Normal, t, chi square distribution confidence intervals

I've been looking into different ways of calculating confidence intervals and I came across methods using the normal distribution, t-distribution and chi square distribution.

I want to find the confidence intervals for a binomial distribution with a high sample size (around N=5000). I am wary of using a normal approximation to the binomial distribution not because it shouldn't be continuous but that it shouldn't have negative values (which the normal distribution does).

I have two questions

1. Is the t-distribution always better than a normal distribution for finite values of N (or rather N less than the population size)? I realise at N=5000 the difference is negligible but I am wondering about this from a theory point of view.

2. Is their any distribution that always gives the best estimate of the confidence interval for a sample smaller than the population?

Re: Normal, t, chi square distribution confidence intervals

Hi Shakarri! :)

The z-distribution (which is the same as the normal distribution) is always better than the t-distribution if and only if you have knowledge of the standard deviation of the population ($\displaystyle \sigma$).

Since you are talking about a binomial distribution that implies you have knowledge about $\displaystyle \sigma$. Therefore the normal distribution is a better approximation than the t-distribution.

There is no single distribution that always gives the best estimate for a confidence interval.

It depends on the (assumed) distribution of the population (normal, binomial, uniform, poisson, ...).

Btw, in the tails of any distribution the probability density becomes very unreliable in practice, since there are always other effects that are unaccounted for.

In practice extreme outcomes are more probable than any normal distribution predicts.

Re: Normal, t, chi square distribution confidence intervals

Thank you for the information. Particularly about tails of the distribution.

Just to confirm, when you say "knowledge of the standard deviation" you mean a having a sample standard deviation suffices? in what I am sampling it would be impossible to get the real standard deviation.

Re: Normal, t, chi square distribution confidence intervals

No, you have omitted the part of my statement that says: the standard deviation *of the population* (usually denoted as $\displaystyle \sigma$).

This is *not* the standard deviation *of the sample* (usually denoted as s).

This is the key difference between the z-test and the t-test.

If you do not have the standard deviation *of the population*, you cannot apply the z-test and are stuck with the t-test that uses the standard deviation *of the sample*.

Since you are talking about a binomial distribution, this implies knowledge of the standard deviation *of the population*: $\displaystyle \sigma = \sqrt{np(1-p)}$.

Re: Normal, t, chi square distribution confidence intervals

I noticed you used sigma not s, I was just unsure what "knowledge" meant. I was confused by you saying I implied I have knowledge of sigma even though I said that my sample size was less than the population, the population size is infinite so I can never get the true vale of the standard deviation.

Anyway, cheers for the advice, I'll sick with t tests unless the distribution is not approximately normal.