, in minutes, for city residents who take only the subway. The p-value rounded to three decimal places is 0.042, the significance level is α=0.05, the null hypothesis is H0:μ1−μ2=0, and the alternative hypothesis is Ha:μ1−μ2≠0.

Reject the null hypothesis that the true difference between the population mean commute time for city residents who take only the bus to commute to work and the population mean commute time for city residents who take only the subway to commute to work is equal to zero. Based on the results of the hypothesis test, there is enough evidence at the α=0.05 level of significance to support the claim that the true difference between the population mean commute time for city residents who take only the bus to commute to work and the population mean commute time for city residents who take only the subway to commute to work is not equal to zero. Compare the p-value, 0.042, to α=0.05. Since the p-value is less than α, reject the null hypothesis H0. Therefore, there is enough evidence at the α=0.05 level of significance to support the claim that the true difference between the population mean commute time for city residents who take only the bus to commute to work and the population mean commute time for city residents who take only the subway to commute to work is not equal to zero.

Cody Addison is a writer for a student magazine at a public university, and he is writing an article about class sizes. Cody would like to make the claim that the mean class size in the school of business is greater than the mean class size in the school of physical sciences at the university. Cody uses the resources at the university's registrar office to select a random sample of classes in the school of business and a random sample of classes in the school of physical sciences, and then he records the class size. The 36 classes in the school of business have a sample mean class size of 33.64 with a sample standard deviation of 9.46. The 38 classes in the school of physical sciences have a sample mean class size of 29.77 with a sample standard deviation of 4.79. Let μ1 be the population mean class size in the school of business, and let μ2 be the population mean class size in the school of physical sciences. Cody tests the alternative hypothesis Ha:μ1−μ2>0 and assumes that the population standard deviations for the two schools are different. If the p-value of the hypothesis test is greater than 0.01 and less than 0.05 and the significance level is α=0.10, what conclusion could be made about the population mean class size between the two schools at the university? Identify all of the appropriate conclusions to the hypothesis test below.

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to suggest that the population mean class size in the school of business is greater than the population mean class size in the school of physical sciences at the university. Compare the p-value that is greater than 0.01 and less than 0.05 to α=0.10. Since the p-value is less than α, reject H0. Therefore, there is sufficient evidence at the α=0.10 level of significance to suggest that the population mean class size in the school of business is greater than the population mean class size in the school of physical sciences at the university.

Caden Carpenter is a professor of entomology who claims that for a certain species of dragonfly, the ones that live by a nearby river have a greater mean wingspan than those that live near a lake that is 10 kilometers away. To test this claim, Caden randomly captures, measures, and releases 45 dragonflies at each site. The sample statistics are provided in the table below. Let μ1 be the population mean wingspan of the dragonflies that live by the river, and let μ2 be the population mean wingspan of the dragonflies that live by the lake. Caden uses the alternative hypothesis Ha:μ1−μ2>0 and assumes that the population standard deviation for both groups of dragonflies is equal. If the p-value of the hypothesis test is less than 0.01 and the significance level is α=0.01, what conclusion could be made about the population mean wingspans of the dragonflies that live by the river and those that live by the lake?

Reject the null hypothesis. The conclusion of the hypothesis test is that there is sufficient evidence to suggest that the population mean wingspan of the dragonflies that live by the river is greater than the population mean wingspan of the dragonflies that live by the lake. Compare the p-value that is less than 0.01 to α=0.01. Since the p-value is less than α, reject H0. Therefore, there is sufficient evidence at the α=0.01 level of significance to suggest that the population mean wingspan of the dragonflies that live by the river is greater than the population mean wingspan of the dragonflies that live by the lake.

