# goodness of fit test to check for normality

bad ability average good ability total
19 64 17 100

It is said that average ability is everything +/- 1sd. This is all data I have. I need to make hypotisses about normal distrubution and do normality distribution test. I am not sure what test? I tried chi squared:

• Ho: distribution is normal
• Ha: distribution is not normal

I know that normal distribution has this characteristic

left tail μ ± 1σ right tail
15.85% 68.3% 15.85%

So I have decided to do chi square test because I can compare observed and expected frequency.

bad ability average good ability total
observed 19 64 17 100
expected 15.85 68.3 15.85
(obs-exp)^2/exp 0.626 0.271 0.0834 0.98

So I got 𝛘2 = 0.98 2crit, .025 = 7.38 and I don't reject Ho.

But, I am nut sure in df. I have used 3-1 = 2 df and 𝛘2 distibution table. Looking great

Real Statistics in Excel

I have seen that maybe I should calculate df = 3-1-2 = 0, but that obviously isn't good result.

And I am not sure if should use 𝛘2crit, .05 = 5.99. Maybe this critical value is for one-sided test.

• Your variable is an ordered categorical variable with 3 distinct values. There is no reason to expect it to be normally distributed and little point in testing such. What are the grounds for expecting (a) average ability to be within 1 SD of the mean (b) even a symmetric distribution for a latent variable hearing ability, let alone that such a variable is normally distributed? Commented Dec 30, 2020 at 18:03
• Is this a homework problem or similar self-study question? If so, please read this page for how such questions are handled on this site, and add the self-study tag to the question.
– EdM
Commented Dec 30, 2020 at 18:11
• Yes I understand you comment but that was the task on university. Assumption about +/- 1SD is given as conditions in task. I saw much comments like yours but that was the task given to me. I don't know what to say
– piga
Commented Dec 30, 2020 at 18:15
• @EdM now I have added self-study tag.
– piga
Commented Dec 30, 2020 at 18:20

Your choice of a chi-square test is standard for this type of problem. With respect to the number of degrees of freedom (df), the Wikipedia page provides guidance for a goodness-of-fit test:

For a test of goodness-of-fit, df = Cats − Parms, where Cats is the number of observation categories recognized by the model, and Parms is the number of parameters in the model adjusted to make the model best fit the observations: The number of categories reduced by the number of fitted parameters in the distribution.

You have 3 categories, so that part is easy. It doesn't seem that you have any parameters that you "adjusted to make the model best fit the observations."* With these tests, however, you always need to reduce the df by 1 because the total of observations must equal the sum of the individual categories; see the Wikipedia entry on a similar test for the uniform distribution.

Unlike the t-test, the chi-square test is inherently one-sided.

*As I read the problem, the SD of the population used to classify into "bad," "average" or "good" ability was determined independently of this particular data sample. Otherwise you would have to correct for the 2 extra parameters estimated from the data (mean and standard deviation), leading to 0 degrees of freedom! Sometimes a simple homework question like this hides some important hidden assumptions, from which you can learn even more.

• Thanks. I agree. So testing and conclusion was correct just at the end I should have used critical value 5.99. At the end it was simple but I really needed 2 days thinking and searching.
– piga
Commented Dec 30, 2020 at 21:45
• @piga those days of thinking and searching are what really help you learn.
– EdM
Commented Dec 30, 2020 at 22:36