Wilcoxon Sign Rank Test rejection region

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SUMMARY

The discussion centers on the application of the Wilcoxon Signed Rank Test to compare two software packages in a manufacturing firm's inventory control department. The null hypothesis (H0) states that the software packages are identical, while the alternative hypothesis (H1) asserts they are not. With a test statistic of min(8.5, 57.5) = 8.5 and a critical value of 14 at α = 0.05, the null hypothesis is rejected, indicating that the software packages differ in performance. The debate also highlights a common misconception regarding the adjustment of the significance level for two-tailed tests.

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Homework Statement


Two computer software packages are being considered for use in the inventory control department of a small manufacturing firm. The firm has selected 12 different computing task that are typical of the kinds of jobs. The results are shown in the table below. At the 0.05 level, can we conclude that those two computer software packages are identical?

1.Ho: those two computer software packages are identical
H1: those two computer software packages are not same
2. Based on the alternative hypothesis, the test is min (〖 T〗^+, 〖 T〗^-) = min(8.5, 57.5) = 8.5
3.α = 0.05, n = 12 – 1 = 11

. From table of Wilcoxon signed rank for two tail test,

α = 0.05, n = 11, then a = 14
We will reject Ho if min (〖 T〗^+, 〖 T〗^-) ≤ a
5. Since min(8.5, 57.5) = 8.5 ≤ 14, thus we reject Ho and conclude that the software packages are not equally rapid in handling computing tasks like those in the sample, or the population median for 〖di=x〗_i-y_i is not equal to zero and that package x is faster than package y in handling computing task like ones sample.

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The Attempt at a Solution



In this question , we already knew that it's 2 tailed test , why the author still use α = 0.05 , not α/2 = 0.025 ? I think it's wrong ... [/B]
I think we should use α/2 = 0.025 when finding the U critical
 

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If you are using a table for the Wilcoxon signed rank for two tail test you should not have to make any adjustments of the α value. The table should already have adjusted for that.
 

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