Chapter 9

In​ randomized, double-blind clinical trials of a new​ vaccine, monkeys were randomly divided into two groups. Subjects in group 1 received the new vaccine while subjects in group 2 received a control vaccine. After the second​ dose, 119 of 662 subjects in the experimental group​ (group 1) experienced fever as a side effect. After the second​ dose, 76 of 543 of the subjects in the control group​ (group 2) experienced fever as a side effect. Does the evidence suggest that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2 at the α=0.10 level of​ significance?

(1) B. n1p11−p1≥10 and n2p21−p2≥10 C. The sample size is less than​ 5% of the population size for each sample. F. The samples are independent. (2) H0: p1 = p2 H1:p1>p2 (3) 1.87 Explanation: x1 = 119 x2 = 76 n1 = 662 n2 = 543 p̂1 = 0.1799 p̂2 = 0.1399 p̂ = 0.1626 (4) .031 (5) equal, greater than, 31 (6) Reject H0. There is sufficient evidence to conclude that a higher proportion of subjects in group 1 experienced fever as a side effect than subjects in group 2 at the α = 0.10 level of significance.

A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show the day of the week for n = 288 randomly selected accidents. Is there reason to believe that the accident occurs with equal frequency with respect to the day of the week at .05 the level of​ significance?

(1) H0​: H1​: At least one proportion is different from the others. (2) 41.14 for all Expected Counts. Explanation: 1/7 = 0.14286 * 288 = 41.14 (3) Test statistic = 16.743 P-value = .01 Explanation: Open StatCrunch with graph → Stat → Goodness-of-fit → Chi-Square Test → Observed set to Frequency → Expected set to All cells in equal proportion → Compute! → Answer is the Chi-Square for the test statistic (also find P-value there) (4) Reject H0​, because the calculated​P-value is less than the given α level of significance. Explanation: .01 < .05 P-value is less than the level of​ significance, reject the null hypothesis.

A book claims that more hockey players are born in January through March than in October through December. The following data show the number of players selected in a draft of new players for a hockey league according to their birth month. Is there evidence to suggest that hockey​ players' birthdates are not uniformly distributed throughout the​ year? Level of significance = .01. Compute the expected counts for each birth month. The total number of hockey players is 178.

(1) H0​: The distribution of hockey​ players' birth months is uniformly distributed. H1​: The distribution of hockey​ players' birth months is not uniformly distributed. (2) 44.5 for all Expected Counts. Explanation: 1/4 = 0.25. 178*0.25 = 44.5 (3) Test statistic = 14.4 P-value = .002 Explanation: Open StatCrunch with graph → Stat → Goodness-of-fit → Chi-Square Test → Observed set to Frequency → Expected set to All cells in equal proportion → Compute! → Answer is the Chi-Square for the test statistic (also find P-value there) (4) Yes, ​because the calculated​ P-value is than the given level of significance. .002 < .01 P-value is less than the level of​ significance, reject the null hypothesis.

According to the manufacturer of​ M&Ms, 13​% of the plain​ M&Ms in a bag should be​ brown, 14​% ​yellow, 13​% ​red, 24​% ​blue, 20​% ​orange, and 16​% green. A student randomly selected a bag of plain​ M&Ms. He counted the number of​ M&Ms of each color and obtained the results shown in the table. Test whether plain​ M&Ms follow the distribution stated by the manufacturer at the α=0.05 level of significance.

(1) H₀: The distribution of colors is the same as stated by the manufacturer. H₁: The distribution of colors is not the same as stated by the manufacturer. (2) 52.78, 56.84, 52.78, 97.44, 81.2, 64.96 Explanation: Add up frequency = 406. Then, multiply total by claimed proportion in chart to get the expected count numbers. (3) Test statistic = 22.16 P-value = .001 Explanation: Open StatCrunch with graph → Stat → Goodness-of-fit → Chi-Square Test → Observed set to Frequency → Expected set to Claimed Proportion → Compute! → Answer is the Chi-Square for the test statistic (also find P-value there) (4) Reject, sufficient, not the same as .001 < .005 Explanation: P-value is less than the level of​ significance, reject the null hypothesis.

Explain the differences between the​ chi-square test for independence and the​ chi-square test for homogeneity. What are the​ similarities?

(1) The difference is that the chi-square test for independence compares 2 characteristics from one population and the chi-square test for homogeneity compares one characteristic from more than one population. (2) B. The assumptions are the same. C. The procedures are the same.

The following table contains the number of successes and failures for three categories of a variable. Test whether the proportions are equal for each category at the α=0.01 level of significance.

(1) ​H0: p1=p2=p3 H1: At least one of the proportions is different from the others. (2) P-value = 0 Explanation: Open StatCrunch with graph → Stat → Tables → Contingency → With Summary → Select all columns → Select any row label → Compute! (3) The​ P-value is less than α​, so reject H0. There is sufficient evidence that the proportions are different from each other.

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below.

(a) -2.8, 1.4, -4.3, -4.8, -1.1, -4.3, -3.3, .2 Explanation: subtract x1 - y1 and so on. (b) d = -2.375 Explanation: -19/8 = -2.375 (b1) sd = 2.294 Explanation: Open graph in StatCrunch, remove Yi column, and add numbers from (a) into Xi column → Stat → Summary Stats → Select column Xi → Highlight Std Dev. → Compute! (c) H0​: μd=0 H1​: μd<0 (c1) P-value = 0.011 Explanation: Still in StatCrunch → Stat → T Stats → One Sample → With Summary → Input numbers, sample size is 8 → Set to H0​: μd=0, H1​: μd<0 → Compute! (c2) Reject the null hypothesis. There is sufficient evidence that μd<0 at the α=0.05 level of significance. (d) -4.29, 0.46 Explanation: (c1) Results Still on StatCrunch → Options → Edit → Change to Confidence interval for Level 0.95 → Compute! → L. Limit and U. Limit are the answers

