Alta - Chapter 9 - Hypothesis Testing

Suppose the null hypothesis, H0, is: the mean age of the horses on a ranch is 6 years. What is the Type I error in this scenario?

Correct answer: You cannot conclude that the mean age of the horses on a ranch is 6 years when, in fact, it is. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is rejecting the null hypothesis that the mean is 6 years when, in fact, it is.

Identify the type of hypothesis test below.H0:X=10.2, Ha:X>10.2

The hypothesis test is right-tailed. Remember the forms of the hypothesis tests. Right-tailed: H0:X=X0, Ha:X>X0. Left-tailed: H0:X=X0, Ha:X<X0. Two-tailed: H0:X=X0, Ha:X≠X0. So, this is a right-tailed test.

Determine the Type II error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250 pounds and less.

Correct answer: You do not reject the null hypothesis that the ladder can withstand weight of 250 pounds and less when, in fact, it cannot. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is stating that there is insufficient evidence to conclude that the ladder can withstand weight of 250 pounds, in fact, it cannot

Determine the critical value or values for a one-mean z-test at the 6% significance level if the hypothesis test is left-tailed (Ha:μ<μ0).

−1.555 For a left-tailed hypothesis test, the rejection region is on the left. Because the test is left-tailed, the critical value is negative. We know that α=0.06, so the area to right of the z-score is equal to 0.06, and we will use z0.06=−1.555 as our critical value.

What is the p-value of a right-tailed one-mean hypothesis test, with a test statistic of z0=1.82? (Do not round your answer; compute your answer using a value from the table below.)

$0.034$0.034​ The p-value is the probability of an observed value of z=1.82 or greater if the null hypothesis is true, because this hypothesis test is right-tailed. This probability is equal to the area under the Standard Normal curve to the right of z=1.82. Using the Standard Normal Table, we can see that the p-value is equal to 0.966, which is the area to the left of z=1.82. (Standard Normal Tables give areas to the left.) So, the p-value we're looking for is p=1−0.966=0.034.

Which of the following results in a null hypothesis μ=7 and alternative hypothesis μ>7?

A study wants to show that the mean number of hours of sleep the average person gets each day is more than 7. Consider each of the options. The scenario in the third option has the alternative hypothesis μ>7 based on the words "more than" and the fact that the null hypothesis is always stated with the equality symbol. Also, remember that the alternative hypothesis is always the statement that is trying to be shown by the study.

Identify the type of hypothesis test below.H0:X=16.9, Ha:X>16.9

Correct answer: The hypothesis test is right-tailed. Remember the forms of the hypothesis tests. Right-tailed: H0:X=X0, Ha:X>X0. Left-tailed: H0:X=X0, Ha:X<X0. Two-tailed: H0:X=X0, Ha:X≠X0. So, this is a right-tailed test.

Determine the Type II error if the null hypothesis, H0, is: the mean price of a loaf of bread is $1.67. And the alternative hypothesis, Ha, states the claim, which is the mean price of a loaf of bread is not $1.67.

Correct answer: You do not conclude that the mean price of a loaf of bread is not $1.67 when, in fact, the mean price of a loaf of bread is not $1.67. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is that you cannot conclude that the mean price of the bread is $1.67 when, in fact, it is not.

Based on the outcome of the company's study regarding their market share, and the following results, conclude whether to reject or not reject H0. H0: μ=16; Ha: μ<16 x¯=15.8 σ=1.8 α=0.01 (significance level) The test statistic isz0=x¯−μ0σn√=15.8−161.821√=−0.51 The critical value is −z0.01=−2.33. Select two responses below.

Do not reject H0. At the 1% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis. The test statistic is NOT in the rejection region. Conclusion: This is a left-tailed test, because Ha would be if the average market share is less than 16 percent. For a left-tailed test, if the test statistic (z0) is less than the critical value (zα), we should reject the null hypothesis.We know that z0=−0.51 and zα=−2.33. Because z0>zα, we do not reject the null hypothesis, which stated that the average market share is greater than or equal to 16 percent.Interpretation: At the 1% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis.

Based on your answers regarding Gavin's study regarding cost of urgent care visits compared to physician office visits, choose the correct conclusion that interprets the results within the context of the hypothesis test.

There is NOT sufficient evidence at the 1% significance level to conclude that the average cost of a visit to an urgent care clinic is more than that of a physician's office of $55. This is a right-tailed test, and test statistic is less than the critical value on the right side, so we fail to reject the null hypothesis. Conclusion: This is a right-tailed test, because Ha states that the average cost of a visit to an urgent care clinic is more than that of a physician's office of $55. For a right-tailed test, if the test statistic (z0) is more than the critical value (zα), we should reject the null hypothesis.We know that z0=1.61 and zα=2.326. Because z0<zα, we fail to reject the null hypothesis, which stated that the mean cost of a visit to an urgent care clinic is equal to that of a physician's office of $55.

Determine the Type I error if the null hypothesis, H0, is: the mean price of a loaf of bread is $1.67.

You state that there is sufficient evidence to conclude that the mean price of bread is not $1.67 , when, in fact, it is. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is that you conclude that the mean price of bread is not $1.67 , when, in fact, it is.

A psychologist is interested in studying the sleeping patterns of individuals diagnosed with depression. She claims that the proportion of depressed patients who sleep 8 hours per night is less than 75%. If the psychologist chooses a 0.5% significance level, what is/are the critical value(s) for the hypothesis test?

move blue down to the 2nd circle. Move purple to the first circle. critical value is correct btw To determine the critical value or values for a one-proportion z-test at the 0.5% significance level when the hypothesis test is left- or right-tailed, we must use the look-up table for zz0.005. Since this is a left-tailed test, our critical value is −2.576.

A mechanic wants to show that more than 44% of car owners do not follow a normal maintenance schedule. Identify the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter p.

Correct answer: H0: p=0.44; Ha: p>0.44 Let the parameter p be used to represent the proportion. Remember that the null hypothesis is the statement already believed to be true and the alternative hypothesis is the statement that is trying to be shown. In this case, the mechanic is trying to show that more than 44% of people do not follow the maintenance schedule. So the alternative hypothesis is p>0.44. The null hypothesis always contains the equality symbol: p=0.44.

If a hypothesis test were to be performed, in which of the following contexts would a t-test be appropriate?

A sample mean of 83 and sample standard deviation of 6 are obtained using a simple random sample of size 15 from a normally distributed population. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the second answer choice.

An airline company claims in a recent advertisement that more than 94% of passenger luggage that is lost is recovered and reunited with the customer within 1 day. Hunter is a graduate student studying statistics. For a research project, Hunter wants to find out whether there is convincing evidence in support of the airline company's claim. He randomly selects 315 passengers of the airline whose luggage was lost by the airline and found that 276 of those passengers were reunited with their luggage within 1 day. Are all of the conditions for this hypothesis test met, and if so, what are the null and alternative hypotheses for this hypothesis test?

All of the conditions to conduct the hypothesis test have been met. The null and alternative hypotheses are {H0:p=0.94Ha:p>0.94. First verify whether all of the conditions have been met. Let p be the population proportion for the airline passengers whose luggage was lost by the airline and were reunited with their luggage within 1 day. Since Hunter is completing a survey where there are two independent outcomes, the proportion follows a binomial model. The question states that Hunter randomly selected the airline passengers whose luggage was lost by the airline. The expected number of successes, np=296.1, and the expected number of failures, nq=n(1−p)=18.9, are both greater than or equal to 5. Since all of the conditions for this hypothesis test have been satisfied, determine the null and alternative hypotheses. Since Hunter is determining whether the proportion for reuniting passengers with their luggage within 1 day is greater than 94%, the null hypothesis is that p is equal to 0.94 and the alternative hypothesis is that p is greater than 0.94. The null and alternative hypotheses are shown below. {H0:p=0.94Ha:p>0.94

Which of the following results in a null hypothesis p=0.47 and alternative hypothesis p>0.47?

An online article claims that at most 47% of internet users participate in social media. A group of researchers think this is incorrect, and they want to show that more than 47% of internet users participate in social media. Remember that the null hypothesis is the statement that the researchers are trying to reject, or show is wrong. The null hypothesis is p=0.47. The alternative hypothesis, p>0.47, should be what the researchers are trying to show. So the fourth answer choice is correct.

A hospital administrator claims that the proportion of knee surgeries that are successful is 87%. To test this claim, a random sample of 450 patients who underwent knee surgery is taken and it is determined that 371 of these patients had a successful knee surgery operation. The following is the setup for this hypothesis test: {H0:p=0.87Ha:p≠0.87 Find the test statistic for this hypothesis test for a proportion. In your calculations, round the value for the sample proportion to 3 decimal places. Round your answer to 2 decimal places.

$\text{Test Statistic=}-2.90$Test Statistic=−2.90​ The proportion of successes is p^=371450=0.824. The test statistic is calculated as follows: z=p^−p0p0⋅(1−p0)n−−−−−−√ z=0.824−0.870.87⋅(1−0.87)450−−−−−−−−√ z≈−2.90

Which of the hypothesis tests listed below is a left-tailed test? Select all correct answers.

Correct answer: H0:μ=18, Ha:μ<18 H0:μ=11.3, Ha:μ<11.3 H0:μ=3.7, Ha:μ<3.7 Remember the forms of the hypothesis tests. Right-tailed: H0:μ=μ0, Ha:μ>μ0. Left-tailed: H0:μ=μ0, Ha:μ<μ0. Two-tailed: H0:μ=μ0, Ha:μ≠μ0. So in this case, the left-tailed tests are: H0:μ=3.7, Ha:μ<3.7 H0:μ=18, Ha:μ<18 H0:μ=11.3, Ha:μ<11.3

A linguistics expert is interested in learning about the amount of time people in his industry spend studying a new language. A random sample of 15 people in his industry were surveyed for a hypothesis test about the mean time people studied a new language last year. He conducts a one-mean hypothesis test, at the 10% significance level, to test the recent publication that the amount of time people in his industry are studying a new language was 30 minutes per week.

1$\pm1.761$±1.761​ The degrees of freedom are found by subtracting 1 from the sample size (n). So n−1=15−1=14 Because this is a two-tailed test ( Ha:μ≠μ0), there are two critical values, and thus α2 is used when finding the t-values. (α2=0.102=0.05) Using the t-table given, t0.05 with 14 degrees of freedom is 1.761. The other critical value is −t0.05=1.761.

Determine the Type I error if the null hypothesis, H0, is Carmin will receive a 95% on her statistics exam and the alternative hypothesis, Ha, is Carmin will not receive a 95% on her statistics exam.

Correct answer: Carmin concludes that she will not receive a 95% on her statistics exam when, in fact, she will. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when Carmin concludes that she will not receive a 95% on her statistics exam when, in fact, she will.

A police officer claims that the proportion of accidents that occur in the daytime (versus nighttime) at a certain intersection is 35%. To test this claim, a random sample of 500 accidents at this intersection was examined from police records it is determined that 156 accidents occurred in the daytime. The following is the setup for this hypothesis test: {H0:p=0.35Ha:p≠0.35 Find the test statistic for this hypothesis test for a proportion. Round your answer to 2 decimal places.

$\text{Test_Statistic=}-1.78$Test_Statistic=−1.78​ The proportion of successes is p^=156500=0.312. The test statistic is calculated as follows: z=p^−p0p0⋅(1−p0)n−−−−−−√ z=0.312−0.350.35⋅(1−0.35)500−−−−−−−−√ z≈−1.78

Suppose the null hypothesis, H0, is: the mean age of the horses on a ranch is 6 years. What is the Type II error in this scenario?