A large drug store chain outsources the manufacturing of its pain relief tablets to two different, independent manufacturers, Manufacturer A and Manufacturer B. The quality control department of the chain would like to ensure that the population mean amount of the active ingredient in the tablets is still the same between the two manufacturers. The department assumes values for the population standard deviation for the amount of the active ingredient in each tablet based on specifications provided by the manufacturers. The quality control department selects a random sample of tablets from both manufacturers across dozens of lots. The tablets are then analyzed using chromatography to determine the amount of the active ingredient in each tablet. The population standard deviations as well as the results of the test are provided in the table below. Let α=0.01, μ1 be the population mean amount of the active ingredient, in milligrams, in tablets produced by Manufacturer A, and μ2 be the population mean amount of the active ingredient, in milligrams, in tablets produced by Manufacturer B. If the test statistic is z≈−2.93 and the rejection region is less than −z0.005≈−2.576 or greater than z0.005≈2.576, what conclusion could be made about the population mean amount of the active ingredient in the tablets from both manufacturers? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the population mean amount of the active ingredient in tablets produced by Manufacturer A is different than the population mean amount of the active ingredient in tablets produced by Manufacturer B. Since z≈−2.93 is less than −z0.005≈−2.576, the z-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the population mean amount of the active ingredient in tablets produced by Manufacturer A is different than the population mean amount of the active ingredient in tablets produced by Manufacturer B.

Marianna is a website designer who wants to determine whether two home page formats lead visitors to click on a different number of links. She writes two scripts: one to randomly display one of the home pages and another to record the page displayed and the number of links clicked. For the sample for home page A, the 132 randomly selected visitors clicked an average of 8.7 links with a standard deviation of 2.3. For the sample for home page B, the 118 randomly selected visitors clicked an average of 7.3 links with a standard deviation of 2.1. Let μ1 be the population mean number of clicks for home page A and μ2 be the population mean number of clicks for home page B. Marianna uses the alternative hypothesis Ha:μ1−μ2≠0, assumes the sample standard deviations are equal, and sets α=0.01 using 248 degrees of freedom. If the t-test statistic is t≈5.01 and the rejection regions are less than −t0.005≈−2.596 and greater than t0.005≈2.596, what conclusion could be made about the population mean number of clicks between the home pages? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the population mean number of clicks for home page A is different than the population mean number of clicks for home page B. Since t≈5.01 is greater than t0.005≈2.596, the t-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the population mean number of clicks for home page A is not equal to the population mean number of clicks for home page B.

Caden Carpenter is a professor of entomology who claims that for a certain species of dragonfly, the ones that live by a nearby river have a greater population mean wingspan than those that live near a lake that is 10 kilometers away. To test this claim, Caden captures, measures, and releases 45 randomly selected dragonflies at each site. The sample statistics are provided in the table below. Let μ1 be the population mean wingspan of the dragonflies that live by the river and μ2 be the population mean wingspan of the dragonflies that live by the lake. Caden assumes that the population standard deviations of the two groups are equal and uses the alternative hypothesis Ha:μ1−μ2>0 with α=0.01. If the t-test statistic is t≈3.76 and the rejection region is greater than t0.01≈2.369 using 88 degrees of freedom, what conclusion could be made about the population mean wingspan of the dragonflies that live by the river and those that live by the lake? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the population mean wingspan of the dragonflies that live by the river is greater than the population mean wingspan of the dragonflies that live by the lake. Since t≈3.76 is greater than t0.01≈2.369, the t-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the population mean wingspan of the dragonflies that live by the river is greater than the population mean wingspan of the dragonflies that live by the lake.

Dylan Rieder is a statistics student investigating whether athletes have better balance than non-athletes for a thesis project. Dylan randomly selected 32 student athletes and 45 students who do not play any sports to walk along a board that was 16 feet long and 2 inches wide, and raised 6 inches off the ground. Dylan recorded the number of times each participant touched the ground. The athletes had a sample mean of 3.7 touches with a sample standard deviation of 2.3. The non-athletes had a sample mean of 4.1 touches with a sample standard deviation of 2.5. Let μ1 be the population mean number of touches for student athletes and μ2 be the population mean number of students who did not play any sports. What type of test is this hypothesis test?

This is a left-tailed test because the alternative hypothesis is Ha:μ1−μ2<0. Since the claim is that athletes have better balance than non-athletes, the population mean number of touches for athletes should be less than the population mean number of touches for non-athletes. Therefore, the alternative hypothesis is Ha:μ1−μ2<0, which means that this hypothesis test is a left-tailed test.

