Thread Subject: glmfit statistics

I am trying to decipher the statistics returned by glmfit.

If I take the fit parameter returned and divide it by the sqrt of the covariance returned for that same parameter, I get a number that equals the t statistic returned. This leads to my first question, which is shouldn't I also have to normalize by the sqrt of n, which n is the number of samples? (Or is n somehow equal to 1 for this fit?)

My second question is that when I work with the t-value returned, I cannot figure out how the p-value is calculated. I am assuming that the hypothesis being tested is the fit parameters returned are equal to zero. But I cannot find a value for the degrees of freedom that will yield the p-values returned given the t-values returned.

Hope that this makes sense. I am probably just missing something simple.

Sincerely,
Dan

> If I take the fit parameter returned and divide it by the sqrt of the
> covariance returned for that same parameter, I get a number that equals
> the t statistic returned. This leads to my first question, which is
> shouldn't I also have to normalize by the sqrt of n, which n is the number
> of samples? (Or is n somehow equal to 1 for this fit?)

Dan, the covariance matrix that you mentioned is an estimate of the
uncertainty in the coefficients, so it already takes the number of samples
into account.

> My second question is that when I work with the t-value returned, I cannot
> figure out how the p-value is calculated. I am assuming that the
> hypothesis being tested is the fit parameters returned are equal to zero.
> But I cannot find a value for the degrees of freedom that will yield the
> p-values returned given the t-values returned.

The d.f. is n-p, where n is the number of observations and p is the number
of estimated coefficients. The p-value is two-sided, so for example if you
have t=2.34 then the p-value is the probablity that the t distribution
assigns to values greater than 2.34 plus the probability of values less
than -2.34. Could that explain the difference you found?

-- Tom

Daniel wrote:
> I am trying to decipher the statistics returned by glmfit.
>
> If I take the fit parameter returned and divide it by the sqrt of the covariance returned for that same parameter, I get a number that equals the t statistic returned. This leads to my first question, which is shouldn't I also have to normalize by the sqrt of n, which n is the number of samples? (Or is n somehow equal to 1 for this fit?)

That cov matrix is an estimate of the covariance of the parameter estimates (estimators, technically), so the number of observations is already figured in. In a simpler context, where x is just a vector of data drawn from a normal dist'n, it's like the difference between std(x), which is an estimate of the population std dev sigma, and std(x)/sqrt(length(x)), which is the std error of muHat = mean(x), i.e. an estimate of the variance of muHat.

> My second question is that when I work with the t-value returned, I cannot figure out how the p-value is calculated. I am assuming that the hypothesis being tested is the fit parameters returned are equal to zero. But I cannot find a value for the degrees of freedom that will yield the p-values returned given the t-values returned.

It's a two-tailed p-value, and the degrees of freedom is given by the dfe field of that same structure. In this case, dfe is (number of observations) - (number of estimated regression coefs).

> Hope that this makes sense. I am probably just missing something simple.

Often the best way to sort this out is to just step through the code in the debugger.

Thanks to both of you for your comments. I had considered the issue of two-sided versus one-sided and had tried it both ways. Hmm, maybe I was just having a bad hair day. I'll go try it again and see if I can figure out what I did wrong yesterday.