Correct answer: You think the mean age of the horses on a ranch is 6 years when, in fact, it is not. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is thinking that the mean age of the horses is 6 years when, in fact, it is not

Lexie, a bowler, claims that her bowling score is more than 140 points, on average. Several of her teammates do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She bowls 18 games. The mean score of the sample games is 155 points. Lexie knows from experience that the standard deviation for her bowling score is 17 points. H0: μ=140; Ha: μ>140 α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

Correct answers:$\text{Test statistic = }3.74$Test statistic = 3.74​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean score, x¯=155. The sample the bowler uses is 18 games, so n=18. She knows the standard deviation of the games, σ=17. Lastly, the bowler is comparing the population mean score to 140 points. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=155−1401718√≈154.007≈3.74 So, the test statistic for this hypothesis test is z0=3.74.

Rebecca is a real estate agent who would like to find evidence supporting the claim that the population mean market value of houses in the neighborhood where she works is greater than $250,000. To test the claim, she randomly selects 35 houses in the neighborhood and finds that the sample mean market value is $259,860 with a sample standard deviation of $24,922. Should Rebecca reject or fail to reject the null hypothesis given the sample data below? H0:μ=250,000 versus Ha:μ>250,000 α=0.05 (significance level) t≈2.34 , which has 34 degrees of freedom 0.01<p-value<0.025

Reject the null hypothesis because 0.01<p-value<0.025 is less than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.01<p-value<0.025 is less than the significance level α=0.05, Rebecca should reject the null hypothesis.

A police office claims that the proportion of people wearing seat belts is less than 65%. To test this claim, a random sample of 200 drivers is taken and its determined that 126 people are wearing seat belts. The following is the setup for this hypothesis test: H0:p=0.65 Ha:p<0.65 In this example, the p-value was determined to be 0.277. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%)

There is NOT sufficient evidence to conclude that the proportion of people wearing seat belts is less than 65%. Compare the p-value to α=0.05. Since the p-value is more than α, fail to reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.05 level of significance to suggest that the proportion of people wearing seat belts is less than 65%.

Suppose a company states that its market share is less than 16 percent, on average. Several employees would like this statement tested, so a director decides to do a hypothesis test, at a 1% significance level He conducts 21 surveys, collects the proper data, and works through the testing procedure: H0: μ=16; Ha: μ<16 x¯=15.8 σ=1.8 α=0.01 (significance level)

This is a left-tailed test, because Ha states company states that its market share is less than 16 percent. "Less than" refers to a less than inequality, which is a left-tailed test. The critical value is z0.01=−2.33

A social media manager for a growing startup claims that exactly 55% of all online sales are made through the company's ads on social media, which draw new customers to the online store. If the social media manager chooses a 10% significance level, what is/are the critical value(s) for this two-tailed hypothesis test (Ha:p≠p0)?

$\text{Smaller Critical Value}=-1.645\text{ Larger Critical Value}=1.645$Smaller Critical Value=−1.645 Larger Critical Value=1.645​ For a two-tailed hypothesis test, the rejection region is on the left and also on the right. Because the test is two-tailed there will be two critical values, one negative and one positive. We know that the significance level is given as α=0.10, so this significance level is divided into two which means there will be an area of 0.05 in the left tail and an area of 0.05 in the right tail. This indicates that the corresponding critical values are −1.645 and 1.645.

Determine the critical value or values for a one-mean z-test at the 20% significance level if the hypothesis test is left-tailed (Ha:μ<μ0).

−0.842 For a left-tailed hypothesis test, the rejection region is on the left. Because the test is left-tailed, the critical value is negative. We know that α=0.2, so the area to right of the z-score is equal to 0.2, and we will use z0.2=−0.842 as our critical value.

Determine the critical value or values for a one-mean z-test at the 10% significance level if the hypothesis test is left-tailed (Ha:μ<μ0).

−1.282 For a left-tailed hypothesis test, the rejection region is on the left. Because the test is left-tailed, the critical value is negative. We know that α=0.1, so the area to right of the z-score is equal to 0.1, and we will use z0.1=−1.282 as our critical value.

A popular charity used 29% of its donations on expenses. An organizer for a rival charity wanted to quickly provide a donor with evidence that the popular charity has expenses that are higher than other similar charities. The organizer randomly selected 20 similar charities and examined their donations. The charity conducts a one-mean hypothesis at the 5% significance level, to test whether the mean is less than 29%. (a) H0:μ=29%; Ha:μ<29%, which is a left-tailed test. (b) The percentages of the donations that those 20 charities spend on expenses are given below. Use Excel to test whether the mean is less than 29%. Identify the test statistic, t, rounding to two decimal places, and p-value, rounding to three decimal places from the Excel output.

1$-1.94$−1.94​ 2$0.034$0.034​ 1. Enter the data into Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. 4. The output gives a mean of x¯¯¯=25.1 and a standard deviation of approximately sx≈8.98. Using these values and the values given in the problem statement, the test statistic, rounding to two decimal places, is t=25.1−298.98/20−−√≈−1.94. 5. The test statistic from step 4 follows a t-distribution with n−1=19 degrees of freedom. The test is left-tailed, so the function T.DIST should be used. Entering the function =T.DIST(−1.94,19,1) into Excel returns a p-value, rounding to three decimal places, of 0.034.

For which of the following is a t-test appropriate for performing a hypothesis test?

A sample mean of 70 and sample standard deviation of 7 are obtained using a simple random sample of size 19 from a normally distributed population. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the third answer choice.

Jacqueline, a medical researcher, would like to make the claim that a newly developed prescription medication for headaches gives patients relief in less than 61 minutes, on average. Jacqueline samples 12 users of the medication and obtains a sample mean of 55 minutes. At the 1% significance level, should Jacqueline reject or fail to reject the null hypothesis given the sample data below? H0:μ=61 minutes; Ha:μ<61 minutes α=0.01 (significance level) test statistic=−1.99 Use the graph below to select the type of test (left-, right-, or two-tailed). Then set the α and the

Do not reject the null hypothesis because the p-value 0.0233 is greater than the significance level α=0.01. Jacqueline is making the claim that the average time it takes the headache medication to work is less than 61 minutes. Therefore, this is a left-tailed test because of "less than". Since this is a left-tailed test, α and p-value will be areas on the left side of the distribution. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since the p-value 0.0233 is greater than the significance level α=0.01, do not reject the null hypothesis.

Doctor Hector claims that the population mean birth weight of babies born at the hospital is exactly 8lb. Sheila, a nurse who assists Hector, wants to test this claim, so she takes a random sample of 55 babies born at the hospital. She determines that the sample mean weight is 7.8lb with a sample standard deviation of 0.9lb. Should Shelia reject or fail to reject the null hypothesis given the sample data below? H0:μ=8 versus Ha:μ≠8 α=0.05 (significance level) t0=−1.65 0.10<p-value<0.20

Fail to reject the null hypothesis because 0.10<p-value<0.20 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.10<p-value<0.20 is greater than the significance level α=0.05, Shelia fails to reject the null hypothesis.

A politician claims that at least 68% of voters support a decrease in taxes. A group of researchers are trying to show that this is not the case. Identify the researchers' null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter p.

H0: p=0.68; Ha: p<0.68 Let the parameter p be used to represent the proportion. The null hypothesis is always stated with the equality symbol: equal (=). Therefore, the null hypothesis H0 is p=0.68. The alternative hypothesis is contradictory to the null hypothesis, so Ha is p<0.68. Also, remember that the alternative hypothesis is the statement that the research or study is trying to show. In this case, they are trying to show that the politician is wrong, so the alternative hypothesis is the opposite of what the politician said, which is at least 0.68. At least means 0.68 or more. Therefore, Ha is p<0.68.

Which graph below corresponds to the following hypothesis test?

The alternative hypothesis, Ha, tells us which area of the graph we are interested in. Because the alternative hypothesis is p>8.1, we are interested in the region greater than (to the right of) 8.1, so the correct graph is the second answer choice

Which graph below corresponds to the following hypothesis test?H0:μ=5.9, Ha:μ<5.9

The alternative hypothesis, Ha, tells us which area of the graph we are interested in. Because the alternative hypothesis is μ<5.9, we are interested in the region less than (to the left of) 5.9, so the correct graph is the first answer choice.

Based on the results of the company's hypothesis test regarding their market share estimate, choose the correct conclusion that interprets the results within the context of the hypothesis test.

There is NOT sufficient evidence 1% significance level to conclude that its market share is less than 16 percent, on average. This is a left-tailed test, and the test statistic is more than the critical value, so we do not reject the null hypothesis. −.51>−2.33 Conclusion: This is a left-tailed test, because Ha would be if the average market share is less than 16 percent. For a left-tailed test, if the test statistic (z0) is less than the critical value (zα), we should reject the null hypothesis.We know that z0=−0.51 and zα=−2.33. Because z0>zα, we do not reject the null hypothesis, which stated that the average market share is greater than or equal to 16 percent.Interpretation: At the 1% significance level, the test results are not statistically significant and at best, provide weak evidence against the null hypothesis.

Vae, a nurse practioner, is interested in determining if her clients with type II diabetes have lower fasting glucose after they started a new nutritional plan. She measures the fasting glucose of 14 random clients, and conducts a hypothesis test about mean of fasting glucose using a significance level of 2.5 %. The mean fasting glucose in these clients was 165 before the nutritional plan.

1$-2.160$−2.160​ The degrees of freedom are found by subtracting 1 from the sample size (n). So n−1=14−1=13 Because this is a left-tailed test, there is one critical value, −2.160. Using the t-table given, t0.025 with 1 degrees of freedom is −2.160.

Which of the following results in a null hypothesis μ=31 and alternative hypothesis μ<31?

A fitness center claims that the mean amount of time that a person spends at the gym per visit is 31 minutes. Some researchers do not think this is correct and want to show that the mean time is less than 31 minutes. Remember that the null hypothesis is the statement that the researchers are trying to reject. The null hypothesis is μ=31, so this should be what the researchers are trying to reject. In other words, that should be the claim of the fitness center. So the fitness center claim (the null hypothesis) is that μ=31, and the researchers are trying to show (the alternative hypothesis) that μ<31, less than 31, which is the second answer choice.

If a hypothesis test were to be performed, in which of the following contexts would a t-test be appropriate?

A sample mean of 30 pounds and a sample standard deviation of 5 pounds are obtained using a simple random sample of 27 dogs from the dogs within a county, which have approximately normally distributed weights. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the second answer choice.

If a hypothesis test were to be performed, in which of the following contexts would a t-test be appropriate?

A simple random sample of 19 students is taken from a school of 1,188 students with approximately normally distributed heights. The sample mean is 61 inches and the sample standard deviation is 2 inches. In order to perform a hypothesis test of a single population mean μ using a Student's t-distribution, the data should be a simple random sample that comes from a population that is approximately normally distributed with an unknown population standard deviation. The only option meeting these criteria is the context in the first answer choice.

A local cable company claims that the proportion of people who have Internet access is less than 63%. To test this claim, a random sample of 800 people is taken and its determined that 478 people have Internet access. The following is the setup for this hypothesis test: H0:p=0.63 Ha:p<0.63 In this example, the p-value was determined to be 0.029. Come to a conclusion and interpret the results for this hypothesis test for a proportion (use a significance level of 5%)

Reject the null hypothesis because the p-value =0.029 is less than the significance level α=0.05. There is enough evidence to suggest that the proportion of people who have Internet access is less than 63%. The p-value is 0.029. Compare the p-value to the level of significance α to make a conclusion for the hypothesis test. In this example, the p-value of 0.029 is less than the level of significance which is 0.05. Since the p-value is less than the level of significance, reject the null hypothesis.

Question Suppose a baker claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him, so the baker decides to do a hypothesis test, at a 5% significance level, to persuade them. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. The baker knows from baking hundreds of loaves that the standard deviation for the height is 0.5 cm. H0: μ=15; Ha: μ>15 α = 0.05 (significance level) What is the test statistic (zz-score) of this one-mean hypothesis test?

