# In the case of sample proportions, why do we not get a $t$-distribution when we estimate the standard deviation $\sigma_{\hat{p}}$

If $$\bar{x}$$ has a normal distribution (or approx normal via CLT), then:

$$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$$ (has a z-distribution)

If we substitute the sample standard deviation $$s$$ for the population standard deviation $$\sigma$$ we get a $$t$$-distribution with n-1 degree's of freedom:

$$t=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$$ (has a z-distribution)

Now, consider the sample proportion random variable $$\hat{p}$$. Then we have that:

$$\mu_{\hat{p}}=p$$ where p is the actual population proportion

$$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$$

If $$\hat{p}$$ has a normal distribution, then:

$$z = \frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}$$ has a z-distribution.

Now, in the former case we estimated the population standard deviation $$\sigma$$ by using the sample standard deviation $$s$$; doing this resulted in going from a $$z$$-distribution to a $$t$$-distribution. In the current case, if we don't know the population proportion $$p$$, we can estimate $$p$$ (and thus estimate $$\sigma_{\hat{p}}$$) by $$\hat{p}$$.

Thus, based on what happens in the former case, one might suspect that the random variable:

$$\frac{\hat{p}-\mu_{\hat{p}}}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}$$

has a t-distribution.

However, this is not the case, and I'd like to know why.

Why, in the first case, when we estimate the population standard deviation we get a $$t$$-distribution, but in the second case, we get a random variable that converges to a $$z$$-distribution (without ever having a chance to be a $$t$$-distribution)

Does this difference have to do that in the former case we have that the numerator and denominator are independent, whilst in the later case they are not?

Here's a fact:

A random variable T has a $$t$$-distribution if $$T = \frac{Z}{\sqrt{V/\nu}}$$, where $$Z$$ is standard normal, and $$V$$ is chi-square distributed with $$\nu$$ degrees of freedom.

Now note that $$Z = \hat{p} - p$$ is not normal. Thus, the ratio cannot be $$t$$-distributed. Furthermore, in large samples, the $$t$$-distribution is arbitrary close to the normal distribution.

• As the sample size increases, $\hat{p}-p$ quickly approaches a normal distribution though.
– user637978
Nov 3, 2022 at 19:09
• @Frogwilldo Sure, but the $t$-distribution result is only relevant in finite samples, since otherwise it is arbitrarily close to the normal distribution, as I mentioned. And for that you need normality (not just asymptotic). Nov 3, 2022 at 21:56
• so, if we are saying $\bar{x}$ is approx normal via CLT, then we should NOT say that $\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$ has a t-distribution?
– user637978
Nov 4, 2022 at 13:09
• @Frogwilldo If you can appeal to a CLT to say that $\bar{x}$ is approximately normal, then the ratio has a distribution that arbitrarily close to the normal and $t$-distribution, since asymptotically they are the same, but it will not have an exact $t$-distribution. The exact $t$-distribution only arises under the assumption of exact normality, not just asymptotic. Nov 5, 2022 at 12:36
• @Frogwilldo No problem! If you have no more questions, please accept the answer. Nov 5, 2022 at 17:54