Why Does a Shapiro-Wilk Test Sometimes Conclude That the Data

Multiple Choice

Why does a Shapiro-Wilk test sometimes conclude that the data are not normally distributed (i.e. significant) when a Kolmogorov-Smirnov test is non-significant?

A) The Shapiro-Wilk test is more powerful.
B) The Shapiro-Wilk test is less powerful.
C) The Shapiro-Wilk test does not work well when there are a large number of cases.
D) The Shapiro-Wilk test is a more conservative test.