stat400-exam3a-Fa23-soln

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University of Illinois, Urbana Champaign *

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400

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Statistics

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Jan 9, 2024

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First Name: SOLUTIONS NetID: __________________________________ Last Name: __________________________________ Signature: __________________________________ Midterm Exam 3 – Version A Friday, December 1, 2023 STAT 400, Fall 2023, D. Unger This exam contains 12 pages which include: this cover page, exam exercises, distribution tables, and scratch paper. There are a total of 4 exercises divided into 16 parts. Check to see if any problems are missing. Enter all requested information on the top of this cover page. Allowable Materials: Pens and/or pencils, a calculator, and one 8.5" x 11" sheet with notes. The Notes Sheet must have your name and NetID written on it, and it will be turned in with the exam. Instructions: Write your final answer in the space provided using at least three significant digits where applicable. Show Your Work: Unsupported answers will not receive full credit. A correct answer, unsupported by calculations, explanation, or algebraic work will receive no credit. An incorrect answer supported by substantially correct calculations and explanations might still receive partial credit. Cheating: If you engage in an act of academic dishonesty, you become liable to severe disciplinary action. Such acts include cheating; falsification or invention of information or citation in an academic endeavor; helping or attempting to help others commit academic infractions; plagiarism; offering bribes, favors, or threats; academic interference; computer related infractions; and failure to comply with research regulations. Do not mark on the exam after time has been called.
Page 2 THIS PAGE IS INTENTIONALLY BLANK We will not grade anything on this page.
Page 3 1. Peoria Charter claims that a bus trip from downtown Champaign to downtown Chicago lasts 200 minutes (i.e., 3 hours and 20 minutes). A student group is conducting an analysis on the accuracy of the claim. A random selection of 24 independent bus trips yields an average duration time of 190 minutes with a sample standard deviation of 12.1 minutes. You are asked to create a confidence interval. (a) What is the parameter of interest in this study? State both the symbol used and its meaning in the context of this study. 𝜇 = the population mean travel time for all bus trips from Champaign to downtown Chicago (a) [3 pts] (b) Identify which distribution you will use to calculate the confidence interval. We will use a t -distribution. [Response does not need to include degrees of freedom.] (b) [3 pts] (c) In one sentence, explain the reason for your choice of distribution. The population standard deviation is unknown and the sample size is relatively small. [Students must provide at least one (but not necessarily both) of the reasons.] (c) [3 pts] (d) Calculate a 95% confidence interval for the population parameter. (Just a typical balanced CI is sufficient. No need to consider a one-sided CI.) With the population standard deviation unknown, a (1 – α )100% confidence interval for μ is 𝑥̅ േ 𝑡 / ሺ𝑛 െ 1 ሻ ∙ 𝑠 √𝑛 190 േ 𝑡 . ଴ଶହ 23 ሻ ∙ 12.1 24 190 2.069 12.1 24 190 5.11 ൌ ሺ 184.9, 195.1 (d) [3 pts] (e) In one sentence, interpret the confidence interval within the context of this situation. We are 95% confident that the population mean travel time for all bus trips from Champaign to downtown Chicago is between 184.9 and 195.1 minutes. (e) [3 pts]