In 1944​, an organization surveyed 1100 adults and​ asked, "Are you a total abstainer​ from, or do you on occasion​ consume, alcoholic​ beverages?" Of the 1100 adults​ surveyed, 429 indicated that they were total abstainers. In a recent​ survey, the same question was asked of 1100 adults and 407 indicated that they were total abstainers. Complete parts​ (a) and​ (b) below. x1 = 429, x2 = 407, n1 = 1100, n2 = 1100

(a) .39, .37 Explanation: 429/1100 = 0.39 407/1100 = 0.37 (b) B. The sample size is less than 5% of the population size for each sample. D. n1p1(1-p1) ≥ 10 and n2p2(1-p2) ≥ 10. F. The samples are independent. (b1) H0: p1 = p2 H1: p1 ≠ p2 (b2) Test statistic: 0.97 P-value: 0.334 Explanation: For test statistic and P value... Open StatCrunch → Stat → Proportion Stats → Two sample → With Summary → input # of Successes (429, 407) and # of trials for each sample (1100, 1100) → Perform: select Hypothesis test for p1 - p2 → H0: p1 - p2 = 0 (p1 = p2, subtract p2 to both sides of the equation, we will obtain p1 - p2 = 0), HA: p1 - p2 ≠ 0 → Compute! → Z-Stat and P-value are the answers (b3) equal, greater than the absolute value of, 33 (b4) Do not reject H0. There is not sufficient evidence at the α=0.05 level of significance to suggest the proportion of adults who totally abstain from alcohol has changed.

Three years​ ago, the mean price of an existing​ single-family home was ​$243,773. A real estate broker believes that existing home prices in her neighborhood are higher. ​(a) State the null and alternative hypotheses in words. ​(b) State the null and alternative hypotheses symbolically. ​(c) Explain what it would mean to make a Type I error. ​(d) Explain what it would mean to make a Type II error.

(a) 1. The mean price of a single family home in the​ broker's neighborhood is $243,773. 2. The mean price of a single family home in the​ broker's neighborhood is greater than ​$243,773. (b) H0: μ = $243,773 H1: μ > $243,773 (c) The broker rejects the hypothesis that the mean price is equal to ​$243,773​, when the true mean price is equal to $243,773. (d) The broker fails to reject the hypothesis that the mean price is equal to ​$243773​, when the true mean price is greater than ​$243773.

Determine ​(a) the χ20 test​ statistic, ​(b) the degrees of​ freedom, ​(c) the​ P-value, and ​(d) test the hypothesis at the α=0.05 level of significance. H0​: The random variable X is binomial with n=4​, p=0.8 H1​: The random variable X is not binomial with n=4​, p=0.8

(a) 14.22 (1-1.6)^2/1.6 = .225 (35-25.2)^2/25.2 = 3.81 (120-151)^2/151 = 6.36424 (439-402.6)^2/402.6 = 3.29101 (388-402.6)^2/402.6 = .529459 (.225) + (3.81) + (6.36424) + (3.29101) + (.529459) = 14.219709 (b) df = 4 Explanation: Number of categories = 5 Degrees of freedom = 5 - 1 = 4 (c) The​ P-value is between 0.005 and 0.01 . https://www.socscistatistics.com/pvalues/chidistribution.aspx (d) Reject the null​ hypothesis, because the​ P-value is less than alpha equals 0.05 .α=0.05. There isis evidence that the random variable X is not binomial with n=4​, p=0.8. Explanation: If the​ P-value is greater than or equal to the level of​ significance, α=0.05​, then do not reject the null hypothesis. If it is less than the level of​ significance, then reject the null hypothesis.

Two researchers conducted a study in which two groups of students were asked to answer 42 trivia questions from a board game. The students in group 1 were asked to spend 5 minutes thinking about what it would mean to be a​ professor, while the students in group 2 were asked to think about soccer hooligans. These pretest thoughts are a form of priming. The 200 students in group 1 had a mean score of 23.7 with a standard deviation of 4.5​, while the 200 students in group 2 had a mean score of 15.7 with a standard deviation of 3.5. Complete parts ​(a) and ​(b) below. ​(a) Determine the 90​% confidence interval for the difference in​ scores, μ1−μ2. Interpret the interval.

(a) 7.335 and 8.665 Explanation: Open StatCrunch → Stat → T Stats → Two samples with summary → Input means (x), standard deviation (s), and sample size (n) → Click on Confidence Interval, change to .90 → Compute! (a1) The researchers are 90​% confident that the difference of the means is in the interval. (b) Since the 90​% confidence interval does not contain​ zero, the results suggest that priming does have an effect on scores.

In one​ month, the national mean price per gallon​ (in dollars) of gasoline was 3.106. The data table here represents a random sample of 20 gas stations in city A. Is gas in city A more expensive than the​ nation? Assume the data come from a normal population with no outliers. Complete parts​ (a) through​ (d) below.

(a) A hypothesis test regarding a single population mean using Student's approximate t (b) Null Hypothesis (H0): μ = 3.106 The null hypothesis states that there is no significant difference between the mean price of gasoline in city A and the national mean price (3.106 dollars per gallon). Alternative Hypothesis (H1): μ > 3.106 The alternative hypothesis states that the mean price of gasoline in city A is greater than the national mean price (3.106 dollars per gallon). (c) 0 (d) Although no level of significance is​ stated, the​ P-value is so small that the null hypothesis can be rejected. There is sufficient evidence to conclude that the gas prices in city A are higher than the national average.

Two professors at a local college developed a new teaching curriculum designed to increase​ students' grades in math classes. In a typical developmental math​ course, 54​% of the students complete the course with a letter grade of​ A, B, or C. In the experimental​ course, of the 14 students​ enrolled, 10 completed the course with a letter grade of​ A, B, or C. Is the experimental course effective at the α=0.1 level of​ significance? Complete parts ​(a) through​ (g). ​(e) Suppose the course is taught with 42 students and 30 complete the course with a letter grade of​ A, B, or C. Verify whether the normal model may now be used to estimate the​ P-value.

