Question

Problem 7- Wilcoxon Signed-Rank test A sample of 10 men was used in a study to test the effects of a relaxant on the time required to fall asleep. Data for 10 subjects showing the number of minutes required to fall asleep with and without the relaxant follows. Calculate the test-statistic (create all necessary columns to the right of the table to show all the working). If the level of significance is 0.05 and the calculated p-value for the test statistic is 3% then determine whether the relaxant reduces the median time required to fall asleep Subject Relaxant Yes 10 10 12 15 12 4 10 9 6 10 9 10 10 14 9 6

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Answer #1
relevant
subject no yes
1 15 10
2 12 10
3 22 12
4 8 11
5 10 9
6 7 5
7 8 10
8 10 7
9 14 11
10 9 6
Wilcoxon's W: 9.5
P-value: 0.05 < P < 0.10

We have calculated both a W-value and z-value. If the size of N is at least 20 - see the Results Details box - then the distribution of the Wilcoxon W statistic tends to form a normal distribution. This means you can use thez-value to evaluate your hypothesis. If, on the other hand, the size of N is low, and particularly if it's below 10, you should use the W-value to evaluate your hypothesis.

You should also note that if a subject's difference score is zero - that is, if a subject has the same score in both treatment conditions - then the test discards the individual from the analysis and reduces the sample size. If you have a lot of ties, this procedure will undermine the reliability of the test (and also suggests that the requirement that the data is continuous has not been met).

Result Details

W-value: 9.5
Mean Difference: 1.5
Sum of pos. ranks: 45.5
Sum of neg. ranks: 9.5

Z-value: -1.8347
Mean (W): 27.5
Standard Deviation (W): 9.81

Sample Size (N): 10

Result 1 - Z-value

The value of z is-1.8347. The p-value is .06724.

The result is not significant at p < .05.

Result 2 - W-value

The value of W is 9.5. The critical value for W at N = 10 (p < .05) is 5.

The result is not significant at p < .05.

There is not enough evidence that the relaxant reduces the median time required to fell asleep

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