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Page 4 2. The student group is conducting a similar analysis on bus trips from downtown Champaign to O’Hare Airport. Peoria Charter claims the travel time is 235 minutes (i.e., 3 hours and 55 minutes). There is historical pre-Covid data that the population standard deviation for travel time is 13.8 minutes. The average travel time for a random selection of 65 independent bus trips is revealed to be 224 minutes. (a) Calculate a 99% confidence interval the parameter of interest. (Just a typical balanced CI is sufficient. No need to consider a one-sided CI.) (No interpretation needed in your final answer; just the numeric interval.) With the population standard deviation known, a (1 – α )100% confidence interval for μ is 𝑥̅ േ 𝑧 / √௡ . 𝑥̅ േ 𝑧 / 𝜎 √𝑛 224 േ 𝑧 . ଴଴ହ 13.8 65 224 2.576 13.8 65 224 4.41 ൌ ሺ 219.6, 228.4 (a) [5 pts] (b) Using the standard deviation above, what is the minimum sample size required to estimate the mean travel time from Champaign to O’Hare to within 3 minutes with 99% confidence? 𝑛 ൌ ቂ 𝑧 / ∙ 𝜎 𝜀 ൌ ൤ 𝑧 . ଴଴ହ 13.8 3 ൌ ൤ 2.576 13.8 3 140.4 141 round up We would need to sample at least 141 bus trips in order to obtain 99% confidence. (b) [5 pts] (c) Now suppose that the sample standard deviation for travel times in this study of 65 bus trips is 16.8 minutes. Calculate a 95% confidence interval for σ 2 , the population variance. (No interpretation needed in your final answer; just the numeric interval.) ሺ𝑛 െ 1 ሻ𝑠 𝜒 ଵିఈ / ሺ𝑛 െ 1 , ሺ𝑛 െ 1 ሻ𝑠 𝜒 / ሺ𝑛 െ 1 ቇ ൌ ቆ 65 1 16.8 𝜒 . ଽ଻ହ 64 , 65 1 16.8 𝜒 . ଴ଶହ 64 ൌ ൬ 18063 83.30 , 18063 40.48 ൌ ሺ 216.8, 446.2 [Other acceptable answers…] The 95% lower bound confidence interval is (228.4, + ). The 95% upper bound confidence interval is (– , 418.2) (0, 418.2). (c) [5 pts]
Page 5 3. You might assume that MWF classes should conclude within their 50-minute time slot. There is a certain (well-intentioned) instructor who gets so excited about statistics that he occasionally (but not always) goes beyond the 50-minute time point. Students want to conduct a test regarding this instructor’s mean class duration time. (a) What is the parameter of interest in this study? State both the symbol used and its meaning in the context of this study. 𝜇 = the true/population mean duration time for all 50-minute MWF class sessions taught by this instructor (a) [2 pts] (b) State an appropriate null and alternative hypothesis for this situation. H 0 : 𝜇 ൌ 50 or H 0 : 𝜇 ൑ 50 or H 0 : “this instructor is like all others” vs. H 1 : 𝜇 ൐ 50 or H 1 : “this instructor takes more time than others” (b) [5 pts] Class archives provide a population standard deviation of 2.1 minutes. Suppose that his students measure a random sample of 6 class periods and find a sample mean duration time of 52 minutes. They test at the 0.05 level of significance. (c) Calculate the test statistic for testing your null hypothesis. With the population standard deviation known, the test statistic is 𝑧 ൌ 𝑥̅ െ 𝜇 𝜎 / √𝑛 52 50 2.1/ 6 2 0.857 2.33 (c) [3 pts] (d) Find the p-value for the test statistic. p-value = P[ Z 2.33] = 0.0099 (d) [3 pts] (e) Make a decision about the null hypothesis. Since p-value = 0.0099 < 0.05, we reject H 0 . (e) [4 pts] (f) In one sentence, interpret your decision in the context of this situation There is significant evidence to suggest that this instructor spends significantly more than 50 minutes for class sessions on average. (f) [3 pts]
Page 6 4. In September 2020, the University Senate passed a resolution to make the belted kingfisher the official mascot of the University. One might assume that senators voted as such because they represent all University students and at least 51% (i.e., a majority) might be in favor of the resolution. You conduct a random sample of 200 University students and find that 94 are in favor of the resolution, 76 are opposed, and 30 do not care at all. (a) Construct a 90% confidence interval for p , the proportion of all University students that favor the resolution. A (1 – α )100% confidence interval for p is 𝑝̂ േ 𝑧 / 𝑝 1 െ 𝑝 𝑛 94 200 േ 𝑧 . ଴ହ 0.5 0.5 200 0.47 1.645 0.035 0.47 0.058 ൌ ሺ 0.412, 0.528 [Response does not need to include interpretation.] We are 90% confident that the true proportion of all University students that favor the resolution is between 0.412 and 0.528. [Other acceptable answers…] The 90% upper bound confidence interval is (0, 0.515). (a) [6 pts] (b) Based on your confidence interval, is it plausible within 90% confidence that a majority of all University students support the resolution? Just answer YES or NO. YES [Response does not need to include explanation. If the response is incorrect but an explanation is provided, give partial credit.] The 90% confidence interval which contains plausible values for p does include values greater than 0.50 (or, “at least 51%”), which would represent a majority. (b) [4 pts]
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