Derby Leicester is a city planner preparing for a meeting with the mayor. He would like to show that the population mean age of the houses on Lincoln Street is less than the population mean age of the houses on Maple Street so that more resources are allotted to repair Maple Street. Derby uses data from a previous study and assumes that the population standard deviation for the ages of the houses on Lincoln Street is 7.72 years and 8.39 years for the houses on Maple Street. Due to limited time, Derby randomly selects houses on Lincoln Street and houses on Maple Street from the city's property records and then records the age of each house in years. The results of the samples are shown in the table below. Let α=0.05, μ1 be the population mean age in years of the houses on Lincoln Street, and μ2 be the population mean age in years of the houses on Maple Street. If the test statistic is z≈−4.56 and the rejection region is less than −z0.05≈−1.645, what conclusion could be made about the population mean age of the houses on the two streets? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to conclude that the population mean age of the houses on Lincoln Street is less than the population mean age of the houses on Maple Street. Since z≈−4.56 is less than −z0.05≈−1.645, the z-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to conclude that the population mean age of the houses on Lincoln Street is less than the population mean age of the houses on Maple Street

A team of engineers tested the shear strength of a new brand of rivets to ensure that they are at least as strong as the brand of rivets that is already approved for the construction of a new bridge. The engineers randomly selected 250 new-brand rivets and 250 approved-brand rivets and used a hydraulic press to apply shear forces on the rivets. The force necessary for failure was recorded, and the sample statistics are shown in the table below. Let μ1 be the population mean force of failure for the new-brand rivets and μ2 be the population mean force of failure for the approved-brand rivets. The engineers used the alternative hypothesis Ha:μ1−μ2<0, assumed the sample standard deviations were equal, and set α=0.05 using 498 degrees of freedom. If the t-test statistic is t≈−3.07 and the rejection region is less than −t0.05≈−1.648, what conclusion could be made about the mean force of failure for the two brands of rivets? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to conclude that the population mean force of failure for the new-brand rivets is less than the population mean force of failure for the approved-brand rivets. Since t≈−3.07 is less than −t0.05≈−1.648, the t-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to conclude that the population mean force of failure for the new-brand rivets is less than the population mean force of failure for the approved-brand rivets.

The quality manager at a breakfast cereal producer claimed that one of the two production lines in the factory, Production Line 1, was over-filling the boxes compared to Production Line 2. A random sample of 40 boxes produced by Production Line 1 had a mean weight of 20.062 ounces with a standard deviation of 0.035 ounce. A sample of 40 boxes produced by Production Line 2 had a mean weight of 20.047 ounces with a standard deviation of 0.033 ounce. The manager assumes that the population standard deviations are equal for both groups. Let α=0.05, μ1 be the population mean weight of boxes filled by Production Line 1, and μ2 be the population mean weight of boxes filled by Production Line 2. If the t-test statistic is t≈1.97 and the rejection region is greater than t0.05≈1.665 using 78 degrees of freedom, which conclusion could be made about the mean weight of the cereal boxes filled by the two production lines? Identify both of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to conclude that the population mean weight of the boxes filled by Production Line 1 is greater than the population mean weight of the boxes filled by Production Line 2. Since t≈1.97 is greater than t0.05≈1.665, the t-test statistic is in the rejection region, and thus, reject H0. Therefore, there is enough evidence at the α=0.05 level of significance to conclude that the population mean weight of the boxes filled by Production Line 1 is greater than the population mean weight of the boxes filled by Production Line 2.

