Question

There are two sections of statistics, one in the afternoon (PM) with 30 students and one in the morning (AM) with 22 students. Each section takes the identical test. The PM section, on average, scored higher than the AM section. The scores from each section are given in the table below. Test the claim that the PM section did significantly better than the AM section, i.e., is the difference in mean scores large enough to believe that something more than random variation produced this difference. Use a 0.01 significance level.

(a) Use software to calculate the test statistic. Do not 'pool' the variance. This means you do not assume equal variances. Round your answer to 2 decimal places. t = (b) Use software to get the P-value of the test statistic. Round to 4 decimal places. P-value = (c) What is the conclusion regarding the null hypothesis? reject H0 fail to reject H0     (d) Choose the appropriate concluding statement. The difference in mean scores is large enough to suggest this difference is due to something more than random variation. There is not a big enough difference in mean scores to suggest that this difference is anything more than a result of random variation.      We have proven that students in PM sections of statistics do better, on average, than students taking AM sections. We have proven that there is no difference between AM and PM sections of statistics.

PM Scores (x1) AM Scores (x2) 80 72 68 61 98 93 93 89 66 61 84 79 57 50 65 61 52 48 80 75 70 64 82 76 88 84 98 93 74 66 76 72 83 77 86 78 82 76 50 45 95 87 65 60 61 65 72 72 100 87 79 72

Answer 1

Answer Date: 11/11/2019 To test the hypothesis is that the PM section did significantly better than the AM section at 1% significance level. The null and alternative hypothesis is, H:1 = 112 H :11 > 12 The hypothesis is right-tailed test. a) The t-test statistics is, By using Excel software, find t-test statistics with the help of following steps: 1) Import the data. 2) Select t-test: Two Sample Assuming Unequal Variances from Data Analysis in Data menu. 3) Select Variable 1 Range and Variable 2 Range. 4) Give Hypothesized Mean Difference. 5) Select Labels and give Alpha.

Choose Output Range from Output options. Click Ok. t-Test: Two-Sample Assuming Unequal Variances PM Scores 76.66666667 180.1609195 30 AM Scores 71.22727273 193.9935065 44 Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 1.412793822 0.082375642 2.414134361 0.164751284 2.692278259 From the Excel output, the t-test statistics is 1.41). b) The p-value for this test is,

From the Excel output in part (a), the p-value for this test is 0.0824). Decision The conclusion is that p-value in this context is greater than 0.01 which is 0.0824, so the null hypothesis is not rejected at 1% level of significance. There is insufficient evidence to indicate that there the PM section did significantly better than the AM section. So, there is not a big enough difference in mean scores to suggest that this difference is anything more than a result of random variation. The result is not statistically significant. The answer option is 2.