Solution The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean height, x¯=17. The sample the baker uses is 10 loaves, so n=10. He knows the standard deviation of the height, σ=0.5. Lastly, the baker is comparing the population mean height to 15 cm. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=17−150.510√≈20.158≈12.65 So, the test statistic for this hypothesis test is z0=12.65.

Question Which graph below corresponds to the following hypothesis test?H0:μ=16.9, Ha:μ>16.9

The alternative hypothesis, Ha, tells us which area of the graph we are interested in. Because the alternative hypothesis is μ>16.9, we are interested in the region greater than (to the right of) 16.9, so the correct graph is the second answer choice.

Based the results from the school psychologist's investigation of the proportion of questions that are related to cultural sensitivity on the aptitude test, choose the correct conclusion that interprets the results within the context of the hypothesis test.

There is NOT sufficient evidence that the proportion of questions related to cultural sensitivity is less than 25%. This is a left-tailed test, because Ha would be if the proportion of questions that are related to cultural sensitivity is less 25%. For a left-tailed test, if the test statistic (z0) is less than the critical value (zα), we should reject the null hypothesis. We know that z0=−1.30 and zα=−2.33. Because z0>zα , we fail to reject the null hypothesis, which stated that the proportion of questions that are related to cultural sensitivity is 25% Interpretation: At the 1% significance level, the data do not provide sufficient evidence to conclude that the proportion of questions that are related to cultural sensitivity is less 25%.

Suppose the null hypothesis, H0, is: a weightlifting bar can withstand weights of 800 pounds and less. And the alternative hypothesis, Ha, is: a weightlifting bar can withstand weights of more than 800 pounds.What is the Type I error in this scenario?

You say there is sufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can not. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is saying there is sufficient evidence to conclude that the weightlifting bar can withstand weights of more than 800 pounds when, in fact, it can not.

Which of the following results in a null hypothesis p=0.69 and alternative hypothesis p>0.69?

Correct answer: A mechanic wants to show that more than 69% of car owners follow a normal maintenance schedule, contrary to a study that found that the percentage was at most 69%. Consider each of the options. Remember that the alternative hypothesis is the statement that the mechanic is trying to show. In this case, the mechanic wants to show that more than 69% of car owners follow a normal maintenance schedule, so the alternative hypothesis, Ha is p>0.69. This is shown in the second option.

Determine the Type I error if the null hypothesis, H0, is: 65% of college students will graduate with debt.And, the alternative hypothesis, Ha, is: that researchers claim more than 65% of college students will graduate with debt.

Correct answer: The researchers conclude that more than 65% of college students will graduate with debt when, in fact, 65% will graduate with debt. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when the researchers conclude that more than 65% of college students will graduate with debt when, in fact, 65% will graduate with debt.

Olivia, a golfer, claims that her drive distance is more than 174 meters, on average. Several of her friends do not believe her, so she decides to do a hypothesis test, at a 10% significance level, to persuade them. She hits 15 drives. The mean distance of the sample drives is 188 meters. Olivia knows from experience that the standard deviation for her drive distance is 14 meters. H0: μ=174; Ha: μ>174 α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

Correct answers:$\text{Test statistic = }3.87$Test statistic = 3.87​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=188. The sample the golfer uses is 15 drives, so n=15. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 174 meters. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=188−1741415√≈143.615≈3.87 So, the test statistic for this hypothesis test is z0=3.87.

Identify the type of hypothesis test below.H0:X=17.9, Ha:X<17.9

The hypothesis test is left-tailed. Remember the forms of the hypothesis tests. Right-tailed: H0:X=X0, Ha:X>X0. Left-tailed: H0:X=X0, Ha:X<X0. Two-tailed: H0:X=X0, Ha:X≠X0. So, this is a left-tailed test.

Colin is a student in a statistics course looking to show whether the proportion of games played that are won by the visiting team in a certain professional lacrosse league is 50%. Colin randomly selects 83 games from the past few seasons and finds that 37 of the games were won by the visiting team. What are the null and alternative hypotheses for this hypothesis test?

{H0:p=0.5Ha:p≠0.5 First verify whether all of the conditions have been met. Let p be the population proportion for the visiting team winning a game played in the league. Since there are two independent outcomes for each trial, the proportion follows a binomial model. The question states that the sample was collected randomly. The expected number of successes, np=41.5, and the expected number of failures, nq=n(1−p)=41.5, are both greater than or equal to 5. Since Colin is testing whether 50% of the games played in the league are won by the visiting team, the null hypothesis is that p is equal to 0.5 and the alternative hypothesis is that p is not equal to 0.5. The null and alternative hypotheses are shown below. {H0:p=0.5Ha:p≠0.5

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.01? (Do not round your answer; compute your answer using a value from the table below.)

$0.312$0.312​ The p-value is the probability of an observed value of z=1.01 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.01, or greater than z=1.01. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.01, or to the right of z=1.01. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.01 is 0.156, which is just the area to the left of z=−1.01. Since the Standard Normal curve is symmetric, the area to the right of z=1.01 is 0.156 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.156)=0.312.

A college administrator claims that the proportion of students that are nursing majors is greater than 40%. To test this claim, a group of 400 students are randomly selected and its determined that 190 are nursing majors. The following is the setup for this hypothesis test: {H0:p=0.40Ha:p>0.40 Find the test statistic for this hypothesis test for a proportion. In your calculations, round the value for the sample proportion to 3 decimal places. Round your answer to 2 decimal places.

$\text{Test Statistic=}3.06$Test Statistic=3.06​ The proportion of successes is p^=190400=0.475. The test statistic is calculated as follows: z=p^−p0p0⋅(1−p0)n−−−−−−√ z=0.475−0.400.40⋅(1−0.40)400−−−−−−−−√ z≈3.06

A fast-food chain claims that the fat content of one of its cheeseburgers is 22 grams. A consumer health group wanted to check this claim, so it obtained a random sample of 20 cheeseburgers from the chain and measured the fat content. The fat contents (in grams) are shown below, rounded to the nearest gram. Use Excel to test whether the mean fat content is different from 22 grams. Identify the p-value from the output, rounding to three decimal places.

$\text{p-value=}0.024$p-value=0.024​ 1. Enter the data into Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. 4. The output gives a mean of x¯¯¯=22.65 and a standard deviation of approximately sx≈1.18. Using these values and the values given in the problem statement, the test statistic, rounding to two decimal places, is t=22.65−221.18/20−−√≈2.46. 5. The test statistic from step 4 follows a t-distribution with n−1=19 degrees of freedom. The test is two-tailed, so the function T.DIST.2T should be used. Entering the function =T.DIST.2T(2.46,19) into Excel returns a p-value, rounding to three decimal places, of 0.024.

A candidate in an election lost by 5.8% of the vote. The candidate sued the state and said that more than 5.8% of the ballots were defective and not counted by the voting machine, so a full recount would need to be done. His opponent wanted to ask for the case to be dismissed, so she had a government official from the state randomly select 500 ballots and count how many were defective. The official found 21 defective ballots. Use Excel to test if the candidate's claim is true and that less than 5.8% of the ballots were defective. Identify the p-value, rounding to three decimal places.

$\text{p-value=}0.063$p-value=0.063​ Step 1: The sample proportion is pˆ=21500=0.042, the hypothesized proportion is p0=0.058, and the sample size is n=500. Step 2: The test statistic, rounding to two decimal places, is z=0.042−0.0580.058(1−0.058)500−−−−−−−−−−−−−−√≈−1.53. Step 3: Since the test is left-tailed, entering the function =Norm.S.Dist(−1.53,1) into Excel returns a p-value, rounding to three decimal places, of 0.063.

A magazine regularly tested products and gave the reviews to its customers. In one of its reviews, it tested two types of batteries and claimed that the batteries from Company A outperformed the batteries from Company B. A representative from Company B asked for the exact data from the study. The author of the article told the representative from Company B that in 200 tests, a battery from Company A outperformed a battery from Company B in 108 of the tests. Company B decided to sue the magazine, claiming that the results were not significantly different from 50% and that the magazine was slandering its good name. Use Excel to test whether the true proportion of times that Company A's batteries outperformed Company B's batteries is different from 0.5. Identify the p-value, rounding to three decimal places.

$\text{p-value=}0.258$p-value=0.258​ Step 1: The sample proportion is pˆ=108200=0.54, the hypothesized proportion is p0=0.5, and the sample size is n=200. Step 2: The test statistic, rounding to two decimal places, is z=0.54−0.50.5(1−0.5)200−−−−−−−−−−√≈1.13. Step 3: Since the test is two-tailed and the test statistic is positive, entering the function =2∗(1−Norm.S.Dist(1.13,1)) into Excel returns a p-value, rounding to three decimal places, of 0.258.

LaNai is a sergeant in the Marines who developed a new training program geared to reduce the time it takes to complete a particular obstacle course. After evaluating 14 random participants of the program, she conducted a hypothesis test about the mean time using a significance level of 1%. Preprogram evaluation revealed a mean of 20 minutes to complete the obstacle course.

1$-2.650$−2.650​ The degrees of freedom are found by subtracting 1 from the sample size (n). So n−1=14−1=13 Because this is a left-tailed test, there is one critical value, −2.650. Using the t-table given, t0.01 with 14 degrees of freedom is −2.650.

Determine the Type I error if the null hypothesis, H0, is: a wooden ladder can withstand weights of 250 pounds and less.

Correct answer: You think the ladder cannot withstand weight of 250 pounds and less when, in fact, it really can. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is thinking the wooden ladder cannot withstand the weights of 250 pounds or less when it really can.

Rachel, a city employee, would like to make the claim that the average amount that residents spend per month on public transit fare is less than $140. Rachel samples 25 city residents and obtains a sample mean of $125.80 spent per month on public transit. At the 5% significance level, should Rachel reject or fail to reject the null hypothesis given the sample data below? H0:μ=$140 per month; Ha:μ<$140 per month α=0.05 (significance level) test statistic=−0.52

Do not reject the null hypothesis because the p-value 0.3015 is greater than the significance level α=0.05. Rachel is making the claim that the average amount that residents spend per month on public transit fare is less than $140.Therefore, this is a left-tailed test because of "less than." Since this is a left-tailed test, α and p-value will be areas on the left side of the distribution. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since the p-value 0.3015 is greater than the significance level α=0.05, do not reject the null hypothesis.

Based on the outcome of Gavin's study on urgent care vs. physician's office visit costs, and the following results, conclude whether to reject or not reject H0. H0: μ=$55; Ha: μ>$55 x¯=$56 σ=$6.2 α=0.01 (significance level) The test statistic isz0=x¯−μ0σn√=56−556.2100√=1.61 The critical value is −z0.01=2.326.

Fail to reject H0. The test statistic of z0=1.61 is less than the critical value of zα/2=2.326 for a right-tailed test; therefore there is NOT enough evidence to reject H0 that the mean cost of urgent care visits is more than physician office visits of $55. The test statistics is NOT in the rejection region. Conclusion: This is a right-tailed test, because Ha states that the average cost of a visit to an urgent care clinic is more than that of a physician's office of $55. For a right-tailed test, if the test statistic (z0) is more than the critical value (zα), we should reject the null hypothesis.We know that z0=1.61 and zα=2.326. Because z0<zα, we fail to reject the null hypothesis, which stated that the mean cost of a visit to an urgent care clinic is more than that of a physician's office of $55.

James was eating a bag of candies that came in eight different colors. He noticed that there appeared to be far less green candies than any of the others and wondered if the true proportion of green candies is lower than the 12.5% that would be expected if all of the candies came in even amounts. For the sake of statistics, he decided that he would need to buy more candy to test his hypothesis. James randomly selected several bags of candies and recorded the color of each piece of candy. He found that out of the first 400 candies that he counted, 37 of them were green. James conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of green candies was lower than 12.5%. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:p=0.125; Ha:p<0.125, which is a left-tailed test. The null hypothesis should be true proportion: H0:p=0.125. James wants to know if the true proportion of green candies was lower than 12.5%. This means that we just want to test if the proportion is less than 12.5%. So, the alternative hypothesis is Ha:p<0.125, which is a left-tailed test because of the less than sign.

