# Justify the use of KS over Shapiro and Wilk normality test

#### skoui

##### New Member
Hello there.
I am testing my data distribution for normality using both Shapiro Wilk and Kolmogorov and Smirnov tests in SPSS.
S&W reveals that the data is not normally distributed whereas K&S shows its approximately normally distributed.
How can i justify the use of K&S over S&W in order to proceed with parametric tests?
My n=24

#### Dason

You are mistaken in your understanding of these tests. KS really shouldn't be used to test for general normality because it assumes you know the parameters of the distribution (and if you're testing for normality then I'm guessing you don't know the parameters). Also - failing to reject a test of normality DOES NOT MEAN THE DATA IS NORMAL. It means you didn't find evidence that there is a lack of normality. That's entirely different than finding evidence for normality. It's like if you never went outside and then declared that the air outside is toxic. You don't have the data to prove your claim wrong - but that doesn't mean you're right.

Honestly if you're looking at this and you did a Shapiro-Wilks test and it rejected the null then that makes you have evidence your data isn't normally distributed. That's how I see it - you don't have normally distributed data. Now whether that matters depends more on what you were planning on doing and your sample size...

#### Mean Joe

##### TS Contributor
Looking at the way the Shapiro Wilk statistic is calculated, maybe it is not so good if you have a small range, eg your x range from 23.05 to 23.07. In such cases the denominator of the statistic will be tiny.

#### Dason

Looking at the way the Shapiro Wilk statistic is calculated, maybe it is not so good if you have a small range, eg your x range from 23.05 to 23.07. In such cases the denominator of the statistic will be tiny.
The numerator will be less than or equal to the denominator.

#### Mean Joe

##### TS Contributor
That's true, and in honesty I was just trying to find something that might spark OP's thinking.
If you do have a small measured range, where the difference is only one significant digit, the precision could be a factor, and compounded by dividing by a small number.