# Thread: Expected values K-S test for Normality

1. ## Expected values K-S test for Normality

Hi all,

I do have a question, given are 12 observed values (see SPSS outprint in attachment)
4,1
4,3
4,6
4,9
5,0
5,1
5,1
5,6
5,6
5,6
6,0
6,4

I have no idea how to calculate the expected values for this K-S test. Can someone help me a bit, this is a test for the normal distribution.

The result is 0,1369

btw:
Just to clarify......the Java right here is solo...so it's Solo/Java

Greetings,
Java/solo

2. ## Re: Expected values K-S test for Normality

About the data: the observed values are storage times (in days) for foodstuff. In that case it is a 2 sided K-S test for normality (Goodness-of-Fit). We did see it with Chi square and the Kolmogorov-Smirnov table.

Note: in SPSS the answer is 0,143, not the result mentioned above which is a result from the textbook and might be wrong.
The question is what kind of formula is used by SPSS to obtain this result ?
Any help is greatly appreciated !!
solo

3. ## Re: Expected values K-S test for Normality

Originally Posted by Java
About the data: the observed values are storage times (in days) for foodstuff. In that case it is a 2 sided K-S test for normality (Goodness-of-Fit). We did see it with Chi square and the Kolmogorov-Smirnov table.

Note: in SPSS the answer is 0,143, not the result mentioned above which is a result from the textbook and might be wrong.
The question is what kind of formula is used by SPSS to obtain this result ?
Any help is greatly appreciated !!
solo
Click on "Help" then search "test of normality" and click on "Examine Algorithm".

Note that SPSS is using the Dallal and Wilkinson (Dallal and Wilkinson, 1986) correction to the critical values for testing normality that was reported by Lilliefors compared to what you are most likely computing based on your textbook.

4. ## The Following User Says Thank You to Dragan For This Useful Post:

Java (10-06-2014)

5. ## Re: Expected values K-S test for Normality

Thanks a lot,

I'm not familiar with Lilliefors test so I did search it on the internet. Not sure whether this is the solution of the problem. The highest value is 0,1424 but this is lower than the critical value 0,242 in the Lilliefors table.

In this case we accept the Null hypothesis. In other words: the values follow the Normal Distribution.

solo

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