For a comparison of two study guides for a mathematics course, 10 student volunt

Question

For a comparison of two study guides for a mathematics course, 10 student volunteers were found. They were randomly assigned, 5 to study guide A, and 5 to study guide B. Following a two-day period of independent study, the students were examined on the material. The students received the following scores (out of 100): Study Guide A: 94 92 91 80 89 Study Guide B: 71 69 80 63 80 Use a permutation test to test the null hypothesis that the distributions of scores are the same for each study guide, against the two-sided alternative that the distributions are different. How many possible ways are there to randomly allocate the 10 students into two groups of 5 students? 252 [2 pt(s)] You are correct. Your receipt no. is 155-4342) Previous Tries Compute the difference in sample means, A-B, as the test statistic. 16.6000 pt(s)] ou are correct. Your receipt no. is 155-1175) PreviousTries How many arrangements of the data would lead to an absolute value of the computed test statistic as great or greater than the absolute value of the test statistic you calculated in the previous question? Note that you don't have to list all possible arrangements in order to answer this. Examine the data carefully and use the trick that was shown in class (also on pages 8-9 of the typed notes) for the permutation problem. 8 [5 pt(s)) Submit Answer Incorrect. Tries 2/3 Previous Tries Suppose that after examining all the possible arrangements, it is found that the absolute value of the difference in means for 12 of the arrangements (including the real one that you observed) are greater than or equal to the absolute value of the computed test statistic. What is the two-sided p-value for the permutation test? Use at least 5 digits to the right of the decimal. [2 pt(s)] Submit Answer Tries 0/3

Explanation / Answer

c) The test statistic value would be greater if we replace a student from A to B if and only if the test score of B is greater than A, thus only student with 80 in test A and the two students with 80 in test B can be interchanged. Thus the possible combinations of these three students is 3 as 3C2 in test B. Subtracting onecase for the original one, we have 2 possible cobinations.

d) All possible arrangements are 12C6 or 12*11*10*9*8*7/6*5*4*3*2=11*2*2*3*7=924, thus p value is 12/924 or 0.0129 or 1.29%

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