A popular charity used 29% of its donations on expenses. An organizer for a rival charity wanted to quickly provide a donor with evidence that the popular charity has expenses that are higher than other similar charities. The organizer randomly selected 20 similar charities and examined their donations. The charity conducts a one-mean hypothesis at the 5% significance level, to test whether the mean is less than 29%. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ=29%; Ha:μ<29%, which is a left-tailed test. The null hypothesis should be the charity's suspected average: H0:μ=29%. The charity wants to test whether the average mean of expenses is less than 29%, so the alternative hypothesis is Ha:μ<29%, which is a left-tailed test because of the less than sign.

Tara is an editor for a weekly magazine. She is testing whether the number of advertisements in the magazine has increased in the past few years. Historically, the magazine has had a mean of 83.22 advertisements per issue with a standard deviation of 6.51 advertisements. The editor randomly selected 30 issues from the past two years and counted the number of advertisements for each issues. The editor conducts a one-mean hypothesis at the 5% significance level, to test if the average number of ads in a magazine is greater than 83.22 ads. (a) Which answer choice shows the correct null and alternative hypotheses for this test?

H0:μ=83.22; Ha:μ>83.22, which is a right-tailed test. The null hypothesis should be the editor's suspected average: H0:μ=83.22. She wants to test if the average number of ads in a magazine is greater than 83.22 ads, so the alternative hypothesis is Ha:μ>83.22, which is a right-tailed test because of the greater than sign.

The null hypothesis, H0, is: no more than 90% of homes in the city are up to the current electric codes.And the alternative hypothesis, Ha, is: an electrician claims more than 90% of homes in the city are up to the current electric codes. What is β, the probability of a Type II error?

The electrician cannot conclude that more than 90% of homes in the city are up to the current electrical codes when, in fact, more than 90% of the homes are up to the current electric codes. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when the electrician cannot conclude that more than 90% of homes in the city are up to the current electrical codes when, in fact, more than 90% of the homes are up to the current electric codes.

A newspaper reports that the average annual salary of a full-time college professor in a certain region is $127,000. Scott, a statistics major, wants to test this claim. He selects a random sample of 75 full-time college professors in the region. The following is the data from this study: The alternative hypothesis Ha:μ≠127,000. The sample mean salary of the 75 college professors is $126,092. The sample standard deviation is $8,509. The test statistic is calculated as −0.92. Values for right-tail areas under the t-distribution curve

The probability of observing a value of t0=−0.92 or less or observing a value of t0=0.92 or more if the null hypothesis is true is greater than 0.20. Notice that the test statistic has 75−1=74 degrees of freedom and that this is a two-tailed test because the alternative hypothesis is Ha:μ≠127,000. Find the p-value for a two-tailed test of a t-distribution with 74 degrees of freedom, where t≈−0.92. Using the table of areas in the right tail for the t-distribution, in the row for 74 degrees of freedom, the absolute value of the t-test statistic, 0.92, is less than 1.293. Since this is a two-tailed test, the p−value is greater than 2∗0.10. Thus, p−value>0.20.

Using the information above, choose the correct conclusion for this hypothesis test.

There is NOT sufficient evidence to conclude that the population meanbirth weight of babies born at the hospital is not equal to 8lb. Compare the p-value to α=0.05. Since the p-value is greater than α, Shelia fails to reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.05 level of significance to support the claim that the population mean birth weight of babies born at the hospital is not equal to 8lb.

Tara is an editor for a weekly magazine. She is testing whether the number of advertisements in the magazine has increased in the past few years. Historically, the magazine has had a mean of 83.22 advertisements per issue with a standard deviation of 6.51 advertisements. The editor randomly selected 30 issues from the past two years and counted the number of advertisements for each issues. The data are provided in the accompanying data table. The editor conducts a one-mean hypothesis at the 5% significance level, to test if the average number of ads in a magazine is greater than 83.22 ads. (a) H0:μ=83.22; Ha:μ>83.22, which is a right-tailed test. (b) z0=1.58, p-value is = 0.057 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We fail to reject H0. At the 5% significance level, the data do not provide sufficient evidence to conclude that the average number of ads in a magazine is greater than 83.22 ads. In this case, p=0.057 and α=0.05, so p>α. Therefore, we fail to reject the null hypothesis. This means that at the 5% significance level, the test results are not statistically significant, and do not provide evidence against the null hypothesis. The magazine editor cannot conclude that at the 5% significance level that the average number of ads per magazine is greater than 83.22 ads.

Using the information above, choose the correct conclusion that interprets the results within the context of the hypothesis test.

We should not reject the null hypothesis because t0>tα. So, at the 10% significance level, the data do not provide sufficient evidence to conclude that the average percentage of tips received by waitstaff in Chicago restaurants is less than 15%. The alternative hypothesis uses a "less than" sign, implying that the test is left-tailed, which means the critical value is negative, and the rejection region is to the left of the critical value. The test statistic is t0=−1.16, and the critical value is −t0.10=−1.310. Since the test statistic is more than (to the right of) the critical value, it is not in the rejection region. So we do not reject the null hypothesis. Therefore, at the 10% significance level, the data do not provide sufficient evidence to conclude that the average percentage of tips received by waitstaff in Chicago restaurants is less than 15%.

A guidance counselor at a university career center is interested in studying the earning potential of certain college majors. He claims that the proportion of graduates with degrees in engineering who earn more than $75,000 in their first year of work is not 15%. If the guidance counselor chooses a 5% significance level, what is/are the critical value(s) for the hypothesis test?

blue dot all the way down (4th) purple dot 2 over (3rd) For a two-tailed hypothesis test, the rejection region is on the left and also on the right. Because the test is two-tailed there will be two critical values, one negative and one positive. We know that the significance level is given as α=0.05, so this significance level is divided into two which means there will be an area of 0.025 in the left tail and an area of 0.025 in the right tail. This indicates that the corresponding critical values are −1.960 and 1.960.

Preston, an official for a state's department of safety, wants to show that the population mean speed of passenger vehicles traveling on a certain expressway is greater than the posted speed limit of 65mph. Preston collects a random sample of 59 passenger vehicles traveling on the expressway and finds that the sample mean speed is 67.1mph with a sample standard deviation of 7.5mph. Should Preston reject or fail to reject the null hypothesis given the sample data below? H0:μ=65 versus Ha:μ>65 α=0.10 (significance level) t≈2.15 with 58 degrees of freedom 0.01<p-value<0.025

Reject the null hypothesis because 0.01<p-value<0.025 is less than the significance level α=0.10. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.01<p-value<0.025 is less than the significance level α=0.10, you reject the null hypothesis.

Suppose a chef claims that her meatball weight is less than 4 ounces, on average. Several of her customers do not believe her, so the chef decides to do a hypothesis test, at a 10% significance level, to persuade them. She cooks 14 meatballs. The mean weight of the sample meatballs is 3.7 ounces. The chef knows from experience that the standard deviation for her meatball weight is 0.5 ounces. H0: μ=4; Ha: μ<4 α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

$\text{Test statistic = }-2.24$Test statistic = −2.24​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean weight, x¯=3.7. The sample the chef uses is 14 meatballs, so n=14. She knows the standard deviation of the meatballs, σ=0.5. Lastly, the chef is comparing the population mean weight to 4 ounces. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=3.7−40.514√≈−0.30.134≈−2.24 So, the test statistic for this hypothesis test is z0=−2.24.

William, a baker, claims that his bread height is more than 13 cm, on average. Several of his customers do not believe him, so he decides to do a hypothesis test, at a 5% significance level, to persuade them. He bakes 17 loaves of bread. The mean height of the sample loaves is 13.9 cm. William knows from experience that the standard deviation for his bread height is 0.7 cm. H0: μ=13; Ha: μ>13 α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test?

Correct answer: 0.90.717√≈5.30 The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean height, x¯=13.9. The sample the baker uses is 17 loaves, so n=17. He knows the standard deviation of the loaves, σ=0.7. Lastly, the baker is comparing the population mean height to 13 cm. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=13.9−130.717√≈0.90.17≈5.30 So, the test statistic for this hypothesis test is z0=5.30.

Which of the following results in a null hypothesis p=0.61 and alternative hypothesis p>0.61?

Correct answer: A study says that at most 61% of students study less than 5 hours per week. A researcher thinks this is incorrect, and wants to show that more than 61% of students study less than 5 hours per week. Remember that the alternative hypothesis is the claim that the researcher is trying to show. The null hypothesis, p=0.61 corresponds to the claim in the study, and the alternative hypothesis, p>0.61 corresponds to what the researcher is trying to show (to reject the null hypothesis). So the third choice is the correct answer

Which of the following results in a null hypothesis p=0.44 and alternative hypothesis p<0.44?

Correct answer: An online article is trying to show that less than 44% of internet users participate in social media, contrary to an established figure saying that 44% of internet users participate in social media. Consider each of the options. The null hypothesis is p=0.44; this must be the established fact that the article is trying to reject. The first answer choice is the correct choice, because the article is trying to show that less than 44% of users participate in social media (p<0.44), which matches the alternative hypothesis given in the question.

A consumer protection company is testing a towel rack to see how much force it can hold. The null hypothesis, H0, is that the rack can hold 100 pounds of force. The alternative hypothesis, Ha, is that the rack can hold less than 100 pounds of force. What is a Type I error in this scenario?

Correct answer: The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds. Remember that a Type I error is rejecting the null hypothesis when the null hypothesis is true. A Type II error is not rejecting the null hypothesis when it is false. We are asked for the Type I error in this scenario. Rejecting the null hypothesis means rejecting the statement that the rack can hold 100 pounds. Therefore, a Type I error is: The researchers states that there is sufficient evidence to conclude that the rack holds less than 100 pounds of force, but the rack actually holds 100 pounds.

A school psychologist would like to investigate whether an aptitude test includes less than 25% questions that are related to cultural sensitivity. Based on the following results, conclude whether to reject or not reject H0. H0 : p=0.25; Ha : p<0.25 α=0.01 (significance level) The test statistic is −1.30. The critical value is z0.01=−2.33.

Fail to reject H0. The test statistic is NOT in the rejection region. Conclusion: This is a left-tailed test, because Ha would be if the proportion of questions that are related to cultural sensitivity is less 25%. For a left-tailed test, if the test statistic (z0) is less than the critical value (zα), we should reject the null hypothesis. We know that z0=−1.30 and zα=−2.33. Because z0>zα , we fail to reject the null hypothesis, which stated that the proportion of questions that are related to cultural sensitivity is less 25%

A district attorney claims the proportion of defendants who are found guilty of robbery is 65%. To test this claim, a random sample of defendants who are accused of robbery are monitored and checked if they were found guilty or not guilty. Assume that the test statistic for this hypothesis test is 1.53. Since this is a two tailed hypothesis test, assume that the critical values for this hypothesis test are −1.645 and 1.645 Come to a decision for the hypothesis test and interpret your results with respect to the original claim.

Fail to reject the null hypothesis.There is not enough evidence to reject the claim that the proportion of defendants who are found guilty of robbery is 65%. In this example this hypothesis test is a two tailed test and thus the rejection region is both on the left and right side. Any values of the test statistic that are less than −1.645 will fall in the rejection region, and also any values of the test statistic that are more than 1.645 will fall in the rejection region The test statistic falls between the critical values and thus the test statistic does not fall in the rejection region, and thus the decision should be to fail to reject the null hypothesis. Since the claim is the null hypothesis for this example, then the conclusion should be that there is not enough evidence to reject the claim that the proportion of defendants who are found guilty of robbery is 65%.

