Question

Q1. (12 points.) In order to compare the size of office in square feet) for full professors (group 1) and assistant professors (group 2), two independent random samples were collected: For full professors, a sample of 5 produced a mean office size of 124 square feet with a standard deviation of 12; For assistant professors, a sample of 9 produced a mean office size of 105 square feet with a standard deviation of 6. Obtain a 99% confidence interval for 44-4. (You may assume both populations are approximately normal and that the population variances are equal.) ☆ Q2. (12 points.) On average, is the number of text messages sent in a day by freshman different from the number sent by seniors? To answer this question, two independent random samples were selected. The results are summarized below. Group Sample size Mean Std. dev. 1-Freshmen 61 12.44 5 2-Seniors 61 10.12 6 Compute the p-value of the test of interest. (You do not actually need to perform the hypothesis test.)

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Answer 1

1 1 + S= 1 + Van (7-7)+ins, (21) From the given information, for Group 1: = 5,27=124, 5 = 12, for Group 2: 7 = 9,73 = 105, and sy = 6. 100(1-a)% confidence interval for difference of true means is (71 - 7)-tas +1, (5-7)+tare, Van2 Vm m where s be the pooled standard deviation and is obtained by (n-1)3* +(n-1) 14(12)* +86) = 8.4853 (m+,-2) 5+9-2 99% confidence interval for 41-42 is 1 1 (7.-)-tams 'm m 1 1 (124–105)-3.055x8.4853, + (124–105)+3.055x8.4853, 59 = (4.5411, 33.4589) 99% confidence interval for 4-6, is 4.541154-4533.4589. (22)Two-Sample T-Test and CI Sample N Mean StDev SE Mean 61 12.44 5.00 0.64 2 61 10.12 6.00 0.77 Difference =mu (1) - mu (2) Estimate for difference: 2.32 95% CI for difference: (0.34, 4.30) T-Test of difference = 0 (vs not =): T-Value = 2.32 P-Value = 0.022 DF = 116. + 1