Cody Addison is a writer for a student magazine at a public university, and he's writing an article about class sizes. Cody would like to make the claim that the population mean class size in the school of business is greater than the population mean class size in the school of physical sciences at the university. Cody uses the resources at the university's registrar office to select a random sample of classes in the school of business and a random sample of classes in the school of physical sciences, and then he records the class sizes. The 36 classes in the school of business have a sample mean class size of 33.64 with a sample standard deviation of 9.46. The 38 classes in the school of physical sciences have a sample mean class size of 29.77 with a sample standard deviation of 4.79. Let μ1 be the population mean class size in the school of business and μ2 be the population mean class size in the school of physical sciences. Cody tests the alternative hypothesis Ha:μ1−μ2>0, with α=0.10, assuming that the population standard deviations of the two groups are not equal and using 35 degrees of freedom. If the t-test statistic is t≈2.20 and the rejection region is greater than t0.10=1.306, what conclusion could be made about how the population mean class size in the school of business compares to the population mean class size in the school of physical sciences? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.10 level of significance to conclude that the population mean class size in the school of business is greater than the population mean class size in the school of physical sciences. Since t≈2.20 is greater than t0.10=1.306, the t-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.10 level of significance to conclude that the population mean class size in the school of business is greater than the population mean class size in the school of physical sciences.

Lara Norris is a poultry farmer who claims that her Rhode Island Red hens on average lay more eggs than her Plymouth Rock hens. A random sample of 60 values of the daily egg production by the Rhode Island Red hens had a mean of 0.765 egg per hen per day and a standard deviation of 0.113. A random sample of 50 values of the daily egg production by the Plymouth Rock hens had a mean of 0.559 egg per hen per day and a standard deviation of 0.092. Let μ1 be the population mean number of eggs laid per hen per day by the Rhode Island Red hens and μ2 be the population mean number of eggs laid per hen per day by the Plymouth Rock hens. What type of test is this hypothesis test?

This is a right-tailed test because the alternative hypothesis is Ha:μ1−μ2>0. Since the claim is that Rhode Island Red hens on average lay more eggs than Plymouth Rock hens, Lara is looking for evidence that μ1 is greater than μ2. Therefore, the alternative hypothesis is Ha:μ1−μ2>0, which means that this hypothesis test is a right-tailed test.

Lara Norris is a poultry farmer who wants to know whether different breeds of hen lay different numbers of eggs, on average. A random sample of 60 values of the daily egg production by the Rhode Island Red hens had a mean of 0.765 egg per hen per day and a standard deviation of 0.113. A random sample of 50 values of the daily egg production by the Plymouth Rock hens had a mean of 0.559 egg per hen per day and a standard deviation of 0.092. Let μ1 be the population mean number of eggs laid per hen per day by the Rhode Island Red hens and μ2 be the population mean number of eggs laid per hen per day by the Plymouth Rock hens. Lara assumes that the population standard deviations of the two groups are equal and uses the alternative hypothesis Ha:μ1−μ2≠0 with α=0.10. If the t-test statistic is t≈10.34 and using 108 degrees of freedom, the rejection regions are less than −t0.05=−1.659 and greater than t0.05=1.659, what conclusion could be made about the population mean number of eggs laid per hen per day for the two types of hens? Identify all of the appropriate conclusions.

Reject the null hypothesis. There is sufficient evidence at the α=0.10 level of significance to conclude that the population mean number of eggs laid per hen per day by the Rhode Island Red hens is different than the population mean number of eggs laid per hen per day by the Plymouth Rock hens. Since t≈10.34 is greater than t0.05=1.659, the t-test statistic is in the rejection region, so reject the null hypothesis. There is sufficient evidence at the α=0.10 level of significance to conclude that the population mean number of eggs laid per hen per day by the Rhode Island Red hens is different than the population mean number of eggs laid per hen per day by the Plymouth Rock hens.

An internet researcher creates a poll with two related questions on an internet website. The researcher places each person's answer to the first question in one sample and each person's answer to the second question in the second sample. Are these samples independent or dependent?

The samples are dependent. The two samples consist of answers from the same group of people. Therefore, the answers can be linked by a person's answer to the first question in the first sample and same person's answer to the second question in the second sample. This makes the two samples dependent.