Using the information above, choose the correct conclusion that interprets the results within the context of the hypothesis test.

We should not reject the null hypothesis because −tα/2<t0<tα/2. So, at the 10% significance level, the data do not provide sufficient evidence to conclude that the average percentage received in tips by waitstaff in New York City is not equal to 14%. The alternative hypothesis uses a "not equal" sign, implying that the test is two-tailed, which means there are two critical values, and the rejection region is to the left of the negative critical value and to the right of the positive critical value. The test statistic is t0=1.489, and the critical values are ±t0.1=±1.697. Since the test statistic is between the critical values, it is not in the rejection region. So we do not reject the null hypothesis. Therefore, at the 10% significance level, the data do not provide sufficient evidence to conclude that the average percentage received in tips by waitstaff in New York City is not equal to 14%.

As a part of her studies, Lexie gathered data on the speeds of 19 vehicles traveling on a highway. She works through the testing procedure: H0:μ=58; Ha:μ<58 α=0.05 The test statistic is t0=x¯−μ0sn√=−0.943. The critical value is −t0.05=−1.734. At the 5% significance level, does the data provide sufficient evidence to conclude that the mean speed of vehicles traveling on the highway is less than 58 miles per hour?

We should not reject the null hypothesis because t0>tα. So, at the 5% significance level, the data do not provide sufficient evidence to conclude that the average speed of vehicles traveling on the highway is less than 58 miles per hour. The alternative hypothesis uses a "less than" sign, implying that the test is left-tailed, which means the critical value is negative, and the rejection region is to the left of the critical value. The test statistic is t0=−0.943, and the critical value is −t0.05=−1.734. Since the test statistic is more than (to the right of) the critical value, it is not in the rejection region. So we do not reject the null hypothesis. Therefore, at the 5% significance level, the data do not provide sufficient evidence to conclude that the average speed of vehicles traveling on the highway is less than 58 miles per hour.

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.73? (Do not round your answer; compute your answer using a value from the table below.)

$0.084$0.084​ The p-value is the probability of an observed value of z=1.73 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.73, or greater than z=1.73. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.73, or to the right of z=1.73. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.73 is 0.042, which is just the area to the left of z=−1.73. Since the Standard Normal curve is symmetric, the area to the right of z=1.73 is 0.042 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.042)=0.084.

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=−1.59? (Do not round your answer; compute your answer using a value from the table below.)

$0.112$0.112​ The p-value is the probability of an observed value of z=1.59 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−1.59, or greater than z=1.59. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−1.59, or to the right of z=1.59. Using the Standard Normal Table given, we can see that the p-value that corresponds with z=−1.59 is 0.056, which is just the area to the left of z=−1.59. Since the Standard Normal curve is symmetric, the area to the right of z=1.59 is 0.056 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.056)=0.112.

What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic of z0=0.27? (Do not round your answer; compute your answer using a value from the table below.)

$0.788$0.788​ The p-value is the probability of an observed value of z=0.27 or greater in magnitude if the null hypothesis is true, because this hypothesis test is two-tailed. This means that the p-value could be less than z=−0.27, or greater than z=0.27. This probability is equal to the area under the Standard Normal curve that lies either to the left of z=−0.27, or to the right of z=0.27. Using the Standard Normal Table, we can see that the p-value that corresponds with z=0.27 is 0.606, which is the area to the left of z=0.27. However, we want the area to the right of 0.27, which is 1−0.606=0.394. Because the Standard Normal curve is symmetric, the area to the left of z=−0.27 is 0.394 as well. So, the p-value of this two-tailed one-mean hypothesis test is (2)(0.394)=0.788.

A manufacturer of home products is designing a new wall mount for large flat-panel televisions. Since some flat-panel televisions weigh up to 100 pounds, the manufacturer wants to make sure that the mean weight of failure for the wall mounts is greater than 180 pounds. Dayton is a researcher for the manufacturer who is conducting the test. He randomly selects 35 wall mounts and tests them by using a machine that pulls down on the mounts until each one fails. The results are shown in the data set provided. Based on the research completed during the development phase, Dayton assumes that the population standard deviation is 6.9 pounds. Use Excel to test whether the mean failure weight is greater than 180 pounds. Calculate the test statistic, z, rounding to two decimal places, and the p-value, rounding to three decimal places.

$z=6.49,\ \text{p-value}=0.000$z=6.49, p-value=0.000​ Step 1: Enter the data into Excel. Step 2: Select Data, then select Data Analysis, and then select Descriptive Statistics. Step 3: In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. Step 4: The output gives a mean of x¯¯¯≈187.57. Using this mean and the given values in the problem statement, the test statistic, rounding to two decimal places, is z=187.57−1806.9/35−−√=6.49. Step 5: The test is right-tailed, so subtract the Excel function Norm.S.Dist from 1 to find the p-value. Entering the function =1−Norm.S.Dist(6.49,1) into Excel returns a p-value, rounding to three decimal places, of 0.000.

James was eating a bag of candies that came in eight different colors. He noticed that there appeared to be far less green candies than any of the others and wondered if the true proportion of green candies is lower than the 12.5% that would be expected if all of the candies came in even amounts. For the sake of statistics, he decided that he would need to buy more candy to test his hypothesis. James randomly selected several bags of candies and recorded the color of each piece of candy. He found that out of the first 400 candies that he counted, 37 of them were green. James conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of green candies was lower than 12.5%. (a) H0:p=0.125; Ha:p<0.125, which is a left-tailed test. (b) Use Excel to test whether the true proportion of green candies is less than 12.5%. Identify the test statistic, z, and p-value from the Excel output, rounding to three decimal places.

1$-1.965$−1.965​ 2$0.025$0.025​ Step 1: The sample proportion is pˆ=37400=0.0925, the hypothesized proportion is p0=0.125, and the sample size is n=400. Step 2: The test statistic, rounding to two decimal places, is z=0.0925−0.1250.125(1−0.125)400−−−−−−−−−−−−−−√≈−1.965. Step 3: Since the test is left-tailed, entering the function =Norm.S.Dist(−1.97,1) into Excel returns a p-value, rounding to three decimal places, of 0.025.

Tara is an editor for a weekly magazine. She is testing whether the number of advertisements in the magazine has increased in the past few years. Historically, the magazine has had a mean of 83.22 advertisements per issue with a standard deviation of 6.51 advertisements. The editor randomly selected 30 issues from the past two years and counted the number of advertisements for each issues. The editor conducts a one-mean hypothesis at the 5% significance level, to test if the average number of ads in a magazine is greater than 83.22 ads. (a) H0:μ=83.22; Ha:μ>83.22, which is a right-tailed test. (b) The number of ads per magazine are given below. Using the historical standard deviation as the population standard deviation, use Excel to test whether the mean number of advertisements has increased. Calculate the test statistic, z, rounding to two decimal places, and the p-value, rounding to three decimal places.

1$1.58$1.58​ 2$0.057$0.057​ Step 1: Enter the data into Excel. Step 2: Select Data, then select Data Analysis, and then select Descriptive Statistics. Step 3: In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. Step 4: The output gives a mean of x¯¯¯=85.1. Using this mean and the given values in the problem statement, the test statistic, rounding to two decimal places, is z=85.1−83.226.51/30−−√=1.58. Step 5: The test is right-tailed, so subtract the Excel function Norm.S.Dist from 1 to find the p-value. Entering the function =1−Norm.S.Dist(1.58,1) into Excel returns a p-value, rounding to three decimal places, of 0.057.

Ross has created a puzzle game that he believes will take 30 minutes for a player to complete. A week after the game's release, he collects data for a random sample of 20 players on how long they took to complete the game. These times, in minutes, are shown in the following table. Use Excel to test whether the mean time to complete the game is different from 30 minutes, and then draw a conclusion in the context of the problem. Use α=0.05.

Fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean is not equal to 30 minutes. 1. Enter the data into Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. 4. The output gives a mean of x¯¯¯=32.25 and a standard deviation of approximately sx≈5.70. Using these values and the values given in the problem statement, the test statistic, rounding to two decimal places, is t=32.25−305.70/20−−√≈1.77. 5. The test statistic from step 4 follows a t-distribution with n−1=19 degrees of freedom. The test is two-tailed, so the function T.DIST.2T should be used. Entering the function =T.DIST.2T(1.77,19) into Excel returns a p-value, rounding to three decimal places, of 0.093. 6. Since the p-value is greater than α=0.05, fail to reject the null hypothesis. There is insufficient evidence to conclude that the mean is not equal to 30.

Key Terms Type I error: the decision to reject the null hypothesis, when it is actually true; the probability of this type of error occurring is denoted by the variable α Type II error: the decision to fail to reject the null hypothesis, when it is actually false; the probability of this type of error occurring is denoted by the variable β

Key Terms Type I error: the decision to reject the null hypothesis, when it is actually true; the probability of this type of error occurring is denoted by the variable α Type II error: the decision to fail to reject the null hypothesis, when it is actually false; the probability of this type of error occurring is denoted by the variable β

Keira is an administrator for a university's department of residential life. She is determining if the meal plan required for first-year students living on campus is adequate. Currently, the cost for the meal plan for first-year students living on campus is $1,495 per semester, but she has received several complaints from several first-year students that they could not possibly use the entire amount. She randomly selects the records of 35 first-year students who live on campus and finds the total amount charged on the meal plan for a semester. Keira will consider proposing a smaller option for next year at the next department meeting if the mean amount used is less than $1,400. Based on past research, Keira assumes that the population standard deviation is $110.24. Use Excel to test if the mean amount first-year students who live on campus charge on the meal plan is less than $1,400, where α=0.01.

Reject the null hypothesis. There is sufficient evidence at the α=0.01 level of significance to conclude that the mean amount first-year students who live on campus charge on the meal plan is less than $1,400. Step 1: Enter the data into Excel. Step 2: Select Data, then select Data Analysis, and then select Descriptive Statistics. Step 3: In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. Step 4: The output gives a mean of x¯¯¯=1,353. Using this mean and the given values in the problem statement, the test statistic, rounding to two decimal places, is z=1,353−1,400110.24/35−−√=−2.52. Step 5: The test is left-tailed, so the Excel function Norm.S.Dist can be used directly to find the p-value. Entering the function =Norm.S.Dist(−2.52,1) into Excel returns a p-value, rounding to three decimal places, of 0.006. Step 6: Since the p-value is less than α=0.01, reject the null hypothesis. There is sufficient evidence to conclude that the mean amount first-year students who live on campus charge on the meal plan is less than $1,400.

A cookie manufacturer claims in a recent advertisement that there are "more than 1,000 chocolate chips in every bag" for its most popular line of cookies. Austin is an investigative reporter for a local TV news station who is testing this claim. He randomly selects 38 bags from various stores and counts the number of chips in each bag. The results are shown in the accompanying data set. Austin contacted the manufacturer and found that the population standard deviation is 27.73. Use Excel to test whether the mean number of chips in each bag of cookies is greater than 1,000, and then draw a conclusion in the context of the problem, where α=0.05.

Reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to suggest that the mean number of chips in each bag of cookies is greater than 1,000. Step 1: Enter the data into Excel. Step 2: Select Data, then select Data Analysis, and then select Descriptive Statistics. Step 3: In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. Step 4: The output gives a mean of x¯¯¯≈1,079.42. Using this mean and the given values in the problem statement, the test statistic, rounding to two decimal places, is z=1,079.42−1,00027.73/38−−√=17.66. Step 5: The test is right-tailed, so subtract the Excel function Norm.S.Dist from 1 to find the p-value. Entering the function =1−Norm.S.Dist(17.66,1) into Excel returns a p-value, rounding to three decimal places, of 0.000. Step 6: Since the p-value is less than α=0.05, reject the null hypothesis. There is sufficient evidence at the α=0.05 level of significance to suggest that the mean number of chips in each bag of cookies is greater than 1,000.

