Suppose you are carrying out a paired -test for a sample of data with sample size , and you wish to check for normality. Further suppose that, using the paired differences, you have carried out the Shapiro-Wilk test with the following results: Shapiro-Wilk normality test data: x
W = 0.97929, p-value = 0.8706 Using the results of the Shapiro-Wilk test alone, can normality be assumed?

Question

BlackTom AI · Accepted Answer

The Shapiro–Wilk test evaluates whether a sample comes from a normal distribution; its null hypothesis (H0) is "data are normally distributed" and the alternative (H1) is "data are not normal."

1) Relevant concepts
- Shapiro–Wilk W ranges 0–1; values close to 1 indicate closer agreement with normality.
- The p-value gives the probability of observing a test statistic at least as extreme as observed if H0 is true.
- The usual decision rule (with a conventional α = 0.05) is: if p ≤ α, reject H0 (conclude non-normal); if p > α, fail to reject H0 (no evidence against normality).
- Note: failing to reject H0 is not proof of normality, only that there is no statistical evidence against it.

2) Step-by-step reasoning
- Null hypothesis: paired differences are normally distributed.
- Given results: W = 0.97929 and p-value = 0.8706.
- Choose significance level α = 0.05 (standard unless otherwise specified).
- Compare p-value to α: 0.8706 > 0.05.

3) Intermediate calculation
- p-value (0.8706) − α (0.05) = 0.8206, which is positive and large, so p ≫ α.
- W distance from perfect normality: 1 − W = 1 − 0.97929 = 0.02071 (very small), consistent with near-normal shape.

4) Why the correct answer is correct
- Because p = 0.8706 is much larger than the conventional α = 0.05, we fail to reject the null hypothesis of normality. The W statistic near 1 also supports this. Under the instruction "Using the results of the Shapiro–Wilk test alone," these results justify treating the paired differences as normally distributed for purposes of the paired t-test.

5) Why the other option is incorrect
- Option b ("No, normality cannot be assumed") would be chosen only if the Shapiro–Wilk test returned a small p-value (p ≤ α). Here p is large, so there is no statistical evidence to reject normality.

Conclusion: Based on the Shapiro–Wilk test results (W = 0.97929, p = 0.8706) and the conventional α = 0.05 decision rule, the correct choice is a. Yes, normality can be assumed.

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