(a) H0: p = 0.54 versus H1: p > 0.54 (b) Because np01−p0=3.5 < ​10, the normal model may not be used to approximate the​ P-value. Explanation: 14(0.54)(1-0.54) = 3.4776 < 10 (c) There is a fixed number of trials with two mutually exclusive outcomes. The trials are independent and the probability of success is fixed at .54 for each trial. (d) 1. P-value = 0.148 Explanation: 10/14 = 0.714 > 0.54 Find P(X≥10) with the probability being 0.54, the number of trials being 14, and number of successes being 10. https://stattrek.com/online-calculator/binomial 2. No​, do not reject the null hypothesis because the​ P-value is greater than α. There is insufficient evidence to conclude that the experimental course is effective. Explanation: Compare the​ P-value with the level of significance α. If the​ P-value is less than α​, reject the null hypothesis.​ Otherwise, do not reject the null hypothesis. Use this information to determine a proper conclusion. (e) Because np01−p0=10.4 > ​10, the sample size is less than 5% of the population​ size, and the sample can be reasonably assumed to be random, the normal model may be used to approximate the​ P-value. Explanation: (42)(0.54)(1-0.54) = 10.4328 (f) 1. z0 = 2.27 Explanation: p̂ = 30/42 = 0.7143 Use test statistic formula in image. 0.7143-0.54 = 0.1743 √0.54(1-0.54)/42 = 0.0769044 0.1743/0.0769044 = 2.26645 2. P-value = 0.012 Explanation: Use P-Value Calculator ("use right-tailed p-value" and Z-score of 2.26645) https://www.omnicalculator.com/statistics/p-value = 0.012 3. Yes​, reject the null hypothesis because the​ P-value is less than α. There is sufficient evidence to conclude that the experimental course is effective. (g) When there are small sample​ sizes, the evidence against the statement in the null hypothesis must be substantial. One should be wary of studies that do not reject the null hypothesis when the test was conducted with a small sample size.

Conduct the following test at the α=0.10 level of significance by determining ​(a) the null and alternative​ hypotheses, ​(b) the test​ statistic, and​ (c) the​ P-value. Assume that the samples were obtained independently using simple random sampling. Test whether p1≠p2. Sample data are x1=28​, n1=254​, x2=38​, and n2=301.

(a) H0: p1 = p2 versus H1: p1 ≠ p2 For (b) and (c)... Open StatCrunch → Stat → Proportion Stats → Two sample → With Summary → input # of Successes (28, 38) and # of trials for each sample (254, 301) → Perform: select Hypothesis test for p1 - p2 → H0: p1 - p2 = 0 (p1 = p2, subtract p2 to both sides of the equation, we will obtain p1 - p2 = 0), HA: p1 - p2 ≠ 0 → Compute! → Z-Stat and P-value are the answers (b) -.58 (c) .562 Do not reject the null hypothesis because there is not sufficient evidence to conclude that p1≠p2.

Conduct a test at the α=0.10 level of significance by determining ​(a) the null and alternative​ hypotheses, ​(b) the test​ statistic, and​ (c) the​ P-value. Assume the samples were obtained independently from a large population using simple random sampling. Test whether p1>p2. The sample data are x1=121​, n1=241​, x2=144​, and n2=307.

(a) H0: p1=p2 versus H1: p1>p2 (b) 0.76 Explanation: 121 + 144 = 265, 241 + 307 = 548 p̂ = 265/548 = 0.4836 p̂1 = 121/241 = 0.5020 p̂2 = 144/307 = 0.4691 0.5020 - 0.4691 = 0.0329 √0.4836(1-0.4836) = 0.499731 √1/241 + 1/307 = 0.0860622 0.499731 * 0.0860622 = 0.043008 0.0329/0.043008 = 0.76497 (c) 0.222 Explanation: https://www.socscistatistics.com/pvalues/normaldistribution.aspx (use 0.76497) (c1) Do not reject the null hypothesis because there is not sufficient evidence to conclude that p1 > p2.

A can of soda is labeled as containing 16 fluid ounces. The quality control manager wants to verify that the filling machine is neither over-filling nor under-filling the cans. Complete parts​ (a) through​ (d) below.

(a) H0: μ = 16 H1: μ ≠ 16 (b) There is sufficient evidence to conclude that the machine is out of calibration. (c) A Type I error has been made since the sample evidence led the​ quality-control manager to reject the null​ hypothesis, when the null hypothesis is true. (d) The level of significance should be 0.01 because this makes the probability of Type I error small.

According to a food​ website, the mean consumption of popcorn annually by Americans is 59 quarts. The marketing division of the food website unleashes an aggressive campaign designed to get Americans to consume even more popcorn. Complete parts​ (a) through​ (c) below.

(a) H0: μ = 59 H1: μ > 59 (b) There is sufficient evidence to conclude that the mean consumption of popcorn has risen. (c) The marketing department committed a Type I error because the marketing department rejected the null hypothesis when it was true. The probability of making a Type I error is 0.1.

Assume that both populations are normally distributed. ​(a) Test whether μ1≠μ2 at the α=0.01 level of significance for the given sample data. (b) Construct a 99​% confidence interval about μ1−μ2.

(a) H0:μ1=μ2 H1:μ1≠μ2 (a1) P-value = 0.042 Explanation: Open StatCrunch → Stat → T Stats → Two samples with summary → Input means (x), standard deviation (s), and sample size (n) → Compute! (a2) Do not reject H0​, there is not sufficient evidence to conclude that the two populations have different means. Explanation: .042>.01 = do not reject since p-value is greater than α. (b) -.68 and 5.28 Explanation: Edit StatCrunch Two Table T Summary from (a1) → Click on Confidence Interval, change to .99 → Compute!

A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the α=0.05 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers. Part 2 State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms.

(a) H0​: μ1=μ2 H1​: μ1>μ2 (a1) P-value = .139 Explanation: Open StatCrunch through graph → Stat → Summary Stats → Columns → Get the mean and standard deviation of carpeted and uncarpeted → Then go to: Stat → T Stats → Two samples with summary → Input means (x), standard deviation (s), and sample size (n/8) → Change hypothesis symbols to match (a) → Compute! (a2) Do not reject H0. There is not significant evidence at the α=0.05 level of significance to conclude that carpeted rooms have more bacteria than uncarpeted rooms. Explanation: 0.139 > .05 = do not reject since p-value is greater than α.

