Suppose you are carrying out a paired $t$-test for a sample of data with sample size $n = 25$, 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 = 0.93963, p-value = 0.1451 Using the results of the Shapiro-Wilk test alone, can normality be assumed?

Question

BlackTom AI · Accepted Answer

Shapiro-Wilk tests whether the sample comes from a normal distribution: the null hypothesis H0 is “data are normally distributed” and the alternative H1 is “data are not normally distributed.” The test returns a W statistic (here W = 0.93963) and a p-value (here p = 0.1451). A decision rule is required, typically using α = 0.05: if p ≤ α, reject H0 (conclude non-normal); if p > α, fail to reject H0 (do not conclude non-normal).

Step-by-step reasoning and intermediate steps
1. State hypotheses:
   - H0: paired differences ∼ Normal
   - H1: paired differences not Normal
2. Choose significance level (typical): α = 0.05.
3. Compare p-value to α: 0.1451 > 0.05.
4. Decision: since p > α, we fail to reject H0. That is, there is not enough evidence from the Shapiro–Wilk test to conclude the data deviate from normality.
5. Interpretation of W: W is below 1 but not extremely small; combined with p > 0.05, it indicates the departures from normality (if any) are not statistically significant at conventional levels.

Why the correct answer is correct
- Using the Shapiro–Wilk result alone and the standard α = 0.05, the p-value 0.1451 indicates we do not reject normality. Thus, for the purpose of applying the paired t-test assumption-check, normality can be assumed based on this test result.

Why the other option is incorrect
- Option b (“No, normality cannot be assumed”) would be chosen only if the test produced a small p-value (≤ α). Here p = 0.1451 is not small, so that conclusion is not supported by the test.

Caveat (contextual note)
- Formal tests cannot “prove” normality; they only fail to detect non-normality. With n = 25 the test has moderate power. Still, the question asks about using the Shapiro–Wilk test alone — and that test supports assuming normality.

Conclusion: The correct choice is a. Yes, normality can be assumed (matches the provided answer).

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