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# Solved: Using the Wilcoxon Signed-Ranks Test.Refer to the

## Problem 7BSC Chapter 13.3

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition

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Problem 7BSC

Using the Wilcoxon Signed-Ranks Test.?Refer to the sample data for the given exercise .Use the Wilcoxon signed-ranks test to test the claim that the matched pairs have differences that come from a population with a median equal to zero. Use a 0.05 significance level. Exercise Use the sign test for the data consisting of matched pairs. Heights of Presidents Refer to Data Set 12 in Appendix B and use the heights of U.S. presidents and their main opponents in the presidential campaigns to test the claim that there is no difference. Use a 0.05 significance level.

Step-by-Step Solution:

Solution 7BSC Step 1 The given problem explain about heights of U.S. presidents. From the given information we use the Wilcoxon signed-ranks test to test the claim that the matched pairs have differences that come from a population with a median equal to zero. Here we are choosing randomly the heights of the presidents of U.S. from the data given in Data Set – 12 of appendix B taken 13 in total , then the first requirement of the test is fulfilled. So,the given data are a simple random sample. we have to check the symmetry. Then we have to find the differences for each of the matched pairs by subtracting the heights of the main opponents from that of the president's heights as shown: .so we subtracting the heights of presidents (cm) and heights of opponents (cm). Here difference is d and the table as shown in below. Step 2 Now the histogram of the differences is shown in below. From the above histogram the mirror image of the right side is nearly to the left side of the histogram.So the differences are symmetric and the second condition of the Wilcoxon signed rank test is satisfied as well. Then the null hypothesis is the claim of no difference between the heights of the U.S. presidents and their main opponents, and the alternative hypothesis is the claim that there is a difference. H :0 T here is no difference. H : There is a difference. 1 Now ignoring the signs of the differences, sort the differences from lowest value to highest value and neglecting those values which are equal to zero then replace the differences by the corresponding rank value. The rank value as shown in below table. From the above table the first least value of d | |1 and it is assigned the rank 1,second least value of d| | 2 ,which appears four times so its assigned the rank equal to the mean of 2 , 3 , 4 , 5 , and 6 . Mean of 2 , 3 , 4 , 5 , and 6 = 4. Thus each value 2 of d | | assigned the rank 4. The other values of d| |are assigned the ranks in the same way. Highest value of d is 19| | which is assigned the rank 11.

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