Question

6. (11 marks). Students sometimes compare different instructors teaching the same course and claim that one instructor is harder than another. The Department of Mathematics at the University of Kazoo decided to test this claim. They asked two different calculus instructors, Dr. Reimann and Dr. Poincaire, to prepare final exams and 20 volunteer students who had already passed calculus with other instructors to write both exams (one in the morning and one in the afternoon). The students grades were then recorded for both exams in the following table (both exams are graded out of 100). 10 11 12 13 14 15 16 17 18 19 20 Student 1 2 3 4 5 6 7 8 9 Dr. Reimann Grade 62 92 77 67 80 78 70 67 90 Dr. Poincaire Grade 75 91 85 64 91 81 71 67 86 70 83 59 75 65 78 82 79 67 90 82 The differences in grades for each student were computed (Dr. Reimann Grade - Dr. Poincaire Grade). The sum of differences is d. = -41 and the sum of the squares of the differences is =991. (a) (b) (e) (3 marks). Compute the mean and standard deviation of the differences. (5 marks). Find the 99% confidence interval for the true mean of differences in grades as described in this experiment. (3 marks). Suppose a hypothesis test was done at the same level of confidence with the null hypothesis as the claim that the mean of the differences is 0 while the alternative hypothesis is that mean of the differences is not o. Using the result from part (b), what would be the conclusion? Briefly explain how you could can make this conclusion. Do nor perform the hypothesis test. Would these results be evidence that one of the two instructors is harder than the other?

Answer 1

a) Mean =ă = Edi = -1 = -2.05, n=20|| SD=sa = V[(dj-)= = Ed – n(d)*]/(n − 1) = V((991 – 20 * 2.05%)/19) = 217.9449 Vold-ud) et sd b) Since the scores are taken for the same students, the appropriate statistic is Pairedt statistic. Thus we use the fact that, T="" Then the 99% Cl is + oosn=3*e] =L-141.4747, 137.3747] c) We observe that the value 0 is contained in the 99% Cl. Thus we fail to reject the null at 1% significance level. Hence the true mean scores by the two instructors do not vary significantly. Therefore, evidence is not sufficient to conclude that one instructor is harder than the other.

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