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:


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?


1 Answer 1


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.

  • $\begingroup$ As the sample size increases, $\hat{p}-p$ quickly approaches a normal distribution though. $\endgroup$
    – user637978
    Nov 3, 2022 at 19:09
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    $\begingroup$ @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). $\endgroup$ Nov 3, 2022 at 21:56
  • $\begingroup$ 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? $\endgroup$
    – user637978
    Nov 4, 2022 at 13:09
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    $\begingroup$ @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. $\endgroup$ Nov 5, 2022 at 12:36
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    $\begingroup$ @Frogwilldo No problem! If you have no more questions, please accept the answer. $\endgroup$ Nov 5, 2022 at 17:54

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