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 = 0.95473, p-value = 0.3196 Using the results of the Shapiro-Wilk test alone, can normality be assumed?

Question

BlackTom AI · Accepted Answer

Shapiro-Wilk tests H0: the sample comes from a normal distribution versus H1: it does not. The p-value quantifies the probability of observing data as extreme as yours if H0 were true.

1) Relevant knowledge points and concepts
- Null hypothesis (H0) for Shapiro-Wilk: data are normally distributed.
- Decision rule (common): choose α = 0.05. If p ≤ α, reject H0 (evidence against normality). If p > α, do not reject H0 (no evidence against normality).
- The test statistic W ranges 0–1; values near 1 support normality.
- For sample size n = 25 the Shapiro-Wilk test is appropriate and has reasonable power, though graphical checks are still recommended in practice.

2) Step-by-step reasoning
- Given: W = 0.95473 and p-value = 0.3196.
- Set conventional significance level α = 0.05 (unless another α is specified).
- Compare p-value to α: 0.3196 > 0.05.

3) Intermediate calculation/comparison
- p-value − α = 0.3196 − 0.05 = 0.2696 (positive), so p-value is substantially larger than α.
- W is fairly close to 1 (0.95473), which is consistent with approximate normality.

4) Why the correct answer is correct
- Because p > α we fail to reject H0. Failing to reject H0 means there is no statistically significant evidence from the Shapiro-Wilk test to conclude the paired differences deviate from normality. Under the instruction “Using the results of the Shapiro-Wilk test alone,” the appropriate practical conclusion is to assume normality for subsequent procedures (for example, using the paired t-test).

5) Why the other option is incorrect
- Option b (“No, normality cannot be assumed”) would be chosen only if the test provided statistically significant evidence against normality (p ≤ α). Here p = 0.3196 is not significant, so option b is not supported by the test results.

6) Conclusion
- Based on the Shapiro-Wilk result alone (p = 0.3196 > 0.05, W ≈ 0.955), normality can be assumed for the purposes of the paired t-test. This confirms answer a.

类似问题

Before running our statistical test, we should test for normality. Which of the following is the correct way to do this on this dataset?

Question 30 Choose a, b, c or d as the best answer. The author’s main argument is that the 100:80:100 model __________.

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单