Steve listens to his favorite streaming music service when he works out. He wonders whether the service's algorithm does a good job of finding random songs that he will like more often than not. To test this, he listens to 50 songs chosen by the service at random and finds that he likes 32 of them. Use Excel to test whether Steve will like a randomly selected song more often than not, and then draw a conclusion in the context of the problem. Use α=0.05.

Reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a randomly selected song more often than not. Step 1: The sample proportion is pˆ=3250=0.64, the hypothesized proportion is p0=0.5, and the sample size is n=50. Step 2: The test statistic, rounding to two decimal places, is z=0.64−0.50.5(1−0.5)50−−−−−−−−−−√≈1.98. Step 3: Since the test is right-tailed, entering the function =1−Norm.S.Dist(1.98,1) into Excel returns a p-value, rounding to three decimal places, of 0.024. Step 4: Since the p-value is less than α=0.05, reject the null hypothesis. There is sufficient evidence to conclude that Steve will like a randomly selected song more often than not.

An automobile manufacturer claims that the gas mileage on its new model of hybrid sport utility vehicle (SUV) is 40 miles per gallon. A consumer group tests this claim by driving a random sample of 20 of these hybrid SUVs and recording the gas mileages. These mileages are shown below. Use Excel to test whether the mean gas mileage is different from 40 miles per gallon, and then draw a conclusion in the context of the problem. Use α=0.05.

Reject the null hypothesis. There is sufficient evidence to conclude that the mean is not equal to 40. 1. Enter the data into Excel. 2. Select Data, then select Data Analysis, and then select Descriptive Statistics. 3. In the Descriptive Statistics dialog box, enter the cells containing the data sets into Input Range, tick the Summary statistics check box, and press OK. 4. The output gives a mean of x¯¯¯=38.76 and a standard deviation of approximately sx≈1.94. Using these values and the values given in the problem statement, the test statistic, rounding to two decimal places, is t=38.76−401.94/20−−√≈−2.86. 5. The test statistic from step 4 follows a t-distribution with n−1=19 degrees of freedom. The test is two-tailed, so the function T.DIST.2T with the absolute value of the test statistic should be used. Entering the function =T.DIST.2T(2.86,19) into Excel returns a p-value, rounding to three decimal places, of 0.010. 6. Since the p-value is less than α=0.05, reject the null hypothesis. There is sufficient evidence to conclude that the mean is not equal to 40 miles per gallon.

Using the information above, choose the correct conclusion for this hypothesis test.

There is NOT sufficient evidence to conclude that the average amount residents spend per month on public transit fare is less than $140. Since Rachel is making the claim that the average amount that residents spend per month on public transit fare is less than $140, this is a left-tailed distribution. If the decision is to reject the null hypothesis, then we can conclude there is enough evidence to support the claim. However, since the p-value 0.3015 is greater than the significance level α=0.05, do not reject the null hypothesis. Rachel does not have sufficient evidence to conclude that the average amount spent per month on public transit fare is less than $140.

Using the information above, choose the correct conclusion for this hypothesis test.

There is NOT sufficient evidence to conclude that the headache medication gives patients relief in less than 61 minutes, on average. Since Jacqueline is making the claim that the headache medication gives patients relief in 61 minutes or less, on average, this is a left-tailed distribution. If the decision is to reject the null hypothesis, then we can conclude there is enough evidence to support the claim. However, since the p-value 0.0233 is greater than the significance level α=0.01, do not reject the null hypothesis. Jacqueline does not have sufficient evidence to conclude that the headache medication gives patients relief in less than 61 minutes, on average.

A popular charity used 29% of its donations on expenses. An organizer for a rival charity wanted to quickly provide a donor with evidence that the popular charity has expenses that are higher than other similar charities. The organizer randomly selected 20 similar charities and examined their donations. The charity conducts a one-mean hypothesis at the 5% significance level, to test whether the mean is less than 29%. (a) H0:μ=29%; Ha:μ<29%, which is a left-tailed test. (b) t0=−1.94, p-value is =0.034. (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We reject H0. At the 5% significance level, the data provide sufficient evidence to conclude that the charity used less than an average of 29% of its donations on expenses. In this case, p=0.034 and α=0.05, so p<α. Therefore, we reject the null hypothesis. This means that at the 5% significance level, the test results are statistically significant, and do provide evidence against the null hypothesis. The charity can conclude that at the 5% significance level that the charity used less than 29% of its donations on expenses.

James was eating a bag of candies that came in eight different colors. He noticed that there appeared to be far less green candies than any of the others and wondered if the true proportion of green candies is lower than the 12.5% that would be expected if all of the candies came in even amounts. For the sake of statistics, he decided that he would need to buy more candy to test his hypothesis. James randomly selected several bags of candies and recorded the color of each piece of candy. He found that out of the first 400 candies that he counted, 37 of them were green. James conducts a one-proportion hypothesis test at the 5% significance level, to test whether the true proportion of green candies was lower than 12.5%. (a) H0:p=0.125; Ha:p<0.125, which is a left-tailed test. (b) z0=−1.965, p-value is = 0.025 (c) Which of the following are appropriate conclusions for this hypothesis test? Select all that apply.

We reject H0. At the 5% significance level, the data provide sufficient evidence to conclude the true proportion of green candies was lower than 12.5%. In this case, p=0.025 and α=0.05, so p<α. Therefore, we reject the null hypothesis. This means that at the 5% significance level, the test results are statistically significant, and do provide evidence against the null hypothesis. James can conclude that at the 5% significance level that the true proportion of green candies was lower than 12.5%.

Shannon is a graduate student studying health economics. One of her projects was to study the effects of tobacco product taxation on the frequency of smoking among college students. She hypothesized that the taxation would decrease the frequency of smoking. Using a survey analysis and a known frequency in a recently reported publication, she conducted a hypothesis test to determine if students were less likely to smoke if the tax rate went up by $0.25 per cigarette pack. She used a significance level of 2%. Determine the critical value or values for a one-mean z-test at the 2% significance level if the hypothesis test is left-tailed (Ha:μ<μ0).

−2.054 For a left-tailed hypothesis test, the rejection region is on the left. Because the test is left-tailed, the critical value is negative. We know that α=0.02, so the area to right of the z-score is equal to 0.02, and we will use z0.02=−2.054 as our critical value.

Olivia gathered data on the average percentage of tips received by waitstaff in 31 restaurants in Chicago. She works through the testing procedure: H0:μ=15; Ha:μ<15 α=0.10 (significance level) The test statistic is t0=x¯−μ0sn√=−1.16. The critical value is −t0.10=−1.310. Conclude whether to reject or not reject H0. Select two responses below.

Fail to reject H0. The test statistic is not in the rejection region. This is a left-tailed test, because Ha would be if waitstaff in Chicago restaurants receive less than 15% in tips, on average. For a left-tailed test, if the test statistic (t0) is less than the negative critical value (−tα), we should reject the null hypothesis. We know that t0=−1.16 and tα=−1.310. Because t0>tα , we do not reject the null hypothesis, which stated that waitstaff in Chicago restaurants receive at least 15% in tips, on average.

Isabella gathered data on the average percentage of tips received by waitstaff in 31 restaurants in New York City. She works through the testing procedure: H0:μ=14%; Ha:μ≠14% α=0.1 (significance level) The test statistic is t0=x¯−μ0sn√=1.489. The critical values are ±t0.05=±1.697. Conclude whether to reject or not reject H0. Select two responses below.

Fail to reject H0. The test statistic is not in the rejection region. This is a two-tailed test, because Ha would be if waitstaff is tipped at anything other than 14%, on average. For a two-tailed test, there are two critical values and the rejection region is to the left of the negative critical value and to the right of the positive critical value. We know that t0=1.489 and tα/2=±1.697. Since the test statistic is between the critical values, it is not in the rejection region and we do not reject the null hypothesis, which stated that waitstaff is tipped at 14%, on average.

A cell phone manufacturer claims that the average battery life of its newest flagship smartphone is exactly 20 hours. Javier believes the population mean battery life is less than 20 hours. He tests this claim by selecting a random sample of 33 phones of this model. Javier found that the sample mean battery life is 19.5 hours with a sample standard deviation of 1.9 hours. Should Javier reject or fail to reject the null hypothesis given the sample data below? H0:μ=20 versus Ha:μ<20 α=0.05 (significance level) t0=−1.51 0.05<p-value<0.10

Fail to reject the null hypothesis because 0.05<p-value<0.10 is greater than the significance level α=0.05. In making the decision to reject or not reject H0, recall the criterion for comparing the p−value to α. If p≤α, reject H0. Since 0.05<p-value<0.10 is greater than the significance level α=0.05, Javier fails to reject the null hypothesis.

A linguistics expert is studying the proportion of work time people in his industry study a new language. He predicts it is more than 10%. To test this prediction, he surveys 500 random linguists and determines that 120 of them spend more than 10% of their work time studying a new language. The following is the setup for this hypothesis test: H0:p=0.10 Ha:p>0.10 The p-value for this hypothesis test is 0.09. At the 2.5% significance level, should she reject or fail to reject the null hypothesis?

Fail to reject the null hypothesis because the p-value =0.09 is more than the significance level α=0.025. The p-value is 0.009. Compare the p-value to the level of significance α to make a conclusion for the hypothesis test. In this example, the p-value of 0.09 is more than the level of significance which is 0.025. Since the p-value is more than the level of significance, fail to reject the null hypothesis.

Austin works in the marketing department for a large cable company. Recently, he heard that 56% of cable TV subscribers are considering dropping their cable TV subscription. He tests the claim by randomly selecting a sample of cable TV subscribers and asking whether they are considering dropping their cable TV subscription. Austin finds that 122 of the 224 he surveyed are considering dropping their cable TV subscription. What are the null and alternative hypotheses for this hypothesis test?

{H0:p=0.56Ha:p≠0.56 First verify whether all of the conditions have been met. Let p be the population proportion for cable TV subscribers who are considering dropping their cable TV subscription. Since there are two independent outcomes for each trial, the proportion follows a binomial model. The question states that the sample was collected randomly. The expected number of successes, np=125.44, and the expected number of failures, nq=n(1−p)=98.56, are both greater than or equal to 5. Since Austin is looking for evidence that supports the claim that 56% of cable TV subscribers are considering dropping their cable TV subscription, the null hypothesis is that p is equal to 0.56 and the alternative hypothesis is that p is not equal to 0.56. The null and alternative hypotheses are shown below. {H0:p=0.56Ha:p≠0.56

Kylie works for a large nursery and is investigating whether to use a new brand of seeds. The new brand of seeds advertises that 93% of the seeds germinate, which is higher than the germination rate of the seeds she is currently using. She will change over to this new brand unless the actual germination rate is less than what is advertised. Kylie conducts an experiment by randomly selecting 76 seeds of the new brand and plants them. She finds that 70 of those seeds germinated. What are the null and alternative hypotheses for this hypothesis test?

{H0:p=0.93Ha:p<0.93 First verify whether all of the conditions have been met. Let p be the population proportion for the germination rate of the new seeds. Since there are two independent outcomes for each trial, the proportion follows a binomial model. The question states that the sample was collected randomly. The expected number of successes, np=70.68, and the expected number of failures, nq=n(1−p)=5.32, are both greater than or equal to 5. Since Kylie is testing whether the germination rate is less than 93%, the null hypothesis is that p is equal to 0.93 and the alternative hypothesis is that p is less than 0.93. The null and alternative hypotheses are shown below. {H0:p=0.93Ha:p<0.93

Rosetta, a pitcher, claims that her pitch speed is not equal to 59 miles per hour, on average. Several of her teammates do not believe her, so she decides to do a hypothesis test, at a 1% significance level, to persuade them. She throws 14 pitches. The mean speed of the sample pitches is 58 miles per hour. Rosetta knows from experience that the standard deviation for her pitch speed is 6 miles per hour. H0: μ=59; Ha: μ≠59 α=0.01 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

$\text{Test statistic = }-0.62$Test statistic = −0.62​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean speed, x¯=58. The sample the pitcher uses is 14 pitches, so n=14. She knows the standard deviation of the pitches, σ=6. Lastly, the pitcher is comparing the population mean speed to 59 miles per hour. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=58−59614√≈−11.604≈−0.62 So, the test statistic for this hypothesis test is z0=−0.62.

