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Homework 12 Due Tuesday, April 19, 11:59pm STAT 400, Spring 2022, D. Unger Each Exercise or lettered part of an Exercise is worth 5 points. Homework assignments are worth 50 points. Exercise 1 A hamburger claiming to be a Quarter Pounder should have a beef patty that weighs a quarter of a pound, i.e., 4 ounces. If it weighs more than that, the restaurant is giving away more beef than they need to. If it weighs less than that, the customer is getting ripped off. In a random sample of 11 Quarter Pounder with Cheese hamburgers at McDowell’s, the average weight of the beef patty was 3.90 ounces with a sample standard deviation of 0.12 ounces. Assume the weights are normally distributed. (a) Define the population parameter of interest. State an appropriate null and alternative hypothesis for this situation. Solution µ = the average weight of the beef in all Quarter Pound burgers at McDowell’s H0: 𝜇4 vs. H1: 𝜇4 (b) Calculate the test statistic for testing the null hypothesis you defined in part a. Solution With the population standard deviation unknown, the test statistic is 𝑡∗/√ ~ 𝑡𝑛1𝑡10. 𝑡𝑥 𝜇𝑠/√𝑛3.90 40.12/√110.10.03622.764 (c) Calculate the p-value for the test statistic. Solution p-value = P[observing a value as or more extreme than –2.764] = 2 · P[T10 ≤ –2.764] = 2 · 0.01 = 0.02(d) Report your decision regarding the null hypothesis with a level of significance of α = 0.05. Provide a concluding statement in the context of this situation. Solution Since p-value = 0.02 < 0.05, we reject H0. There is significant evidence to suggest that the average weight of the beef in all Quarter Pound burgers at McDowell’s is not 4 ounces. (e) In Homework 9, you constructed a 95% (two-sided) confidence interval for the overall average weight of beef in a Quarter Pound burger at McDowell’s. Using that as evidence for the hypothesis test in part a, do you come to the same conclusion that you did in part d? Explain. Solution From the earlier assignment, we found that we are 95% confident that the overall average weight of beef in a Quarter Pound burger at McDowell’s is between 3.82 and 3.98 ounces. Since the null hypothesis value of 4 ounces is not in the 95% confidence interval, then it is not a plausible value for µ, and we would similarly reject H0 at the complementary 0.05 significance level. Exercise 2 A nationwide study showed that undergraduate students in the United States spend about 7.1 hours on average studying on weekends. The study also revealed the population standard deviation to be 2.3 hours. A random sample of 60 Illinois undergraduate students are selected to answer the following question: “How many hours do you spend on studying during the weekend?” The sample average of weekend studying time is 6.8 hours. Assume the number of hours each student spends studying on the weekends is independent from others. Conduct a hypothesis test to determine whether the weekend study times at Illinois are at least as many as their fellow students nationwide, or not. (a) State the null and alternative hypotheses, calculate the test statistic, and provide the p-value. Solution Step 1: µ = the average hours spent studying on weekends by Illinois undergraduate students H0: µ ≥ 7.1 vs. H1: µ < 7.1Step 2: 𝑋 = the sample mean of study hours; 𝑋 ~ Normal(µ, σ2/n) with σ known. We will use 𝑍/√. Step 3: 𝑧𝑥 𝜇𝜎/√𝑛6.8 7.12.3/√600.30.2971.01 p-value = P[Z < –1.01] = 0.1562 (b) Using a level of significance of α = 0.10, make a decision about the null hypothesis and provide a concluding statement in the context of this situation. Solution Step 4: Since p-value = 0.1562 < 0.10, we fail to reject H0. Step 5: There is not significant evidence to suggest that the average time spent studying on weekends by Illinois undergraduate students is less than 7.1 hours. (c) State the rejection region for this test using a level of significance of α = 0.10. Based on your test statistic, would you make the same decision as you did with the p-value in part b? Explain. Solution We would reject H0 for any test statistic z* such that z* < –z0.10 –1.28. Since our test statistic is z = –1.01, we would we fail to reject H0. This is the same decision in part b. Exercise 3 A little girl named Sami is normally not allowed to have soda. But for her 7th birthday, she is allowed to have as much as she wants. The restaurant serves drinks in pint glasses (16oz), but Sami won’t always consume a full pint of soda due to ice and refills. On her birthday, in one sitting, she consumed the following quantities of a pint (16 oz) of soda. 15.4 oz, 15.0 oz, 15.1 oz, 15.3 oz, 15.2 oz Suppose that the amount of soda in a pint glass follows a Normal(μ, σ2) distribution and that Sami’s five drinks represent a random sample. (a) Construct a 95% one-sided upper bound confidence interval for the true mean volume (in ounces) of a pint of soda, μ. Interpret the confidence interval within the context of the exercise.Solution A (1 – α)100% one-sided upper bound confidence interval for µ is ∞, 𝑥 𝑡𝑛1∙√. The 95% upper bound is 𝑥 𝑡𝑛1∙𝑠√𝑛 15.2 𝑡.4∙0.1581√5 15.2 2.132 ∙0.1581√5 15.2 0.151 15.351 We are 95% confident that the true mean volume of a pint of soda is no more than 15.351 ounces. (b) A one-sided upper bound confidence interval can be used as evidence in hypothesis tests when the alternative hypothesis suggests that the population parameter is less than a claimed value. For example, suppose the restaurant claims that they put at least 15.5 oz of soda in each pint glass. Thus, we wish to test H0: µ ≥ 15.5 versus H1: µ < 15.5. Using the confidence interval you constructed in part a, make a decision about this test at the α = 0.05 level of significance. Provide a concluding statement in the context of this situation. Solution Since the null hypothesis value of 15.5 ounces is not in the 95% one-sided upper bound confidence interval, then it is not a plausible value for µ. We reject H0 at the 0.05 significance level. There is significant evidence to suggest that the restaurant’s claims that they put at least 15.5 oz of soda in

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