Shapiro-Wilk Normality Test Calculator

How to Use the Shapiro-Wilk Normality Test Calculator

1. 1. Paste your data. Enter between 3 and 5,000 numbers separated by commas, spaces, or new lines. If you are checking a model's assumption, paste its residuals rather than the raw scores.
2. 2. Choose α. The default is .05; the decision line updates as you change it.
3. 3. Read W, p, and the plot together. A small p says the sample departs from normality; the Q–Q plot shows where — in one tail, both tails, or through skew across the whole range.
4. 4. Report it with context. Copy the APA line and pair it with the plot or a histogram, and with what the departure means for the analysis you plan.

What the W Statistic Measures

Sort the data from smallest to largest. If the population were normal, the ordered values would line up with the expected order statistics of a normal sample, and W measures how well they do: W = (Σ aᵢ x₍ᵢ₎)² / Σ(x − x̄)², where the coefficients aᵢ are derived from those expected normal order statistics. W is in effect the squared correlation between the data and normal scores, so it runs from 0 to 1, with values near 1 meaning the sample looks normal. Small departures matter: W = .95 can be a clear rejection in a large sample.

The exact distribution of W is not known in closed form, so p-values come from Royston's (1995) algorithm AS R94, which computes the coefficients and transforms W to an approximately normal statistic for 4 ≤ n ≤ 5,000 (for n = 3 the p-value is exact). It is the algorithm behind R's shapiro.test and SciPy's scipy.stats.shapiro, and this calculator matches SciPy to many decimal places.

Worked Example: Reaction Times With One Slow Participant

Click Try an example to load reaction times, in milliseconds, for 11 participants: 148, 154, 158, 160, 161, 162, 166, 170, 182, 195, and 236. The mean is 172 and the SD 24.95.

In APA style: reaction times departed from normality, W(11) = .79, p = .007. The Q–Q plot shows why: the lower values hug the reference line, the upper tail bends away from it, and the 236 ms response sits far above everything else — a right skew driven mostly by one slow participant. Remove that one value and the test gives W(10) = .91, p = .268 — but deleting data to pass a normality test is not a fix. Ask whether 236 ms is an error or a real slow response, and if it is real, consider a method that does not assume normality, such as a rank-based test.

Reading the Result Honestly

• A non-significant result does not prove normality. It means the test could not detect a departure. With ten values, even clearly skewed data often pass.
• A significant result in a large sample may not matter. With several hundred values, W detects departures far too small to affect a t-test or ANOVA, which tolerate moderate non-normality as samples grow.
• Test what the method assumes. ANOVA and regression assume normal residuals (or normal data within each group), not a normal pooled sample. Testing all groups together can reject normality just because the group means differ.
• Look at the picture. The Q–Q plot shows the kind of departure — skew bends the points into an arc, heavy tails into an S-shape — which a p-value cannot. Choosing a test based only on a preliminary normality test also distorts the final test's error rates.

Frequently Asked Questions

What does the Shapiro-Wilk test tell you?

Whether a sample is consistent with having come from a normal distribution. The null hypothesis is normality, so a p-value below α is evidence that the population is not normal. A p-value above α only says the test found no clear departure, which depends heavily on the sample size.

What is a good W value in the Shapiro-Wilk test?

There is no fixed cut-off, because what counts as unusual depends on n. In a sample of 11, W = .79 is significant (p = .007); in a sample of 500, even W = .99 can be. Always read W together with its p-value and the Q–Q plot rather than against a single threshold.

Should I test the raw data or the residuals for normality?

Test what your method assumes. A t-test assumes each group (or the paired differences) is roughly normal; ANOVA and regression assume the residuals are. Pooling groups with different means into one sample can make perfectly normal groups look non-normal.

Why is my data non-normal with a large sample?

Because the test's power grows with n. With hundreds of values, tiny, harmless departures — slight skew, a few rounded values — become statistically significant. That is a reason to rely on the Q–Q plot and on the robustness of your planned method, not to transform or trim the data automatically.

How do I report a Shapiro-Wilk test in APA style?

Give W with the sample size and the p-value: W(11) = .79, p = .007. W cannot exceed 1, so it takes no leading zero; italicise W and p. Mention what you did with the result — for example, that you used a rank-based test instead of a t-test.

Is Shapiro-Wilk better than the Kolmogorov-Smirnov test?

For testing normality, generally yes. Simulation studies show the Shapiro-Wilk test has more power than the Kolmogorov-Smirnov test (and its Lilliefors correction) across most sample sizes and kinds of departure. The Kolmogorov-Smirnov test without the Lilliefors correction is also miscalibrated — far too conservative — when the mean and SD are estimated from the same data.