Ch18-HTforMu (2)

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Practice Problems – Chapter 18 Hypothesis Test for a Mean and Mean Difference Chapter 18 Exercises, #11 Pizza A researcher tests whether the mean cholesterol level among those who eat frozen pizza exceeds the value considered to indicate a health risk. She gets a P-value of 0.07. Explain in this context what the “7%” represents. Chapter 18 Exercises, #45 Normal temperature again From the measurements of body temperature in Chapter 17, Exercise 29, you created a confidence interval for the true mean body temperature of healthy adults. Summary Temperature Count 52 Mean 98.285 Median 98.200 Midrange 98.600 Standard Deviation 0.6824 Range 2.800 Interquartile Range 1.050 a) 98.6°F is commonly assumed to be “normal.” Set up the null and alternative hypotheses for testing this. b) Check the conditions for performing the test. c) What do the data say about the test in part a?

Chapter 18 Exercises, #49 More Ruffles Recall from Chapter 17, Exercise 55 that students investigated the packaging of potato chips . They purchased 6 bags of Lay’s Ruffles marked with a net weight of 28.3 grams. They carefully weighed the contents of each bag, recording the following weights (in grams). The following summary statistics are found: ? = 6, 𝑥̅ = 28.78, ? = 0.40 . a) Do these data satisfy the assumptions for inference? Explain. b) Test the hypothesis that the net weight is as claimed using a 5% significance level. Chapter 18 Exercises, #53 Chips Ahoy! again As we learned in Chapter 17, Exercise 59, in 1998, as an advertising campaign, the Nabisco Company announced a “1000 Chips Challenge,” claiming that every 18 -ounce bag of their Chips Ahoy! cookies contained over 1000 chocolate chips. Dedicated statistics students at the Air Force Academy purchased some randomly selected bags of cookies and counted the chocolate chips. Summary statistics are provided: ? = 16, 𝑥̅ = 1238.2, ? = 94.3 . The data appear unimodal and symmetric. a) Check the assumptions and conditions for inference. Comment on any concerns you have. b) Test their claim by performing an appropriate hypothesis test. Limit the type I error rate to 1%. Other Problems: 1. What is the difference between 𝜇 and 𝜇 𝑑 ? Use your own words. 2. The level of significance is equal to the probability of a (Type I, Type II) error. Choose one. 3. Answer T for true or F for false. A chemical plant gets extra funding from the government if on average they produce less than 400 pounds of mercury pollution per year. We conduct an 𝛼 = 0.01 level hypothesis test to determine whether this chemical plant should receive funding and decide to reject 𝐻 0 . a)____ This is a hypothesis test for the mean. b)____ The test statistic for this test follows a binomial distribution. c)____ Based on our conclusion, we can suggest that this company get funded from the government. 4. Without using any calculator, draw and estimate the p-value for a test of µ ≠ 10 , where ? ∗ = 2.5 and ?? = 33 . Then use StatCrunch to calculate the p-value for this test. Compare your estimate to the true p-value. How close were you?

5. Mark each of the following as True or False: a. (True / False) The p-value is the probability that the null hypothesis is false. b. (True / False) A low p- value indicates that the sample outcome (or one more extreme) would be very unlikely if the null hypothesis were true. c. (True / False) If we reject the null hypothesis, we have proven that the alternative is true. d. (True / False) A type I error is worse than a type II error. e. (True / False) A non-significant outcome means that the data do not conclusively demonstrate that the null hypothesis is false. 6. Thirty subjects followed the Weight Watchers diet for a year. Their weight changes are summarized by a mean of -6.6 lbs. and standard deviation of 10.8 lbs. (based on results published in the Journal of the American Medical Association). Use a significance level of 0.01 to test the claim that diet has no effect. 7. In 2021, a pop-culture blogger believed that young adults spent too much time watching Netflix. In one article, he wrote, “On average, young adults today spend more than 10 hours per week watching programming on Netflix.” A local newspaper journalist read the blog article and decided to do an in-depth study on the issue. She hired a statistician to test the blogger’s claim. The statistician conducted a study and performed a hypothesis test. The results of the study and hypothesis test are summarized below. α = 0.05, 𝑥̅ = 10.4, s = 1.750, n = 64, t* = 1.829, Decision: Reject null hypothesis a. State the null and alternative hypotheses that were used to test this claim. Label which is the claim. Indicate the direction of the test. b. Verify the value of the standardized test statistic. What degrees of freedom are appropriate? c. Draw a picture of the sampling distribution and indicate the area that represents the p- value. Then, calculate the p - value that was used to arrive at the statistician’s decision. d. Write an interpretation for the statistician’s decision. Word it in terms of the original claim. e. A few months after this study, Netflix released its viewership data. One part of their records included weekly viewing habits by age category. For young adults, the true mean weekly viewing time was, in reality, 9.6 hours. Which type of error did the statistician make, a Type I or a Type II error? f. Explain the meaning of this type of error in this situation, as though you were talking to the journalist at the local newspaper.

