Do students reduce study time in classes where they achieve a higher midterm score? In a Journai' ofEconomic Education article (Winter 2005), Gregory Krohn and Catherine O'Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that \"students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course." Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table 11.5. Assume that the population of all possible paired differences is normally distributed. Table 11.5 Weekly Study Time Data for Students Who Perform Well on the Mid'l'erm Students 1 2 3 4 5 6 7 8 Before 11 12 18 l7 13 12 18 11 After 4 10 6 9 2 10 9 9 Paired T-Test and Cl: StudyBefore, StudyAfter Paired '1' for StudyBefore StudyAfter N Mean StDev SE Mean Studynefore 8 14.0000 3.1168 1.1019 studynfter 3 7 .3750 3. 0203 1. 0630 Difference 3 6.62500 4.13824 1.46309 95% CI for mean difference: (3.16535, 10.08465) T'Test of mean difference = 0 (vs not = 0): T-Value = 4.53, P-Value = .0027 (a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam. 'HO: pd = I versus Ha: pd # I (b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.) '1 = , We have evidence. During 2017 a company implemented a number of policies aimed at reducing the ages of its customers' accounts. In order to assess the effectiveness ofthese measures, the company randomly selects 10 customer accounts. The average age of each account is determined for the years 2016 and 2017. These data are given in Table 11.4. Assuming that the population of paired differences between the average ages in 2017 and 2016 is normally distributed: TA B L E 11.4 AverageAcoountAgesln2015and2016 for 10 Randorriy Selected Accounts (for Exercise 11.20) 1% AcctAge Average Age of Average Age of Account in 2017 Account In 2016 Account (Days) (Days) 1 27 35 2 19 24 3 40 47 4 3O 28 5 33 41 6 25 33 7 31 35 8 29 51 9 15 18 10 21 28 FIG U R E 11.10 JMPOutputofa Paired leferenceAnalyslsoftheAccountAgeData (tor Exercise 11.20) Matched Pairs leference: Average age of account In 2016 (days)-Average age of account In 2017 Average age of account in 2016 (days) 34 t-Ratio 3.61211 Average age of account in 2017 27 DP 9 Mean Difference 7 Prob > ltl 0.0056' Std Error 1.93793 Prob > t 0.0028' Upper 95% 11.3839 Prob 022 with a = .05. What do you conclude? (Round your answers to 2 decimal places.) F = F005 = H0: 0'21 S 0'22 Standard deviation of returns is oen used as a measure of a mutual fund's volatility (risk). A larger standard deviation of returns is an indication of higher risk. According to Morningstar.com (June 17, 2010), the American Century Equity Income Institutional Fund, a large cap fund, has a standard deviation of returns equal to 14.11 percent. Morningstar.com also reported that the Fidelity Small Cap Discovery Fund has a standard deviation of returns equal to 28.44 percent. Each standard deviation was computed using a sample of size 36. (3) Calculate the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.) Test statistic (b) Calculate the critical value by using excel. (Round your answer to 2 decimal places.) (c) At the 0.05 signicance level, what it the conclusion? 0 Reject HO O Fail to reject Ho Suppose we have taken independent. random samples of sizes n1 = 7 and n2 = 7 from two normally distributed populations having means H1 and ,uz, and suppose we obtain 361 = 240, 362 = 210, 51 = 5, 52 = 6. Assuming equal variances calculate a 95 percent condence interval for p1 #2. Can we be 95 percent condent that #1 #2 is greater than 20? (Round your answers to 3 decimal places.) The condence interval = [ , ]. :I . the entire interval is 20. Why we can use the equal variances procedure here? 0 s1 and 52 very close and n1 = n2 0 s1 and 52 very close and n1 at n2