Arianna Estefan manages Salmon Falls Park, which has two flagship roller coasters, the Flyer and the Destroyer. When Arianna walks around the park during hours of operation, she notices that there are fewer people waiting in line for the Flyer than for the Destroyer. Since both roller coasters have similar capacities and similar run times, both roller coasters should have about the same number of riders. Based on the anecdotal evidence, Arianna claims that the population mean number of riders per hour on the Flyer is less than the population mean number of riders per hour on the Destroyer. Arianna reviews data collected in the past and assumes that the population standard deviation is 48.29 for the Flyer and 51.86 for the Destroyer. Arianna randomly selects hours during which the attendant of each ride counts the number of riders. The sampling occurs over the course of several weeks. The results are provided in the table shown below. Let μ1 be the population mean number of riders per hour on the Flyer and μ2 be the population mean number of riders per hour on the Destroyer. What type of test is this hypothesis test?

This is a left-tailed test because the alternative hypothesis is Ha:μ1−μ2<0. Since Arianna is testing whether there are fewer riders per hour on the Flyer than on the Destroyer, she is looking for evidence that supports μ1 being less than μ2. Therefore, the alternative hypothesis is Ha:μ1−μ2<0, which means that this hypothesis test is a left-tailed test.

Austin Clemens is writing a report about the city's climate for his high school environmental science class. To impress his teacher, Austin would like to show evidence that the population mean daily low temperature was lower during the five-year period 1998-2002 than it was for the 5-year period 2013-2017. He researches the claim using a website that records all of the weather data observed at a local airport. After conducting the research, Austin assumes that the population standard deviation is 10.48∘F for 1998-2002 and 11.29∘F for 2013-2017. Due to the large amount of data in each five-year period, Austin randomly selects the daily low temperatures for each group. The results are shown in the table below. Let μ1 be the population mean daily low temperature during the five-year period 1998-2002 and μ2 be the population mean daily low temperature during the five-year period 2013-2017. What type of test is this hypothesis test? What is/are the critical value(s) of the z-test statistic for this hypothesis test, where α=0.10? Use the appropriate value(s) from the table. Use a comma and a space to separate answers as needed.

This is a left-tailed test because the alternative hypothesis is Ha:μ1−μ2<0. Since Austin Clemens is testing whether the mean daily low temperature was lower during the five-year period 1998-2002 than it was for the 5-year period 2013-2017, he is looking for evidence that supports μ1 being less than μ2. Therefore, the alternative hypothesis is Ha:μ1−μ2<0, which means that this hypothesis test is a left-tailed test. $\text{critical value(s)}=-1.282$critical value(s)=−1.282 Since this is a left-tailed test and α=0.10, the critical value occurs at −zα=−z0.10. To find the critical value, use a normal distribution table to find the z-score that corresponds to α=0.10 for a left-tailed test. z0.10z0.05z0.025z0.01z0.0051.2821.6451.9602.3262.576 Therefore, −z0.10=−1.282 is the critical value for this hypothesis test.

This is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0. Since Brennan wants to determine whether the population mean concentrations of the ponds are different, he is looking for evidence that μ1 is different than μ2. Therefore, the alternative hypothesis is Ha:μ1−μ2≠0, which means that this hypothesis test is a two-tailed test.

Quinn Napoli is looking to purchase a vehicle and is deciding between two models, the Glacier and the Ravine. One of Quinn's friends suggests to her that perhaps the annual maintenance costs are different for the two models, so she obtains random samples of the annual maintenance costs for 50 owners of each model of car. The results are provided in the table shown below. Let μ1 be the population mean annual maintenance cost for the Glacier, and let μ2 be the population mean annual maintenance cost for the Ravine. Quinn assumes the population standard deviations are not equal and tests the alternative hypothesis Ha:μ1−μ2≠0. If the test statistic is t≈−1.55 and the number of degrees of freedom is 49, what is the p-value for this hypothesis test?

p-value >0.10 Notice that this is a two-tailed test because the alternative hypothesis is Ha:μ1−μ2≠0, so the p-value is twice the area in one tail. The test statistic is t≈−1.55 with 49 degrees of freedom. The t-distribution table gives the values for a right-tailed test, so to find twice area in the left tail, use the absolute value of the test statistic. Using the table of areas in the right tail for the t-distribution, in the row for 49 degrees of freedom, 1.55 is less than 1.677. So, the p-value is greater than 2(0.05)=0.10.