William, a chef, claims that his meatball weight is not equal to 3 ounces, on average. Several of his customers do not believe him, so he decides to do a hypothesis test, at a 1% significance level, to persuade them. He cooks 19 meatballs. The mean weight of the sample meatballs is 2.9 ounces. William knows from experience that the standard deviation for his meatball weight is 0.5 ounces. H0: μ=3; Ha: μ≠3 α=0.01 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

$\text{Test statistic = }-0.87$Test statistic = −0.87​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean weight, x¯=2.9. The sample the chef uses is 19 meatballs, so n=19. He knows the standard deviation of the meatballs, σ=0.5. Lastly, the chef is comparing the population mean weight to 3 ounces. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=2.9−30.519√≈−0.10.115≈−0.87 So, the test statistic for this hypothesis test is z0=−0.87.

Jamie, a chef, claims that her meatball weight is more than 3 ounces, on average. Several of her customers do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She cooks 13 meatballs. The mean weight of the sample meatballs is 3.6 ounces. Jamie knows from experience that the standard deviation for her meatball weight is 0.5 ounces. H0: μ=3; Ha: μ>3 α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

$\text{Test statistic = }4.33$Test statistic = 4.33​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean weight, x¯=3.6. The sample the chef uses is 13 meatballs, so n=13. She knows the standard deviation of the meatballs, σ=0.5. Lastly, the chef is comparing the population mean weight to 3 ounces. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=3.6−30.513√≈0.60.139≈4.33 So, the test statistic for this hypothesis test is z0=4.33.

A researcher is investigating a government claim that the unemployment rate is less than 5%. To test this claim, a random sample of 1500 people is taken and its determined that 92 people are unemployed. The following is the setup for this hypothesis test: {H0:p=0.05Ha:p<0.05 Find the test statistic for this hypothesis test for a proportion. Round your answer to 2 decimal places.

$\text{Test_Statistic=}2.01$Test_Statistic=2.01​ The proportion of successes is p^=921500=0.0613¯¯¯. The test statistic is calculated as follows: z=p^−p0p0⋅(1−p0)n−−−−−−√ z=0.0613¯¯¯−0.050.05⋅(1−0.05)1500−−−−−−−−√ z≈2.01

A software developer claims the proportion of companies that use JavaScript as their primary software technology is different than 61%. To test this claim, a random sample of 720 companies is taken and it is determined 460 of them use JavaScript as their primary software technology. The following is the setup for this hypothesis test: H0:p=0.61 Ha:p≠0.61 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve:

$\text{p-value=}0.112$p-value=0.112​ Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: Determine if the hypothesis test is left tailed, right tailed, or two tailed. Compute the value of the test statistic. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a two tailed test and the test statistic, rounding to two decimal places, is z=0.638889−0.610.61(1−0.61)720−−−−−−−−−−−−√≈1.59. Thus the p-value is the area under the Standard Normal curve to the left of a z-score of −1.59, plus the area under the Standard Normal Curve to the right of a z-score of 1.59. From a lookup table, the area under the Standard Normal curve, the area to the left of 1.59 is 0.944. This means the area in one tail will be 1−0.944=0.056. Therefore, the corresponding area is then 2(0.056)=0.112.

A medical assistant claims the proportion of people who take a certain medication and develop serious side effects is different than 12%. To test this claim, a random sample of 900 people taking the medication is taken and it is determined 93 of them have experienced serious side effects. The following is the setup for this hypothesis test: H0:p=0.12 Ha:p≠0.12 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve:

$\text{p-value=}0.124$p-value=0.124​ Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: Determine if the hypothesis test is left tailed, right tailed, or two tailed. Compute the value of the test statistic. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a two tailed test and the test statistic, rounding to two decimal places, is z=0.1033−0.120.12(1−0.12)900−−−−−−−−−−−−√≈−1.54. Thus the p-value is the area under the Standard Normal curve to the left of a z-score of −1.54, plus the area under the Standard Normal Curve to the right of a z-score of 1.54. From a lookup table, the area under the Standard Normal curve, the corresponding area is then 2(0.062)=0.124.

The chief economist of a busy city would like to test the claim that the proportion of housing units that are owner-occupied is greater than 58%. To test this claim, a random sample of 420 housing units is taken and 252 of them are determined to be owner-occupied. The following is the setup for this hypothesis test: H0:p=0.58 Ha:p>0.58 Find the p-value for this hypothesis test for a proportion and round your answer to 3 decimal places. The following table can be utilized which provides areas under the Standard Normal Curve:

$\text{p-value=}0.203$p-value=0.203​ Here are the steps needed to calculate the p-value for a hypothesis test for a proportion: Determine if the hypothesis test is left tailed, right tailed, or two tailed. Compute the value of the test statistic. If the hypothesis test is left tailed, the p-value will be the area under the standard normal curve to the left of the test statistic z0If the test is right tailed, the p-value will be the area under the standard normal curve to the right of the test statistic z0If the test is two tailed, the p-value will be the area to the left of −|z0| plus the area to the right of |z0| under the standard normal curve For this example, the test is a right tailed test and the test statistic, rounding to two decimal places, is z=0.60−0.580.58(1−0.58)420−−−−−−−−−−−−√≈0.83. Thus the p-value is the area under the Standard Normal curve to the right of a z-score of 0.83. From a lookup table, the area under the Standard Normal curve to the left of 0.83 has area 0.797. We want the area to the right of 0.83, so subtract this from 1 to get a p-value of 1−0.797=0.203.

Dr. da Vinci would like to determine if a new drug to treat high blood pressure has had an effect on clients that he diagnosed with high blood pressure. He gathers the appropriate data on 16 random clients, and conducts a hypothesis test about the mean blood pressure after taking the medication using a significance level of 1 %. The mean blood pressure of these clients was 165/96 before taking the medication.

1$\pm2.947$±2.947​ The degrees of freedom are found by subtracting 1 from the sample size (n). So n−1=16−1=15 Because this is a two-tailed test ( Ha:μ≠μ0), there are two critical values, and thus α2 is used when finding the t-values. (α2=0.012=0.005) Using the t-table given, t0.005 with 15 degrees of freedom is 2.947. The other critical value is −t0.005=−2.947.

Suppose the null hypothesis, H0, is: a surgical procedure is successful at least 80% of the time. What is the Type I error in this scenario?

Correct answer: Doctors think the surgical procedure is successful less than 80% of the time when, in fact, it is successful at least 80% of the time. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when the doctors believe the procedure is successful less than 80% of the time when it is actually successful at least 80% of the time.

A city wants to show that the mean number of public transportation users per day is more than 5,575. Identify the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the parameter μ.

Correct answer: H0: μ=5,575; Ha: μ>5,575 Let the parameter μ be used to represent the mean. Remember that the null hypothesis is the statement already believed to be true, and the alternative hypothesis is the statement trying to be shown. In this case, the city is trying to show that there is more, so μ>5,575, so this is the alternative hypothesis. The null hypothesis is : μ=5,575.

Suppose the null hypothesis, H0, is a surgical procedure is successful at least 80% of the time. And the alternative hypothesis, Ha, states the doctors' claim, which is a surgical procedure is successful less than 80% of the time.Which of the following gives β, the probability of a Type II error?

Correct answer: The doctors do not conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful less than 80% of the time. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when the doctors do not conclude that the surgical procedure is successful less than 80% of the time when, in fact, the surgical procedure is successful less than 80% of the time.

Is the test below left-, right-, or two-tailed?H0:p=0.39, Ha:p≠0.39

Correct answer: The hypothesis test is two-tailed. Remember the forms of the hypothesis tests. Right-tailed: H0:p=p0, Ha:p>p0. Left-tailed: H0:p=p0, Ha:p<p0. Two-tailed: H0:p=p0, Ha:p≠p0. So, this is a two-tailed test.

Suppose the null hypothesis, H0, is: no more than 70% of customers at a sporting goods store do not shop at any other sporting goods stores.And the alternative hypothesis, Ha, is: the sporting goods store claims more than 70% of its customers do not shop at any other sporting goods stores.What is β, the probability of a Type II error in this scenario?

Correct answer: The probability that the sporting goods store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% of its customers do not shop at any other sporting goods stores. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when the store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% do not shop at any other sporting goods stores.

A consumer protection company is testing a seat belt to see how much force it can hold. The null hypothesis, H0, is that the seat belt can hold at least 5000 pounds of force. The alternative hypothesis, Ha, is that the seat belt can hold less than 5000 pounds of force. What is a Type II error in this scenario?

Correct answer: The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds. Remember that a Type I error is rejecting the null hypothesis when the null hypothesis is true. A Type II error is not rejecting the null hypothesis when it is false. We are asked for the Type II error in this scenario. Failing to reject the null hypothesis means failing to reject the statement that the seat belt can hold at least 5000 pounds. Therefore, a Type II error is: The researchers states that there is insufficient evidence to conclude that the seat belt holds less than 5000 pounds of force, but the seat belt actually holds less than 5000 pounds.

Suppose the null hypothesis, H0, is at least 70% of customers, who shop at a particular sporting good store, do not shop at any other sporting goods stores.And the alternative hypothesis, Ha, is less than 70% of customers, who shop at a particular sporting good store, do not shop at any other sporting goods stores.What is the Type I error in this scenario?

Correct answer: The sporting goods store concludes that less than 70% of its customers do not shop at any other sporting goods stores when, in fact, at least 70% of its customers do not shop at any other sporting goods stores. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when the store concludes that less than 70% of its customers only shop at their sporting goods store when, in fact, it is at least 70%.

Determine the Type I error if the null hypothesis, H0, is: the percentage of homes in the city that are not up to the current electric codes is no more than 10%. And, the alternative hypothesis, Ha, is: the percentage of homes in the city that are not up to the current electric codes is more than 10%.

Correct answer: There is sufficient evidence to conclude that more than 10% of homes in the city are not up to the current electrical codes when, in fact, there are no more than 10% that are not up to the current electric codes. A Type I error is the decision to reject the null hypothesis when it is true. In this case, the Type I error is when there is sufficient evidence to conclude that more than 10% of homes in the city are not up to the current electrical codes when, in fact, there are no more than 10% not up to the current electric codes.

Suppose the null hypothesis, H0, is a weightlifting bar can withstand weights of 800 pounds and less.And the alternative hypothesis, Ha, is a weightlifting bar can withstand weights of greater than 800 pounds.What is α, the probability of a Type I error in this scenario?

Correct answer: You conclude the weightlifting bar can withstand weights of greater than 800 pounds when, in fact, the weightlifting bar can withstand weights of less than or equal to 800 pounds.

Jolyn, a golfer, claims that her drive distance is not equal to 222 meters, on average. Several of her friends do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She hits 11 drives. The mean distance of the sample drives is 218 meters. Jolyn knows from experience that the standard deviation for her drive distance is 14 meters. H0: μ=222; Ha: μ≠222 α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

Correct answers:$\text{Test statistic = }-0.95$Test statistic = −0.95​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=218. The sample the golfer uses is 11 drives, so n=11. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 222 meters. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=218−2221411√≈−44.221≈−0.95 So, the test statistic for this hypothesis test is z0=−0.95.