Use the given statistics to complete parts​ (a) and​ (b). Assume that the populations are normally distributed. ​(a) Test whether μ1>μ2 at the α=0.01 level of significance for the given sample data. ​(b) Construct a 99​% confidence interval about μ1−μ2.

(a) H0​: μ1=μ2 H1​: μ1>μ2 (a1) Test statistic = 1.67 P-value = 0.054 Explanation: Open StatCrunch → Stat → T Stats → Two samples with summary → Input means (x), standard deviation (s), and sample size (n) → Change hypothesis symbols to match (a) → Compute! →T-Stat and P-value are the answers (a2) Do not reject H0. There is not sufficient evidence at the α=0.01 level of significance to conclude that μ1>μ2. Explanation: .054>.01 = do not reject since p-value is greater than α. (b) -3.303 and 13.103 Explanation: Edit StatCrunch Two Table T Summary from (a1) → Click on Confidence Interval, change to .99 → Compute!

Several years​ ago, the mean height of women 20 years of age or older was 63.7 inches. Suppose that a random sample of 45 women who are 20 years of age or older today results in a mean height of 63.9 inches. ​(a) State the appropriate null and alternative hypotheses to assess whether women are taller today. ​(b) Suppose the​ P-value for this test is 0.05. Explain what this value represents. ​(c) Write a conclusion for this hypothesis test assuming an α=0.10 level of significance.

(a) H0​: μ=63.7 in. versus H1​: μ>63.7 in. (b) There is a 0.05 probability of obtaining a sample mean height of 63.9 inches or taller from a population whose mean height is 63.7 inches. (c) Reject the null hypothesis. There is sufficient evidence to conclude that the mean height of women 20 years of age or older is greater today.

To test H0: μ=103 versus H1: μ≠103 a simple random sample of size n=35 is obtained. ​(a) Does the population have to be normally distributed to test this​ hypothesis? Why? ​(b) If x=100.0 and s=5.7​, compute the test statistic. ​(e) If the researcher decides to test this hypothesis at the α=0.01 level of​ significance, will the researcher reject the null​ hypothesis?

(a) No, because n≥30. (b) -3.11 Explanation: The formula for the test statistic (t0) is in the photo provided. 5.7/√35 = 0.964 Now, calculate the test statistic... t0 ≈100.0−103/0.964 ≈ −30.964 ≈ −3/0.964 = −3.113 (c) Graph with small bit of rid on both ends. (d) 1. 0.002 < P-value < 0.005 2. If 1000 random samples of size n=35 are​ obtained, about 4 samples are expected to result in a mean as extreme or more extreme than the one observed if μ=103. (e) Yes

Determine​ (a) the χ2 test​ statistic, (b) the degrees of​ freedom, (c) the critical value using α=0.05​, and​ (d) test the hypothesis at the α=0.05 level of significance.

(a) Test statistic = 0.16 Explanation: (26-25)^2/25 = .04 (26-25)^2/25 = .04 (24-25)^2/25 = .04 (24-25)^2/25 = .04 Add up outcome = 0.16 (b) There are 3 degrees of freedom. Explanation: Number of categories = 4 Degrees of freedom = 4 - 1 = 3 (c) The critical value is 7.815. Explanation: Since the degrees of freedom is 3, you can use a chi-square table or calculator to find the critical value for the given α = 0.05 and df = 3. The critical value is approximately 7.815. (d) No because χ20 is less than or equal to χ20.05. Since 0.16 < 7.815, we do not reject the null hypothesis.

The table to the right contains observed values and expected values in parentheses for two categorical​ variables, X and​ Y, where variable X has three categories and variable Y has two categories. Use the table to complete parts ​(a) and ​(b) below.

(a) Test statistic = 1.759 Explanation: Open StatCrunch with graph → Stat → Tables → Contingency → With Summary → Select all columns → Select any row label → Compute! → The Value is the test statistic (also find P-value there) (a1) ​H0: The Y category and X category are independent. H1: The Y category and X category are dependent. (a2) P-value = .415 (a3) No​, do not reject H0. There is not sufficient evidence at the α = .01 level of significance to conclude that X and Y are dependent because the ​P-value > α. Explanation: .415 > .01 P-value is greater than the level of​ significance, do not reject the null hypothesis.

A researcher with the Department of Education followed a cohort of students who graduated from high school in a certain​ year, monitoring the progress the students made toward completing a​ bachelor's degree. One aspect of his research was to determine whether students who first attended community college took longer to attain a​ bachelor's degree than those who immediately attended and remained at a​ 4-year institution. The data in the table attached below summarize the results of his study. Complete parts​ a) through​ e) below.

(a) The response variable is the time to graduate. The explanatory variable is the use of community college or not. (b) B. The sample sizes are not more than​ 5% of the population. C. The sample sizes are large​ (both greater than or equal to​ 30). D. The samples are independent. E. The samples can be reasonably assumed to be random. (c) H0​: μcommunity college=μno transfer​, H1​: μcommunity college>μno transfer (d) Test statistic = 13.53 P-value = 0 Explanation: Open StatCrunch → Stat → T Stats → Two samples with summary → Input means (x), standard deviation (s), and sample size (n) → Change hypothesis symbols to match (c) → Compute! →T-Stat and P-value are the answers (d1) reject, does Explanation: 0 <.05 = reject since p-value is less than α. (d) .897 and 1.203 Explanation: Edit StatCrunch Two Table T Summary → Click on Confidence Interval, change to .95 → Compute! (e) No

Determine whether the following sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitative. A researcher wishes to compare annual salaries of married lawyers and their spouses. She obtains a random sample of 880 such couples who work and determines each spouse's annual salary.

(a) The sampling is dependent because an individual selected for one sample does dictate which individual is to be in the second sample. (b) The variable is quantitative because it is a numerical measure.

