goodness of fit test to check for normality I have task about hearing ability and here is frequency table




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.
So I am suspicious about my test. And IComments and answers are welcome. Thanks
 A: 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.