A hypothesis test is performed at the 10% significance level to determine if the mean time studying (in hours per week) of students at private universities and public universities are different. The hypotheses were determined and data was collected in the table below for both samples. H0:μ1≥μ2, Ha:μ1<μ2 α=0.1

t=17.6−13.6(3.7)218+(1.8)226√≈4.25 Sample 1 is the private universities: n1 (sample size) = 18 x¯1 (sample mean) = 17.6 s1 (sample standard deviation) = 3.7 Sample 2 is the public universities: n2 (sample size) = 26 x¯2 (sample mean) = 13.6 s2 (sample standard deviation) = 1.8 Since both the the samples, n1 and n2, are independent, we can calculate the test statistic with the following formula: t=x¯1−x¯2(s1)2n1+(s2)2n2−−−−−−−−−√=17.6−13.6(3.7)218+(1.8)226−−−−−−−−−−√≈40.76+0.12−−−−−−−−−√≈4.25 So, the test statistic is t=4.25.

A hypothesis test is performed at the 1% significance level to determine if the mean time studying (in hours per week) of students at private universities and public universities are different. The hypotheses were determined and data was collected in the table below for both samples. H0:μ1=μ2, Ha:μ1≠μ2 α=0.01

t=21.8−12.1(3.8)218+(2.5)230√≈9.65 Sample 1 is the private universities: n1 (sample size) = 18 x¯1 (sample mean) = 21.8 s1 (sample standard deviation) = 3.8 Sample 2 is the public universities: n2 (sample size) = 30 x¯2 (sample mean) = 12.1 s2 (sample standard deviation) = 2.5 Since both the the samples, n1 and n2, are independent, we can calculate the test statistic with the following formula: t=x¯1−x¯2(s1)2n1+(s2)2n2−−−−−−−−−√=21.8−12.1(3.8)218+(2.5)230−−−−−−−−−−√≈9.70.8+0.21−−−−−−−−√≈9.65 So, the test statistic is t=9.65.

A hypothesis test is performed at the 5% significance level to determine if the mean credit card debt (in thousands of dollars) of graduates of college and high school are different. The hypotheses were determined and data was collected in the table below for both samples. H0:μ1≤μ2, Ha:μ1>μ2 α=0.05

t=25.6−0.8(22.1)225+(4.4)228√≈5.51 Sample 1 is the college graduates: n1 (sample size) = 25 x¯1 (sample mean) = 25.6 s1 (sample standard deviation) = 22.1 Sample 2 is the high school graduates: n2 (sample size) = 28 x¯2 (sample mean) = 0.8 s2 (sample standard deviation) = 4.4 Since both the the samples, n1 and n2, are independent, we can calculate the test statistic with the following formula: t=x¯1−x¯2(s1)2n1+(s2)2n2−−−−−−−−−√=25.6−0.8(22.1)225+(4.4)228−−−−−−−−−−−√≈24.819.54+0.69−−−−−−−−−−√≈5.51 So, the test statistic is t=5.51.

A hypothesis test is performed at the 1% significance level to determine if the mean time of faculty contact (in hours per week) with students at research universities and non-research universities are different. The hypotheses were determined and data was collected in the table below for both samples. H0:μ1≤μ2, Ha:μ1>μ2 α=0.01

t=9.3−16.8(2.4)229+(3.9)215√≈−6.81 Sample 1 is the research universities: n1 (sample size) = 29 x¯1 (sample mean) = 9.3 s1 (sample standard deviation) = 2.4 Sample 2 is the non-research universities: n2 (sample size) = 15 x¯2 (sample mean) = 16.8 s2 (sample standard deviation) = 3.9 Since both the the samples, n1 and n2, are independent, we can calculate the test statistic with the following formula: t=x¯1−x¯2(s1)2n1+(s2)2n2−−−−−−−−−√=9.3−16.8(2.4)229+(3.9)215−−−−−−−−−−√≈−7.50.2+1.01−−−−−−−−√≈−6.81 So, the test statistic is t=−6.81.

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