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For each of the following scenarios in questions 8-10, complete parts a-c below: a. From the description, determine if the data that would be collected is paired data. Explain your reasoning. b. State the parameter of interest. c. Set up the null and alternative hypotheses. 8. Researchers are interested in determining if birth order makes a difference on the age at which a child begins to read. They randomly select 15 families with more than one child and record the age at which the youngest sibling began to read and the age at which the oldest sibling began to read. 9. A quilter wants to know if the average time to construct a 50’’ by 55’’ quilt is more than 18 hours. She takes a random sample of 20 quilters and asks them how long it takes them to complete a 50’’ by 55’’ quilt. 10. Acupuncture is said to provide relief for chronic pain. To test the claim a random sample of 50 patients in chronic pain was taken. Each patient was asked to provide a pain score prior to acupuncture treatment and after acupuncture treatment. 11. A student frequently shops at two different grocery stores, Store A and Store B. She wants to know which store has cheaper organic produce. She took a random sample of 25 foods that were available at both stores and recorded their prices. The resulting values are given below along with a summary of the difference (Store A – Store B). Mean Std. Dev. Store A 2.658 1.18 Store B 3.215 1.39 Differences -0.557 0.94 a. Test the claim at the 𝛼 = 0.01 significance level that Store A has cheaper organic produce than Store B. b. Based on your results in part (a), can you recommend from which store the student should purchase her organic produce? Explain your reasoning.

Key Chapter 18 Exercises, #11 If, in fact, the mean cholesterol level of pizza eaters does not indicate a health risk, then only 7 of every 100 people sampled, on average, would have a mean cholesterol level as high as (or higher than) was observed in this sample. Chapter 18 Exercises, #45 a) H0:μ=98.6 ∘ F;HA:μ≠98.6 ∘ F b) The sample was randomly selected. n=52>30. c) t=−3.33 with 51 df, P -value<0.01. Because the P-value is low, we reject the null hypothesis. We have sufficient evidence at the 1% level to suggest that the mean body temperature of adults is not 98.6 ∘ F. Chapter 18 Exercises, #49 a) Probably a representative sample; we do not know the shape of the distribution and will have to assume it is approximately Normal. b) t=2.94, df=5, P-value is between 0.02 and 0.05. Because the P-value is lower than 0.05, we reject H0. We have convincing evidence at the 5% significance level to suggest that the mean weight of bags of Ruffles potato chips is not 28.3 grams. Chapter 18 Exercises, #53 a) Random sample; The histogram is roughly unimodal and symmetric with no outliers. b) H0:μ=1000,HA:μ>1000, where μ is the mean number of chips per bag; t=10.1, df=15, P-value < 0.005. Because the P-value is so low, we reject the null hypothesis. We have enough evidence at the 1% significance level to support that the mean number of chips per bag is greater than 1000. Note: T heir statement isn’t about the mean ! They claim that all bags have at least 1000 chips. So this test doesn’t really answer the question. Other Questions: 1. 𝜇 is the true mean for one population. 𝜇 𝑑 is the true mean difference between two dependent populations (i.e. the mean of the differences for all paired measurements in the population). 2. The level of significance is equal to the probability of a ( Type I , Type II) error. Choose one. 3. Answer T for true or F for false. A chemical plant gets extra funding from the government if on average they produce less than 400 pounds of mercury pollution per year. We conduct an 𝛼 = 0.01 level hypothesis test to determine whether this chemical plant should receive funding and decide to reject 𝐻 0 . a)__ T __ This is a hypothesis test for the mean. b)__ F __ The test statistic for this test follows a binomial distribution.

