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As far as I have studied, given a normal random sample, we can build the confidence interval of the mean $\mu$ if we know the variance through the relation \begin{align*} \frac{\sqrt{n}(\overline{X}-\mu)}{\sigma}\sim\mathcal{N}(0,1) \end{align*}

where $n$ stands for the size of the sample. On the other hand, if we do not know the variance, we can make use of the relation \begin{align*} \frac{(n-1)S^{2}}{\sigma}\sim\chi^{2}_{n-1} \end{align*} to build the confidence interval for it, where $S^{2}$ stands for the sample variance. And, finally, if we do not know the mean neither the variance, we can build the confidence interval for the mean according to the relation \begin{align*} \frac{\sqrt{n}(\overline{X}-\mu)}{S}\sim t_{n-1}(0,1) \end{align*}

Here it is my question: can we convert the last relation into a normal distribution when $n$ is big enough? That is to say, $n \geq 30$?

I still have one more question. Given information about the sample mean and variance as well as its sample size, how can I decide the best pivot to use? Thanks in advance!

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All relationships below are based on normal data.

If $\mu$ is unknown and $\sigma^2$ is known, the first relationship can be used to get a confidence interval for $\mu.$

If neither $\mu$ nor $\sigma$ is known, the second one can be used to get a CI for $\sigma^2.$ Notice that $\bar X$ estimating $\mu$ is required in the computation of $S^2.$

If $\sigma^2$ is unknown and $\mu$ is known, then $V = \frac 1 n \sum_i (X_i - \mu)^2$ estimates $\sigma^2,$ and the relationship

$$\frac{nV}{\sigma^2} \sim \mathsf{Chisq}(n)$$

can be used to find a CI for $\sigma^2.$

If neither $\mu$ nor $\sigma^2$ is known then the third relationship, corrected to read

$$\frac{\sqrt{n}(\bar X - \mu)}{S} \sim \mathsf{T}(n-1),$$ based on Student's t distribution with $n-1$ degrees of freedom, can be used to find a CI for $\mu.$

For sufficiently large $n,$ the distribution $\mathsf{T}(n-1)$ is very similar to standard normal. In particular, $n \ge 30$ is large enough to get serviceable 95% CIs, but this 'rule of 30' should not be used for other confidence levels without checking how well the tails of standard normal and Student's t distribution match for the appropriate quantiles. For example, at quantile $0.975$ standard normal and $\mathsf{T}(30)$ tail probabilities are similar, but not so close at quantile $0.975.$ [Computations in R.]

qnorm(.975);  qt(.975, 30)  # both near 2.0 for 95% CI
[1] 1.959964
[1] 2.042272

qnorm(.995);  qt(.995, 30)  # not so near each other for 99% CI
[1] 2.575829
[1] 2.749996
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  • $\begingroup$ In the first place, thanks for the contribution. Thus, If I am given the sample mean and the sample variance, I can substitute the t-Student distribution on the third relation by the standard normal distribution to determine the confidence interval of $\mu$, if $n$ is large enough. Is it right? $\endgroup$ – user1337 Apr 13 '19 at 1:33
  • $\begingroup$ Maybe you mean substitute std normal for t. Would be OK, if you're making a 95% CI and don't mind dealing with an approximation. $\endgroup$ – BruceET Apr 13 '19 at 1:37
  • $\begingroup$ As far as I have understood, we use the t-Student distribution when we do not know the mean neither the variance. In this case, we substitute the variance by its sample counterpart. However, when we do this, the new pivotal quantity has a t-Student distribution. My question is: does the t-Student distribution still appropriate when $n\geq 30$? Or do we need to exchange it by the standard normal distribution? $\endgroup$ – user1337 Apr 13 '19 at 1:40
  • $\begingroup$ I have set out conditions where this might be OK near then end of my answer, but only because you raised the issue in your question. I wish this often-misapplied 'rule of 30' (a relic of the pre-computer era) would just quietly go away. (I might use it if making CIs on a desert island with an abacus for computation.) $\endgroup$ – BruceET Apr 13 '19 at 1:44
  • $\begingroup$ Ok, I got it. Thanks for the contribution. $\endgroup$ – user1337 Apr 13 '19 at 1:47
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HiL The first relation only holds when $X$ is normally distributed. The third relation also only holds when $X$ is normally distributed. These statements are true regardless of the value of $n$.

Now, the third relation can also hold through the CLT holding, if $n$ is large enough. But, if the CLT was holding due to $n$ being large enough, then the first relation holds also, so you probably wouldn't use the third relation anyway. Whether $n$ is large enough is a difficult question and is probably best decided on through the use of histograms, normal probability plot etc. Stay away from tests for normality.

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