Suppose a pitcher claims that his pitch speed is less than 43 miles per hour, on average. Several of his teammates do not believe him, so the pitcher decides to do a hypothesis test, at a 10% significance level, to persuade them. He throws 19 pitches. The mean speed of the sample pitches is 35 miles per hour. The pitcher knows from experience that the standard deviation for his pitch speed is 6 miles per hour. H0: μ=43; Ha: μ<43 α=0.1 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

Correct answers:$\text{Test statistic = }-5.81$Test statistic = −5.81​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean speed, x¯=35. The sample the pitcher uses is 19 pitches, so n=19. He knows the standard deviation of the pitches, σ=6. Lastly, the pitcher is comparing the population mean speed to 43 miles per hour. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=35−43619√≈−81.376≈−5.81 So, the test statistic for this hypothesis test is z0=−5.81.

Floretta, a pitcher, claims that her pitch speed is less than 46 miles per hour, on average. Several of her teammates do not believe her, so she decides to do a hypothesis test, at a 5% significance level, to persuade them. She throws 24 pitches. The mean speed of the sample pitches is 37 miles per hour. Floretta knows from experience that the standard deviation for her pitch speed is 5 miles per hour. H0: μ=46; Ha: μ<46 α=0.05 (significance level) What is the test statistic (z-score) of this one-mean hypothesis test, rounded to two decimal places?

Correct answers:$\text{Test statistic = }-8.82$Test statistic = −8.82​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean speed, x¯=37. The sample the pitcher uses is 24 pitches, so n=24. She knows the standard deviation of the pitches, σ=5. Lastly, the pitcher is comparing the population mean speed to 46 miles per hour. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=37−46524√≈−91.021≈−8.82 So, the test statistic for this hypothesis test is z0=−8.82.

Correct answers:$\text{Test statistic = }3.87$Test statistic = 3.87​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean distance, x¯=188. The sample the golfer uses is 15 drives, so n=15. She knows the standard deviation of the drives, σ=14. Lastly, the golfer is comparing the population mean distance to 174 meters. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=188−1741415√≈143.615≈3.87 So, the test statistic for this hypothesis test is z0=3.87.

Correct answers:$\text{Test statistic = }4.44$Test statistic = 4.44​ The hypotheses were chosen, and the significance level was decided on, so the next step in hypothesis testing is to compute the test statistic. In this scenario, the sample mean height, x¯=13.8. The sample the baker uses is 25 loaves, so n=25. She knows the standard deviation of the loaves, σ=0.9. Lastly, the baker is comparing the population mean height to 13 cm. So, this value (found in the null and alternative hypotheses) is μ0. Now we will substitute the values into the formula to compute the test statistic: z0=x¯−μ0σn√=13.8−130.925√≈0.80.18≈4.44 So, the test statistic for this hypothesis test is z0=4.44.

A store owner claims that the proportion of shoppers who use coupons is less than 25%. To test this claim, a random sample of shoppers are monitored and checked for use of coupons. Assume that the test statistic for this hypothesis test is −1.17. Assume the critical value for this hypothesis test is −1.282. Come to a decision for the hypothesis test and interpret your results with respect to the original claim.

Fail to reject the null hypothesis.There is not enough evidence to support the claim that the proportion of shoppers who use coupons is less than 25%. In this example this hypothesis test is a left tailed test and thus the rejection region is on the left. The test statistic is more than the critical value and thus the test statistic does not fall in the rejection region, and thus the decision should be to fail to reject the null hypothesis. Since the claim is the alternative hypothesis for this example, then the conclusion should be that there is not enough evidence to support the claim that the proportion of shoppers who use coupons is less than 25%.

The mean math SAT score for seniors at Blue Moon High School is 560. An investigator believes that this score is different than the national average. After obtaining all relevant data, she conducted a hypothesis test using a significance level of 5%. Determine the critical value or values for a one-mean z-test at the 5% significance level if the hypothesis test is two-tailed (Ha:μ≠μ0). Use the curve below to show your answer. Select the appropriate test by dragging the blue point to a right-, left- or two-tailed diagram. The shaded area represents the rejection region. Then, set the critical value(s) on the x-axis by moving the purple slider on the right.

For a two-tailed hypothesis test, the rejection regions are on the left and the right. Because the test is two-tailed, there will be 2 critical values (one positive, one negative), which split the area under the normal curve into a middle 95%. We know that α=0.05, and so the area of the rejection region is equal to 0.05. But this is split between 2 areas, so instead of using zα, we need to use zα5. Since zα=z0.05, we know that: zα2=z0.052=z0.025 So the critical values are −z0.025=−1.960 and z0.025=1.960.

Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly selects 61 employees who work in Yarmouth and records their annual salaries. The following is the data from this study: The alternative hypothesis Ha:μ>55,000. The sample mean salary of the 61 employees is $56,500 per year. The sample standard deviation is $3,750. The test statistic is calculated as 3.12. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=3.12 or more if the null hypothesis is true is less than 0.005. Notice that the test statistic has 61−1=60 degrees of freedom and that this is a right-tailed test because the alternative hypothesis is Ha:μ>55,000. Find the p-value for right-tailed test of a t-distribution with 60 degrees of freedom, where t=3.12. Using the table of areas in the right tail for the t-distribution, in the row for 60 degrees of freedom, the t-test statistic, 3.12, is greater than 2.660. Since this is a right-tailed test, the p-value is less than 0.005.

Using the information regarding the proportion of work time linguists study a new language, choose the correct conclusion for this hypothesis test. H0:p=0.10 ; Ha:p>0.10 The p-value for this hypothesis test is 0.09. The level of significance is α=0.025

There is NOT sufficient evidence to conclude the proportion of work time linguists study a new language is more than 10%. Compare the p-value to α=0.025. Since the p-value is more than α, reject the null hypothesis H0. Therefore, there is NOT enough evidence at the α=0.025 level of significance to suggest that the proportion of work time linguists study a new language is more than 10%.

A car manufacturer claims that the company's newest model has a peak power of 329 horsepower. Katherine is a reporter for an automotive magazine who is testing the manufacturer's claim and says that the peak horsepower is less than the advertised amount. She measures the peak horsepower of 42 randomly selected cars of the new model. The following is the data from this study: The alternative hypothesis Ha:μ<329. The sample mean peak power of the 42 cars is 328.279 horsepower. The sample standard deviation is 2.41 horsepower. The test statistic is calculated as −1.94. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=−1.94 or less if the null hypothesis is true is between 0.025 and 0.05. Notice that the number of degrees of freedom is 42−1=41 and that this is a left-tailed test because the alternative hypothesis is Ha:μ<329. Find the p-value for a left-tailed test of a t-distribution with 41 degrees of freedom, where t≈−1.94. Using the table of areas in the right tail for the t-distribution, in the row for 41 degrees of freedom, the absolute value of the t-test statistic, 1.94, is greater than 1.683 and less than 2.020. So, the p-value is between 0.025 and 0.05.

The mean body temperature of the human body under normal conditions is accepted to be 98.6∘F. Jason is a pre-med student who is skeptical of this claim. To test this claim, he selects a random sample of 47 people to record their body temperatures. The following is the data from this study: The alternative hypothesis Ha:μ≠98.6. The sample mean body temperature of the 47 people is 98.3∘F. The sample standard deviation is 0.80∘F. The test statistic is calculated as −2.57. Using the information above and the portion of the t− table below, choose the correct p− value and interpretation for this hypothesis test.

The probability of observing a value of t0=−2.57 or less or observing a value of t0=2.57 or more if the null hypothesis is true is between 0.01 and 0.02. Notice that the test statistic has 47−1=46 degrees of freedom and that this is a two-tailed test because the alternative hypothesis is Ha:μ≠98.6. Find the p-value for a two-tailed test of a t-distribution with 46 degrees of freedom, where t=−2.57. Using the table of areas in the right tail for the t-distribution, in the row for 46 degrees of freedom, the absolute value of the t-test statistic, 2.57, is greater than 2.410 and less than 2.687. Since this is a two-tailed test, the p−value is less than 2∗0.01 and greater than 2∗0.005. Thus, 0.01<p-value<0.02.

The following data was calculated during a study about medication side effects. Use the following information to choose the correct interpretation for this hypothesis test: The alternative hypothesis is Ha:p≠0.12 The test statistic is calculated as z0=−1.54. Based on the z− table, the p− value is approximately 0.124. Use the curve below to help visualize your answer. Select the appropriate test by dragging the blue point to a right-, left- or two-tailed diagram, then set the test statistic on the x-axis to visualize the p-value.

The probability of observing a value of z0=±1.54 or more extreme if the null hypothesis is true is 12.4%. The medical assistant is interested in testing the claim that the proportion of people who take a certain medication and develop serious side effects is different than 12%. "Different" means this is a two-tailed test. Since this is a two-tailed test, the p−value will have a probability that is to the left of z0=−1.54 and to the right of z0=1.54. When dragging the blue point to the two-tailed diagram and setting the test statistic as z0=−1.54, the p−value is approximately 0.124. This p−value means the probability of observing a value of z0=±1.54 or more extreme if the null hypothesis is true is 12.4%.

Suppose the null hypothesis, H0, is: no more than 70% of customers at a sporting goods store do not shop at any other sporting goods stores.And the alternative hypothesis, Ha, is: the sporting goods store claims more than 70% of its customers do not shop at any other sporting goods stores.What is β, the probability of a Type II error in this scenario?

The probability that the sporting goods store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% of its customers do not shop at any other sporting goods stores. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is when the store cannot conclude that more than 70% of its customers do not shop at any other sporting goods stores when, in fact, more than 70% do not shop at any other sporting goods stores.

Determine the Type II error if the null hypothesis, H0, is: the percentage of college students that will graduate with debt is no more than 65%. And, the alternative hypothesis, Ha, is: the percentage of college students that will graduate with debt is more than 65%

There is insufficient evidence to conclude that more than 65% of college students will graduate with debt when, in fact, more than 65% of college students will graduate with debt. A Type II error is the decision not to reject the null hypothesis when, in fact, it is false. In this case, the Type II error is that there is insufficient evidence to conclude that more than 65% of college students will graduate with debt when, in fact, more than 65% of college students will graduate with debt.

Gavin is a healthcare economist who expects that the average cost of a visit to an urgent care clinic is more than that of a physician's office of $55. After calculating the mean and standard deviation for the cost of 100 of urgent care visits, and using a significance level 1% and works through the testing procedure: H0: μ=$55; Ha: μ>$55 x¯=$56 σ=$6.2 α=0.01 (significance level) The test statistic isz0=x¯−μ0σn√=56−556.2100√=1.61 The critical value is z0.01=2.326. Conclude whether to reject or not reject H0, and interpret the results. Based on this context and the null and alternative hypothesis, what type of hypothesis test should be used? Use the curve below to show your answer. Select the appropriate test by dragging the blue point to a right-, left- or two-tailed diagram. The shaded area represents the rejection region. Then, set the critical value(s) on the x-axis by moving the purple slider on the right.

This is a right-tailed test, because Ha states the average cost of a visit to an urgent care clinic is more than that of a physician's office of $55. A "More than" refers to a greater than inequality, which is a right-tailed test. The critical value is z0.01=2.326

An economist is interested in studying unemployment. He claims that the proportion of people who are unemployed for more than six months is not 20%. If the economist chooses a 1% significance level, what is/are the critical value(s) for the hypothesis test?

blue down to 4th (up to down) and purple to the first one (R to L) For a two-tailed hypothesis test, the rejection region is on the left and also on the right. Because the test is two-tailed there will be two critical values, one negative and one positive. We know that the significance level is given as α=0.01, so this significance level is divided into two which means there will be an area of 0.005 in the left tail and an area of 0.005 in the right tail. This indicates that the corresponding critical values are −2.576 and 2.576.