I have an issue for how a sample of the population can be used to infer about the population parameters.

For example, see the following questions:

The average weekly earnings for female social workers is $\$670$. Do men in the same positions have average weekly earnings that are higher than those for women? A random sample of $n = 40$ male social workers showed $\bar{x}=\$725$ and $s= \$102$. Test the appropriate hypothesis using a $\alpha = 0.01$

This is an example in the book and in order to solve this problem we first identify the alternative hypothesis which is $H_a : \mu > 670$ where $\mu$ is the parameter mean of the men population. Thus implies that the null hypothesis is $H_0: \mu \leq 670$.

Now, the difficulty I'm facing is the following: he compute the z-score of the test statistics $\mu = 670$ as following:

$$z = \frac{\bar{x} - 670}{s/\sqrt{n}}$$

Now, this is because CLT. But, wait! CLT states the following:

Given $\sigma$ and $\mu$ then the sampling distribution of any sample with any probability distribution of the population when n is large (n>30), then the sampling distribution converges on a normal distribution where the sample mean equals the population mean (i.e. $\bar{x}_* = \mu$ ), and the sample standard deviation equals the population standard deviation divided by root square n (i.e. $s_*=\sigma/\sqrt n$).

Thus, in the question above, we don't know about neither the population standard deviation nor population mean.

So what he does is the following**: since n>30, then it is an appropriate to estimate the population standard deviation by the sample standard deviation (i.e. $s = \sigma$) and the population mean by the sample mean (i.e. $\bar{x} = \mu$). Then we have the following:

$$z = \frac{\bar{x} - 670}{s/\sqrt{n}}$$

So, my question is: I don't know how he said that since n>30, then it is possible to estimate the population standard deviation by the sample standard deviation and the population mean by the sample mean, even though this is not what CLT said. Is there anything that I am missing?

Thanks in advance

  • 1
    $\begingroup$ Does this answer your question? $\endgroup$
    – Dayne
    Dec 2, 2020 at 2:11

1 Answer 1


The sample standard deviation and sample mean are all consistent estimators of population parameters. This means "for large sample sizes", they should be pretty good approximations of the population parrameters.

The whole $n > 30$ distinction is rather arbitrary; it's what some people consider a "large sample size". Of course, this rule of thumb is not appropriate in all situations.

Also, the Central Limit Theorem you have in your question is incorrect. There are a few "central limit theorems" out there. The one I think that you are trying to reference makes an asymptotic statement about the distribution of the sample mean of a particular I.I.D sample. This is opposed to your "CLT" which makes a nonasymptotic statement about the distribution of your sample itself.

  • $\begingroup$ Thank you for your answer. Let try to understand what you said: Are you saying that the issue that I have in ** have nothing to do with CLT and what they want to say is that with large sample sizes then $s=\sigma$ and $\bar{x}=\mu$? I rewrite the theorem of CLT. Thanks for pointing this out. $\endgroup$
    – user777
    Dec 2, 2020 at 2:34
  • 1
    $\begingroup$ First, your comment. $s$ and $\bar{x}$ are both random variables. So, for any fixed sample size, you are not guaranteed equality to the population parameters. Only as the sample size approaches infinity do you get convergence in probability (still not "equality" in the conventional sense) to your population parameters. You are right that we don't need the CLT to say our estimators are approximately equal to our parameters for large $n$. Your rewritten CLT still has problems. For one, the distribution of the sample mean (not the distribution of the sample) is approximately normal for large $n$. $\endgroup$
    – user303375
    Dec 2, 2020 at 2:50
  • $\begingroup$ It seems like you have a very vague idea of what the CLT guarantees; look at the exact statement of the theorem (maybe on Wikipedia, under Lindeberg-Levy CLT) to clear up your misconceptions. $\endgroup$
    – user303375
    Dec 2, 2020 at 2:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.