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Normality
Shapiro-Wilk Test — Normal & Non-Normal Worked Examples
Two examples: systolic BP confirming normality W=0.972, p=.895; ER waiting times detecting skew W=0.825, p=.030.
📊 11a: W=0.972 ✓ | 11b: W=0.825 ✗
Step-by-step solution
📐 11 · Normality Testing (Shapiro-Wilk)
11a · Systolic BP — Confirming Normality
Normality
Research Question:
Before running a t-test, verify that systolic blood pressure readings (n=10) are normally distributed.
| i | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| BP (mmHg) | 118 | 120 | 122 | 124 | 124 | 126 | 128 | 130 | 133 | 135 |
Shapiro-Wilk W Statistic
$$W = \frac{b^2}{SS},\quad b = \sum_{i=1}^{\lfloor n/2\rfloor} a_i\bigl(x_{(n+1-i)}-x_{(i)}\bigr),\quad SS=\sum(x_i-\bar{x})^2$$
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1Sort data ascending; compute mean and SS:$$\bar{x}=126.0,\quad SS=(118{-}126)^2+\cdots+(135{-}126)^2=274$$
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2Apply SW coefficients (n=10): a₁=0.5739, a₂=0.3291, a₃=0.2141, a₄=0.1224, a₅=0.0399$$x_{(10)}-x_{(1)}=17,\; x_{(9)}-x_{(2)}=13,\; x_{(8)}-x_{(3)}=8,\; x_{(7)}-x_{(4)}=4,\; x_{(6)}-x_{(5)}=2$$ $$b=0.5739(17)+0.3291(13)+0.2141(8)+0.1224(4)+0.0399(2)=9.756+4.278+1.713+0.490+0.080=16.317$$
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3$$W=\frac{(16.317)^2}{274}=\frac{266.2}{274}=\mathbf{0.972}$$
0.972
W
.895
p
10
n
⚪ Fail to Reject H₀ — Data is consistent with normality. Proceed with t-test. W(10) = 0.97, p = .895.
APA-7
Shapiro-Wilk testing confirmed that blood pressure scores were approximately normally distributed, W(10) = 0.97, p = .895.
11b · Emergency Waiting Times — Non-Normal (Right-Skewed)
Non-Normal
Research Question:
Are emergency department waiting times (minutes, n=10) normally distributed?
| i | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| Minutes | 5 | 8 | 10 | 12 | 15 | 20 | 35 | 52 | 78 | 95 |
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1$$\bar{x}=33.0,\quad SS=9{,}106$$ $$\text{Differences: }x_{(10)}-x_{(1)}=90,\;70,\;42,\;23,\;5$$ $$b=0.5739(90)+0.3291(70)+0.2141(42)+0.1224(23)+0.0399(5)=51.65+23.04+8.99+2.82+0.20=86.70$$ $$W=\frac{(86.70)^2}{9106}=\frac{7516.9}{9106}=\mathbf{0.825}$$
0.825
W
.030
p
<.05
Violation
🔴 Reject H₀ — Waiting times are NOT normally distributed, W(10) = 0.83, p = .030. Use non-parametric alternatives (e.g., Mann-Whitney U).
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Decision rule: p > .05 → normality assumed (safe to proceed with parametric tests). p ≤ .05 → use non-parametric tests or apply log/square-root transformation. For large samples (n > 50), prefer Kolmogorov-Smirnov or visual Q-Q plots.