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Chapter 15: Problem 21

Suppose you wish to detect a difference in the locations of two population distributions based on a paired-difference experiment consisting of \(n=30\) pairs. a. Give the null and alternative hypotheses for the Wilcoxon signed-rank test. b. Give the test statistic. c. Give the rejection region for the test for \(\alpha=.05\). d. If \(T^{+}=249,\) what are your conclusions? [NOTE: \(T^{+}+T^{-}=n(n+1) / 2 .\)

Short Answer

Expert verified
Answer: We cannot conclude that there is a significant difference in the locations of the two population distributions.

Step by step solution

01

a. Null and alternative hypotheses

For the Wilcoxon signed-rank test, the null hypothesis (H鈧�) states that the median difference between the paired observations is zero, implying no significant difference in the locations of the two population distributions. The alternative hypothesis (H鈧�) states that the median difference is not zero, which means there is a significant difference in the locations of the two population distributions. H鈧�: The median difference between the paired observations is 0. H鈧�: The median difference between the paired observations is not 0.
02

b. Test statistic

The test statistic for the Wilcoxon signed-rank test is the smaller of the two sums of the ranks of the positive and negative differences, denoted as T鈦� and T鈦�. The test statistic T is defined as: T = min(T鈦�, T鈦�)
03

c. Rejection region for the test at 伪 = 0.05

To determine the rejection region for the test at a significance level of 伪 = 0.05, we need to find the critical values of the test statistic (T) from a Wilcoxon signed-rank table or an appropriate statistical software. For n = 30 pairs and 伪 = 0.05, the critical values are: Lower critical value = 58 Upper critical value = 367 Thus, the rejection region for the test at 伪 = 0.05 is: T < 58 or T > 367
04

d. Conclusions based on the given value of T鈦�

We are given that T鈦� = 249. To find the value of T鈦�, we can use the formula: T鈦� + T鈦� = n(n+1) / 2 Plugging in the values, we get: 249 + T鈦� = 30(31) / 2 T鈦� = 465 - 249 T鈦� = 216 Now, we need to find the test statistic T = min(T鈦�, T鈦�): T = min(249, 216) T = 216 Since the test statistic T is not in the rejection region (T 鈮� 58 and T 鈮� 367), we fail to reject the null hypothesis (H鈧�). Therefore, we cannot conclude that there is a significant difference in the locations of the two population distributions.

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