Question

Some statistics instructors have students use an online homework system and some do not. Final exam scores were recorded for a sample of statistic students who used an online homework system and were compared to the final exam scores from a sample of statistic students who did not use an online homework system. Is there evidence that the students who used an online homework system did better on the final? a) Test at the 3% level. b) Compute a 94% confidence interval for the difference. Final Exam Scores for Students Who Used an Online Homework System (in %) Final Exam Scores for Students Who Did Not Use an Online Homework System (in %) 74 78 60 56 76.85 70 75

Answer 1

first we will find the mean of the two sample

A formula B formula 76.85 72 20 mean 21 S.Dev. 22 78.94444 AVERAGE(B2:319) 70.60278 AVERAGE(D2:019) 3.316132 STDEV.S(B2:B19) 7.333976 STDEV.S(D2:019)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Hopi = 12 Ha: 1 > 119 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.03, and the degrees of freedom are df = 23.693. In fact, the degrees of freedom are computed as follows, assuming that the Hence, it is found that the critical value for this right-tailed test is te = 1.975, for a = 0.03 and df = 23.693. The rejection region for this right-tailed test is R = {t:t > 1.975). (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: t= 11 - X VA 78.94 - 70.60 = 4.397 3.372 + 7332

(4) Decision about the null hypothesis Since it is observed that t = 4.397 > t = 1.975, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0.0001, and since p=0.0001 <0.03, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean y is greater than ply, at the 0.03 significance level.

94% confidence interval for the difference between mean is

Based on the informationa provided, we assume that the population variances are unequal, so then the number of degrees of freedom is computed as follows: (sini + s/n2 df = $ini + ng) = 24.481 The critical value for a = 0.06 and df = 24.481 degrees of freedom is te = tı-a/2:1-1 = 1.972. The corresponding confidence interval is computed as shown below: Since we assume that the population variances are unequal, the standard error is computed as follows: = 1.897 Now, we finally compute the confidence interval: CI = (X1 - X2 - te x se, X1 - X2 + te x se) = (78.94 - 70.6 - 1.972 x 1.897, 78.94 - 70.6 +1.972 x 1.897) = (4.6, 12.08)