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# Solved: Appendix B Data Sets.Refer to the sample data for

ISBN: 9780321836960 18

## Solution for problem 10BSC Chapter 13.3

Elementary Statistics | 12th Edition

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Problem 10BSC

Appendix B Data Sets.?Refer to the sample data for the given exercise. Use the Wilcoxon signed-ranks test for the claim about the median of a population. Exercise Refer to the indicated data set in Appendix B and use the sign test for the claim about the median of a population. Earthquake Magnitudes Refer to Data Set 16 in Appendix B for the earthquake magnitudes. Use a 0.01 significance level to test the claim that the median is equal to 1.00.

Step-by-Step Solution:

Solution 10BSC Step 1 The given problem explain about Earthquake Magnitudes. From the given information we use the Wilcoxon signed rank test. We use the Wilcoxon signed-ranks test for the claim about the median of a population. We know that the given data are a simple random sample. Then the differences should be symmetrical . Choose a random sample from the Date Set 16 of the appendix B Clearly the given data are a simple random sample. Here we have to check the symmetry. Then we have to find the differences and it is denoted by d. first find the differences for each of the matched pairs by subtracting the ages of the best actors from that of best actresses as shown: Now the histogram of the differences is given 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. Step 2 Then the null hypothesis is the claim that the median of the magnitude of the earthquake is 1.00, while the alternative hypothesis is the claim that itâ€™s not equal to 1.00. Then the hypothesis is H : edian is equal to 1.00 median = 1.00) 0 H :1M edian is not equal to 1.00 (median 1.00) Now ignoring the signs of the differences, sort the differences from the lowest value to the highest value.and replace the differences by the corresponding rank value as shown in below table. From the above table values we know that. Then the the first least value of d| | 0.01 which appears twice so each is given the rank equal to the mean of 1 and 2 that is 1.5. econd least value of d is 0.07, so its assigned | | the rank 3, and so on the other least values of d are assigned the ranks. | | Here the highest value of d | |0.6 which is assigned the rank 46.

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