A chi square test on truncated normal distribution? Result interpretation

Hi everyone,

I have some data which looks to be a truncated normal distribution. So I want to do a chi square test to see if that's true.

I found that chi2gof allows me to use a custom cdf, so I prepared the following codes:

m = 10000;
mu = 65;
sigma = 20;
x = normrnd(mu,sigma,m,1);
%remove data above and below truncation points
b = 110;
x(x >= b) = [];
a = 30;
x(x < a) = [];

First I fit data into a truncated normal distribution, as posted before. http://www.mathworks.com/matlabcentral/newsreader/view_thread/100921

pdf_truncnorm = @(x,mu,sigma) normpdf(x,mu,sigma) ./ (normcdf(b,mu,sigma)-normcdf(a,mu,sigma));
start = [mean(x),std(x)]
[paramEsts,paramCIs] = mle(x, 'pdf',pdf_truncnorm, 'start',start, 'lower',[-Inf 0])

Now I perform the chi square test:

cdf_truncnorm = @(x,mu,sigma) ( normcdf(x,mu,sigma)- normcdf(a,mu, sigma) )./ (normcdf(b,mu,sigma)-normcdf(a,mu, sigma))

[h,p] = chi2gof(x,'cdf',{cdf_truncnorm,paramEsts(1),paramEsts(2)})

*Is this the right way to do it? *

One other thing that gets me curious is the confidence interval from mle. It is pretty tight:

paramEsts =

     65.1106   19.9608

paramCIs =

     64.6396   19.5126
     65.5816   20.4089

But the p-value from chi-square test is actually low:

h =

           0

    p =

        0.2361

Is there a necessary relation between confidence interval form MLE and p-value from chi2gof? Actually I found that if I fit a uniform data into normal distribution using MLE, the confidence interval is not wide either. How should that be interpreted? Thanks a lot!

Rui