The following data represent the level of health and the level of education for a random sample of residents. Complete parts​ (a) and​ (b) below.

(a) ​H0: Level of education and health are independent. H1: Level of education and health are dependent. (a1) Test statistic = 23.217 P-value = .006 Explanation: Open StatCrunch with graph → Stat → Tables → Contingency → With Summary → Select all columns expect Education because it has letters → Select any row label → Compute! → The Value is the test statistic (also find P-value there) (a2) Reject H0. There is sufficient evidence that level of education and health are associated. Explanation: .006 < .05 P-value is less than the level of​ significance, reject the null hypothesis. (b) Shown below Explanation: Add the numbers in each row... Not a H.S. graduate: 450 H.S. graduate: 468 Some college: 380 Bachelor Degree or higher: 347 Then divide each by the total (example: 96/450 = 0.213)

If the consequences of making a Type I error are​ severe, would you choose the level of​ significance, α​, to equal​ 0.01, 0.05, or​ 0.10?

0.01

Explain what a​ P-value is. What is the criterion for rejecting the null hypothesis using the​ P-value approach?

1. A​ P-value is the probability of observing a sample statistic as extreme or more extreme than the one observed under the assumption that the statement in the null hypothesis is true. 2. If P-value < α, reject the null hypothesis.

Suppose a researcher is testing the hypothesis H0​: p=0.4 versus H1​: p≠0.4 and she finds the​ P-value to be 0.26. Explain what this means. Would she reject the null​ hypothesis? Why?

1. If the​ P-value for a particular test statistic is 0.26​, she expects results at least as extreme as the test statistic in about 26 of 100 samples if the null hypothesis is true. 2. Since this event is not​ unusual, she will not reject the null hypothesis.

Test the hypothesis using the​ P-value approach. Be sure to verify the requirements of the test. H0: p=0.88 versus H1: p≠0.88 n=500, x=430, α=0. Is np01−p0≥10​?

1. Yes, 52.8 Explanation: 500(0.88)(1 - 0.88) 52.8≥10 2. p̂ = 0.86 Explanation: https://www.omnicalculator.com/statistics/p-hat 3. z0 = -1.38 Explanation: Use test statistic formula in image. .86-.88 = -.02 √.88(1-.88)/500 = 0.0145327 -.02/0.0145327 = 1.37621 4. P-value = 0.168 Explanation: The null hypothesis (H0) is "p = 0.88," and the alternative hypothesis (H1) is "p ≠ 0.88." The "≠" symbol in the alternative hypothesis indicates a two-tailed test. Use P-Value Calculator ("use two-tailed p-value" and Z-score of -1.37621) https://www.omnicalculator.com/statistics/p-value = 0.168 5. Do not reject the null hypothesis, because the ​P-value is greater than α. Explanation: Compare the​ P-value with the level of significance α. If the​ P-value is less than α​, reject the null hypothesis.​ Otherwise, do not reject the null hypothesis. Use this information to determine a proper conclusion.

Compute the critical value zα/2 that corresponds to a 85​% level of confidence.

1.44 Explanation: .15/2 = .075 → 1 - 0.075 = 0.925 The z-score for a cumulative probability of 0.925 is approximately 1.44. Level of confidence: (1 - α) * 100%

Construct a confidence interval for p1−p2 at the given level of confidence. x1=379​, n1=534​, x2=437​, n2=591​, 99​% confidence

99%, -.098 and .039 Explanation: Open StatCrunch → Stat → Proportion Stats → Two Sample →With Summary → Input # of successes (379, 437) and # of trials (534, 591) for each sample → Perform: select Confidence interval for p1 - p2 → Level: 0.99 → Compute! → L. Limit and U. Limit are the answers

Explain the difference between an independent and dependent sample.

A sample is independent when an individual selected for one sample does not dictate which individual is to be in the second sample. A sample is dependent when an individual selected for one sample dictates which individual is to be in the second sample.

In a survey conducted by the Gallup​ Organization, 1100 adult Americans were asked how many hours they worked in the previous week. Based on the​ results, a​ 95% confidence interval for the mean number of hours worked had a lower bound of 42.7 and an upper bound of 44.5. Provide two recommendations for decreasing the margin of error of the interval.

A. Decrease the confidence level. C. Increase the sample size.

The headline reporting the results of a poll conducted by the Gallup organization​ stated, "Majority of Americans at Personal Best in the​ Morning." The results indicated that a survey of 1100 adults resulted in 55​% stating they were at their personal best in the morning. The​ poll's results were reported with a margin of error of 3​%. Explain why the Gallup​ organization's headline is accurate.

All the values within the margin of error are greater than​ 50%.

For the study given​ below, explain which statistical procedure would most likely be used for the research objective. Assume all model requirements for conducing the appropriate procedure have been satisfied. Does hotel chain A charge more than hotel chain B for a​ one-night stay?

A​ matched-pairs t-test on the difference of means is most likely appropriate because the mean price is a good measure of how much a hotel chain​ charges, the research objective involves a comparison of two​ things, and one would likely select hotels paired by location.

A national survey of 1000 adult citizens of a nation found that 24​% dreaded​ Valentine's Day. The margin of error for the survey was 1.9 percentage points with 85​% confidence. Explain what this means.

B. There is 85% confidence that the proportion of the adult citizens of the nation that dreaded Valentine's Day is between 0.221 and 0.259. D. There is 85​% confidence that the proportion of the adult citizens of the nation that did not dread​ Valentine's Day is between 0.741 and 0.779.

State the requirements to perform a​ goodness-of-fit test.

C. all expected frequencies≥1 F. at least​ 80% of expected frequencies ≥5

Sample evidence can prove that a null hypothesis is true.

False

If we do not reject the null hypothesis when the statement in the alternative hypothesis is​ true, we have made a Type​ _______ error.

II

Explain what ​"95​% ​confidence" means in a 95​% confidence interval.

If 100 different confidence intervals are constructed, each based on a different sample of size n from the same population, then we expect 95 of the intervals to include the parameter and 5 to not include the parameter.

