Example: Shapiro-Wilk W Test

The following example is taken from Conover (1980, pp. 195, 364). The data consists of 50 two-digit numbers taken from a telephone book. The W test fails to reject the null hypothesis of normality at the .05 level of significance.

using System;
using Imsl;
using Imsl.Stat;

public class NormalityTestEx1
{
	public static void  Main(String[] args)
	{
		double[] x = new double[]{
									 23.0, 36.0, 54.0, 61.0, 73.0, 23.0,
									 37.0, 54.0, 61.0, 73.0, 24.0, 40.0,
									 56.0, 62.0, 74.0, 27.0, 42.0, 57.0,
									 63.0, 75.0, 29.0, 43.0, 57.0, 64.0,
									 77.0, 31.0, 43.0, 58.0, 65.0, 81.0,
									 32.0, 44.0, 58.0, 66.0, 87.0, 33.0,
									 45.0, 58.0, 68.0, 89.0, 33.0, 48.0,
									 58.0, 68.0, 93.0, 35.0, 48.0, 59.0, 
									 70.0, 97.0};
		
		NormalityTest nt = new NormalityTest(x);

		Console.Out.WriteLine
			("p-value = " + nt.ShapiroWilkWTest().ToString("0.0000"));
		Console.Out.WriteLine("Shapiro Wilk W Statistic = " + 
			nt.ShapiroWilkW.ToString("0.0000"));
	}
}

Output

p-value = 0.2309
Shapiro Wilk W Statistic = 0.9642