c)__ T __ Based on our conclusion, we can suggest that this company get funded from the government. 4. Let us first consider what we know: the t-distribution is near in shape to the standard Normal. The test statistic for this test is 2.5, which is pretty far from 0 (in terms of standard deviations). This is a two-tailed test. As an estimate, I would say the probability beyond 2.5 is roughly 0.025/2, or 0.0125 (since the area to the right of 𝑧 = 2 is about 0.025). Then I would estimate the p-value to be roughly 0.025 (multiplied by two because the test is two-sided). The actual p-value is 2 ∗ 𝑃(? 33 > 2.5) = 0.0176 . My estimate is a bit larger than the true p- value. Why? Maybe because the t distribution will not be the exact same shape as the standard normal when the df = 33. But more likely because my estimate simply divided the area to the right of 𝑧 = 2 by 2, when the curve of the distribution does not distribute the area beneath it evenly. There is more area (i.e. probability) near 2 than near 3, since the distribution curves downward as it extends. 5. Mark each of the following as True or False: a. (True / False ) The p-value is the probability that the null hypothesis is false. b. ( True / False) A low p- value indicates that the sample outcome (or one more extreme) would be very unlikely if the null hypothesis were true. c. (True / False ) If we reject the null hypothesis, we have proven that the alternative is true. d. (True / False ) A type I error is worse than a type II error. e. ( True / False) A non-significant outcome means that the data do not conclusively demonstrate that the null hypothesis is false. 6. 𝐻 0 : 𝜇 = 0 (??𝑎𝑖? − "?? ??????") 𝐻 𝐴 : 𝜇 ≠ 0 𝛼 = 0.01 ?? = ? − 1 = 29 Test Statistic: ? ∗ = −6.6 − 0 10.8 √30 ⁄ ≈ −3.347 P-value (two-tailed): 2 ∙ 𝑃(? 29 < −3.347) = 0.0023 The P-value is less than the level of significance. Reject 𝐻 0 . Interpretation: At a significance level of 0.01, there is sufficient evidence to reject the claim that the mean weight change of Weight Watchers followers is 0 lbs. (that the diet has no effect). 7. a. Hypotheses: 𝐻 0 : 𝜇 = 10 𝐻 1 : 𝜇 > 10 (??𝑎𝑖?)

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Right-tailed test (t-distribution) b. Verify the value of the standardized test statistic. What degrees of freedom are appropriate? ? ∗ = 10.4 − 10 1.750 √64 = 1.829 ?? = ? − 1 = 64 − 1 = 63 c. Draw a picture of the sampling distribution and indicate the area that represents the p- value. Then, calculate the p -value that was used to arrive at the statistician’s decision. ? − ?𝑎??? = 𝑃(? > 1.829) = 0.036 The P-value is less than the level of significance. Decision: Reject 𝐻 0 . d. At a significance level of 0.05, there is sufficient evidence to support the claim that the mean amount of time spent watching Netflix by young adults today is more than 10 hours per week. e. Type I Error (rejected H 0 in favor of H 1 when H 0 was true). f. The journalist’s claim that young adults watch more than 10 hours per week of Netflix was supported, but the true mean is actually less than 10 hours and the claim should not have been supported. 8. a. This is paired because the data are matched by family. b. 𝜇 𝑑 = the mean of the differences (where d=youngest siblings age – oldest siblings age) Note : You can define the differences either way! c. 𝐻 0 : 𝜇 𝑑 = 0 𝐻 𝑎 : 𝜇 𝑑 ≠ 0 9. a. This is not paired because the observations are not matched in any way. b. 𝜇 = the mean time it takes to complete a 50’’ by 55’’ quilt c. 𝐻 0 : 𝜇 = 18 𝐻 𝑎 : 𝜇 > 18

10. a. This is paired because the data are matched on patient (before and after measurement). b. 𝜇 𝑑 = the mean of the differences (where d = before pain score – after pain score) c. 𝐻 0 : 𝜇 𝑑 = 0 𝐻 𝑎 : 𝜇 𝑑 > 0 Note : If you define the differences as d = after – before, switch the alternative to a left-tailed test. 11. a. 𝐻 0 : 𝜇 𝑑 = 0 𝐻 𝑎 : 𝜇 𝑑 < 0, ??𝑎𝑖? ? = −0.557 − 0 0.94 √25 = −2.96 ? = 𝑃(? 24 < −2.96) = 0.0034 Decision: Reject the null hypothesis Conclusion: At the 0.01 significance level there is enough evidence to support the claim that Store A has cheaper organic produce than Store B. b. The student should purchase her organic produce from Store A because we have evidence to conclude it is cheaper.

Ch18-HTforMu (2)

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