A survey​ asked, "How many tattoos do you currently have on your​ body?" Of the 1206 males​ surveyed, 191 responded that they had at least one tattoo. Of the 1066 females​ surveyed, 130 responded that they had at least one tattoo. Construct a 95​% confidence interval to judge whether the proportion of males that have at least one tattoo differs significantly from the proportion of females that have at least one tattoo. Interpret the interval.

Lower bound ≈ 0.006509 Upper bound ≈ 0.066691 There is 95​% confidence that the difference of the proportions is in the interval. Conclude that there is a significant difference in the proportion of males and females that have at least one tattoo.

Construct a 99​% confidence interval of the population proportion using the given information. x=240, n=300

Lower bound: .741 Upper bound: .859 Explanation: 240/300 = 0.8 (2.575 is the critical value of 99% in next problem) Lower bound: .8 - 2.575 * √(0.8 * (1 - .8) / 300) = .741 Upper bound: .8 + 2.575 * √(0.8 * (1 - .8) / 300) = .859

Two​ researchers, Jaime and​ Mariya, are each constructing confidence intervals for the proportion of a population who is​ left-handed. They find the point estimate is 0.29. Each independently constructed a confidence interval based on the point​ estimate, but​ Jaime's interval has a lower bound of 0.219 and an upper bound of 0.361​, while​ Mariya's interval has a lower bound of 0.265 and an upper bound of 0.317. Which interval is​ wrong? Why?

Mariya's interval is wrong because it is not centered on the point estimate.

A trade magazine routinely checks the​ drive-through service times of​ fast-food restaurants. A 90​% confidence interval that results from examining 753 customers in one​ fast-food chain's​ drive-through has a lower bound of 175.7 seconds and an upper bound of 179.3 seconds. What does this​ mean?

One can be 90% confident that the mean drive-through service time of this fast-food chain is between 175.7 seconds and 179.3 seconds.

Explain what​ "statistical significance" means.

Statistical significance means that the result observed in a sample is unusual when the null hypothesis is assumed to be true.

Explain the difference between statistical significance and practical significance.

Statistical significance means that the sample statistic is not likely to come from the population whose parameter is stated in the null hypothesis. Practical significance refers to whether the difference between the sample statistic and the parameter stated in the null hypothesis is large enough to be considered important in an application.

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.53 ​hours, with a standard deviation of 2.29 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.09 ​hours, with a standard deviation of 1.69 hours. Construct and interpret a 95​% confidence interval for the mean difference in leisure time between adults with no children and adults with children μ1−μ2. Let μ1 represent the mean leisure hours of adults with no children under the age of 18 and μ2 represent the mean leisure hours of adults with children under the age of 18.

The 95​% confidence interval for μ1−μ2 is the range from 0.54 hours to 2.34 hours. There is 95​% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

Assuming all model requirements for conducting the appropriate procedure have been​ satisfied, what proportion of registered voters is in favor of a tax increase to reduce the federal​ debt? Explain which statistical procedure would most likely be used for the research objective given.

The correct procedure is a confidence interval for a single proportion. The goal is to determine the proportion of the population that favors a tax increase. There is no comparison being made and there is only one​ population, so rather than hypothesis​ testing, it is appropriate to use a confidence interval.

Determine the expected count for each outcome.

The expected count for outcome 1 is 114.27. The expected count for outcome 2 is 395.55. The expected count for outcome 3 is 167.01. The expected count for outcome 4 is 202.17. Explanation: Multiply n by pᵢ.

Suppose there are n independent trials of an experiment with k>3 mutually exclusive​ outcomes, where p Subscript i represents the probability of observing the ith outcome. What would be the formula of an expected count in this​ situation?

The expected counts for each possible outcome are given by Eᵢ = npᵢ.

A group conducted a poll of 2053 likely voters just prior to an election. The results of the survey indicated that candidate A would receive 49​% of the popular vote and candidate B would receive 48​% of the popular vote. The margin of error was reported to be 2​%. The group reported that the race was too close to call. Use the concept of a confidence interval to explain what this means.

The margin of error suggests candidate A may receive between 47% and 51% of the popular vote, and candidate B may receive between 46% and 50% of the popular vote. Because the poll estimates overlap when accounting for the margin of error, the poll cannot predict the winner.

For the following​ study, explain which statistical procedure​ (estimating a single​ proportion; estimating a single​ mean; hypothesis test for a single​ proportion; hypothesis test for a single​ mean; hypothesis test or estimation of two​ proportions, hypothesis test or estimation of two​ means, dependent or​ independent) would most likely be used for the research objective given. Assume all model requirements for conducting the appropriate procedure have been satisfied. Is the mean IQ of the students in Professor​ Dang's statistics class higher than that of the general​ population, 100?

The most appropriate procedure is a hypothesis test for a single mean. The comparison is between the mean IQ of the class and the national average IQ. The class is a sample of the​ population, so it is not a comparison between two population means. The objective is to find whether the sample mean is higher than the population​ mean, so a hypothesis test is more appropriate.

Determine the point estimate of the population mean and margin of error for the confidence interval. Lower bound is 20​, upper bound is 30.

The point estimate of the population mean is 25. The margin of error for the confidence interval is 5. Explanation: 20+30 = 50/2 = 25 25+E (margin of error) = 30 E = 5

What happens to the probability of making a Type II​ error, β​, as the level of​ significance, α​, ​decreases? Why?

The probability increases. Type I and Type II errors are inversely related.

If a hypothesis is tested at the α=0.05 level of​ significance, what is the probability of making a type I​ error?

The probability of making a type I error is 0.05.

Explain why​ chi-square goodness-of-fit tests are always right tailed.

The​ chi-square goodness-of-fit tests are always right tailed because the numerator in the test statistic is​ squared, making every test​ statistic, other than a perfect​ fit, positive.

Explain why the​ t-distribution has less spread as the number of degrees of freedom increases.

The​ t-distribution has less spread as the degrees of freedom increase​ because, as n​ increases, s becomes closer to σ by the law of large numbers.

