Question

Using R studio:

1. We will test whether the Shapiro-Wilk test is resistant to outliers. Run the code below and decide whether the presence of a single outlier [the 5] changes the ability of the test to determine normality. What conclusions can you make from this experiment?

shapiro.test(c(rnorm(100), 6))

shapiro.test(c(rnorm(1000), 6))

shapiro.test(c(rnorm(4000), 6))

Answer 1

> shapiro.test(c(rnorm(100), 6))

Shapiro-Wilk normality test

data: c(rnorm(100), 6)
W = 0.90901, p-value = 3.489e-06

> shapiro.test(c(rnorm(1000), 6))

Shapiro-Wilk normality test

data: c(rnorm(1000), 6)
W = 0.9936, p-value = 0.0002736

> shapiro.test(c(rnorm(4000), 6))

Shapiro-Wilk normality test

data: c(rnorm(4000), 6)
W = 0.99848, p-value = 0.0007784

If outlier is far away from original data then presence of single outlier can change the result of Shapiro wilk test.

But if outlier is not far away from original data then Shapiro -Wilk test give correct result even outlier is present in the data.

Consider following result:

> shapiro.test(c(rnorm(100), 4))

Shapiro-Wilk normality test

data: c(rnorm(100), 4)
W = 0.9749, p-value = 0.05094

> shapiro.test(c(rnorm(1000), 4))

Shapiro-Wilk normality test

data: c(rnorm(1000), 4)
W = 0.99789, p-value = 0.2399

> shapiro.test(c(rnorm(4000), 4))

Shapiro-Wilk normality test

data: c(rnorm(4000), 4)
W = 0.99965, p-value = 0.7486

Here 4 is outlier. Then also Shapiro Wilk test gives correct result.