Determine if the following statement is true or false.​ Why? The expected frequencies in a​ chi-square test for independence are found using the formula below.

True. It is a simplification of multiplying the proportion of a row variable by the proportion of the column variable to find the proportion for a​ cell, then multiplying by the table total.

If the expected count of a category is less than​ 1, what can be done to the categories so that a​ goodness-of-fit test can still be​ performed?

Two of the categories can be​ combined, or the sample size can be increased. D.

The null and alternative hypotheses are given. Determine whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. What parameter is being​ tested? H0​: p = 0.3 H1​: p ≠ 0.3 What type of test is being conducted in this​ problem?

Two-tailed test Population proportion

Complete parts ​(a) through ​(c) below. ​(a) Determine the critical​ value(s) for a​ right-tailed test of a population mean at the α=0.10 level of significance with 20 degrees of freedom. ​(b) Determine the critical​ value(s) for a​ left-tailed test of a population mean at the α=0.01 level of significance based on a sample size of n=10. ​(c) Determine the critical​ value(s) for a​ two-tailed test of a population mean at the α=0.01 level of significance based on a sample size of n=14.

Use: https://www.omnicalculator.com/statistics/critical-value (a) + 1.325 Explanation: Find .10 at 20 degrees. (b) -2.821 Explanation: n-1 = 10-1= 9 degrees at .01. (c) ±3.012 Explanation: n-1 = 14-1= 13 degrees at .01.

A simple random sample of size n is drawn. The sample​ mean, x​, is found to be 18.9​, and the sample standard​ deviation, s, is found to be 4.1. ​(a) Construct a 95​% confidence interval about μ if the sample​ size, n, is 35. ​(b) Construct a 95​% confidence interval about μ if the sample​ size, n, is 51. ​(c) Construct a 99​% confidence interval about μ if the sample​ size, n, is 35. ​(d) If the sample size is 12​, what conditions must be satisfied to compute the confidence​ interval?

a) 17.49, 20.31 Explanation: 35-1=34. (100% - 95%) / 2 = 0.025 Critical value for T calculator: https://www.socscistatistics.com/tests/criticalvalues/default.aspx = 2.032 E ≈ 2.032 * (4.1 / √35) ≈ 1.40823 Lower bound: 18.9 - 1.40823 = 17.49177 Upper bound: 18.9 + 1.40823 = 20.30823 b) 17.75, 20.05 Explanation: 51-1=50 Critical value for T calculator: https://www.socscistatistics.com/tests/criticalvalues/default.aspx = 2.009 E ≈ 2.009 * (4.1 / √51) ≈ 1.1534 Lower bound: 18.9 - 1.1534 = 17.7466 Upper bound: 18.9 + 1.1534 = 20.0534 b2) The margin of error decreases. c) 17.01, 20.79 Explanation: 35-1=34 (100% - 99%) / 2 = 0.005 Critical value for T calculator (0.005) https://www.omnicalculator.com/statistics/critical-value = 2.728 E ≈ 2.728 * (4.1 / √35) ≈ 1.89058 Lower bound: 18.9 - 1.89058 = 17.00942 Upper bound: 18.9 + 1.89058 = 20.79058 c2) The margin of error increases. d) The sample data must come from a population that is normally distributed with no outliers.

A simple random sample of size n is drawn. The sample​ mean, x​, is found to be 19.2​, and the sample standard​ deviation, s, is found to be 4.7. ​(a) Construct a 95​% confidence interval about μ if the sample​ size, n, is 34. ​(b) Construct a 95​% confidence interval about μ if the sample​ size, n, is 61. ​(c) Construct a 99​% confidence interval about μ if the sample​ size, n, is 34. (d) If the sample size is 15​, what conditions must be satisfied to compute the confidence​ interval?

a) 17.56, 20.84 Explanation: Step 1: Determine the degrees of freedom (df) = n - 1 = 34 - 1 = 33. Step 2: Find the critical t-value for a 95% confidence level and 33 degrees of freedom. You can refer to the t-distribution table or use a statistical calculator to find the value. Let's say the critical t-value is approximately 2.035. Step 3: Calculate the margin of error (E) using the formula: E = tα/2 * (s / √n) E ≈ 2.035 * (4.7 / √34) ≈ 1.782 Step 4: Construct the confidence interval: Lower bound = x - E Lower bound ≈ 19.2 - 1.782 ≈ 17.56 Upper bound ≈ 19.2 + 1.782 ≈ 20.84 b) 18.02, 20.38 b2) The margin of error decreases. c) 17.00, 21.40 c2) The margin of error increases. d) The sample data must come from a population that is normally distributed with no outliers.

Determine the​ t-value in each of the cases. ​(a) Find the​ t-value such that the area in the right tail is 0.01 with 17 degrees of freedom. ​(b) Find the​ t-value such that the area in the right tail is 0.02 with 17 degrees of freedom. (c) Find the​ t-value such that the area left of the​ t-value is 0.20 with 18 degrees of freedom.​ [Hint: Use​ symmetry.] ​(d) Find the critical​ t-value that corresponds to 80​% confidence. Assume 11 degrees of freedom.

a) 2.567 Explanation: View table, find row 17, then locate column 0.01. b) 2.224 c) -0.862 Explanation: Same as explanation in (a), but make it negative. d) 1.363 Explanation: Determine the area in each tail: (100% - 80%) / 2 = 10% ... Therefore the area to the right tail is 0.10, find row 11.

A television sports commentator wants to estimate the proportion of citizens who​ "follow professional​ football." Complete parts​ (a) through​ (c). ​(a) What sample size should be obtained if he wants to be within 4 percentage points with 94​% confidence if he uses an estimate of 48​% obtained from a​ poll? ​(b) What sample size should be obtained if he wants to be within 4 percentage points with 94​% confidence if he does not use any prior​ estimates? ​(c) Why are the results from parts​ (a) and​ (b) so​ close?

a) 551 Explanation: Identify the confidence level: 94% Determine the area in each tail: (100% - 94%) / 2 = 3% Use a standard normal distribution table or https://statscalculator.com/zcriticalvaluecalculator?x1=0.03 to find the z-score corresponding to the area of 3% in the tails. The critical value is approximately 1.88 0.04 is the 4 percentage points 0.48(1-0.48)(1.88/0.04)^2 = 552 [you round up always] b) 553 n = 0.25(1.88/0.04)^2 = 553 [idk where 0.25 came from] c) The results are close because 0.48(1−0.48)=0.2496 is very close to 0.25.

People were polled on how many books they read the previous year. Initial survey results indicate that s=19.5 books. Complete parts ​(a) through ​(d) below. ​(a) How many subjects are needed to estimate the mean number of books read the previous year within four books with 90​% ​confidence? ​(b) How many subjects are needed to estimate the mean number of books read the previous year within two books with 90​% ​confidence? ​(c) What effect does doubling the required accuracy have on the sample​ size? ​(d) How many subjects are needed to estimate the mean number of books read the previous year within four books with 99​% ​confidence? (d2) Compare this result to part ​(a). How does increasing the level of confidence in the estimate affect sample​ size? Why is this​ reasonable?

a) 65 Explanation: Critical value in table is 1.645 for 95%. When multiplying in next problems, make sure to have the 2 first numbers that are multiplying as the numerator, then the number of books as the denominator. (1.645 * 19.5/4)^2 = 64.3 [always round up, even if not meant to] b) 258 Explanation: (1.645 * 19.5/2)^2 = 257.24 [always round up] c) Doubling the required accuracy nearly quadruples the sample size. d) Explanation: 158 Critical value in table is 2.575 for 99%. (2.575 * 19.5/4)^2 = 157.58 [round up] d2) Increasing the level of confidence increases the sample size required. For a fixed margin of​ error, greater confidence can be achieved with a larger sample size.

In a survey conducted by a reputable marketing​ agency, 267 of 1000 adults 19 years of age or older confessed to bringing and using their cell phone every trip to the bathroom​ (confessions included texting and answering phone​ calls). Complete parts​ (a) through​ (f) below. ​(a) What is the sample in this​ study? What is the population of​ interest? ​(b) What is the variable of interest in this​ study? Is it qualitative or​ quantitative? ​(c) Based on the results of this​ survey, obtain a point estimate for the proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. ​(d) Explain why the point estimate found in part​ (c) is a statistic. Explain why it is a random variable. What is the source of variability in the random​ variable? ​(e) Construct and interpret a 95​% confidence interval for the population proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. Select the correct choice below and fill in any answer boxes within your choice. (f) What ensures that the results of this study are representative of all adults 19 years of age or​ older?

a) The sample is the 1000 adults 19 years of age or older. The population is all adults 19 years of age or older. b) The variable of interest is bringing one's cell phone every trip to the bathroom. This variable is qualitative with two outcomes because individuals are classified based on a characteristic. c) p̂ = 0.267 267/1000 = 0.267 d) 1. Its value is based on a sample. 2. Its value may change depending on the individuals in the survey. 3. The individuals selected to be in the study. e) We are 95​% confident the proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom is between 0.240 and 0.294. Use calculator: https://statscalculator.com/zcriticalvaluecalculator?x1=0.025 (right tailed) = 1.96 Lower bound: 0.267 - 1.96 * √(0.267 * (1 - 0.267) / 1000) = 0.240 Upper bound: 0.267 + 1.96 * √(0.267 * (1 - 0.267) / 1000) = 0.294 f) Random sampling

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 with 99​% confidence if ​(a) she uses a previous estimate of 0.32​? ​(b) she does not use any prior​ estimates?

a) n = 1603 0.32(1-0.32)(2.575/0.03)^2 = 1603 2.575 is the critical value associated with 99%. b) n = 1842 0.25(2.575/0.03)^2 = 1842 [idk why had to do 0.25]

A survey of 2284 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 412 have donated blood in the past two years. Complete parts​ (a) through​ (c) below. ​(a) Obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. ​(b) Verify that the requirements for constructing a confidence interval about p are satisfied. ​(c) Construct and interpret a 90​% confidence interval for the population proportion of adults in the country who have donated blood in the past two years. Select the correct choice below and fill in any answer boxes within your choice.

a) p̂ = .180 Explanation: x = 412 n = 2284 412/2284 = 0.180 b) can be assumed to be, np̂ (1-p̂), 337.118, greater than or equal to, sample size, can be assumed to be, population size Explanation: np̂ (1-p̂) → 2284*.180(1-.180) = 337.1184 c) We are 90​% confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between .167 and .193. Explanation: 1-.9 = .1 → .1/2 = .05 Use calculator: https://statscalculator.com/zcriticalvaluecalculator?x1=0.05 (right tailed) to find 1.65 Lower bound: 0.180 - 1.65 * √(0.180 * (1 - 0.180) / 2284) = 0.167 Upper bound: 0.180 + 1.65 * √(0.180 * (1 - 0.180) / 2284) = 0.193

A sampling method is ___ when an individual selected for one sample does not dictate which individual is to be in the second sample.

independent

The​ ___ represents the expected proportion of intervals that will contain the parameter if a large number of different samples of size n is obtained. It is denoted​ ___.

level of confidence, (1 - α) * 100%

A ___ ___ is the value of a statistic that estimates the value of a parameter.

point estimate

The head of institutional research at a university believed that the mean age of​ full-time students was declining. In​ 1995, the mean age of a​ full-time student was known to be 27.4 years. After looking at the enrollment records of all 4934​ full-time students in the current​ semester, he found that the mean age was 27.1​ years, with a standard deviation of 7.3 years. He conducted a hypothesis of H0​: μ=27.4 years versus H1​: μ<27.4 years and obtained a​ P-value of 0.0020. He concluded that the mean age of​ full-time students did decline. Is there anything wrong with his​ research?

​Yes, the head of institutional research has access to the entire​ population, inference is unnecessary. He can say with​ 100% confidence that the mean age has decreased.