# Sampling Distribution of the Sample Mean

Concerning estimations/intervals:

• True or False: if the three conditions Random, Normal, and Independent for using a confidence interval for a population mean are not met, then the sampling distribution of $\bar x$ is unknown.

I'm unsure about the criteria - would you not know the distribution of the sample mean even if these criteria are unfilled? Are there specific conditions for doing so?

Thanks!

• Technically speaking, if you don't know the underlying distribution of the data, then you cannot know the distribution of any statistic calculated from that data. However, practically speaking, many iid samples will have roughly normal sample means, provided your data do not come from a heavy tailed or highly skewed distribution. – user237392 Mar 23 '16 at 5:16

## 1 Answer

It is difficult to make valid true-false questions about general statistical principles, especially ones as detailed as this one. In making such questions, the danger is that one might not have thought of a situation beyond the material in the current chapter. My guess is that the authors of this one intended the answer to be True, but I don't think so.

Let $X_1, X_2, \dots, X_{10}$ be a random sample from $Exp(rate=\lambda),$ the exponential distribution with rate $\lambda$ and mean $\mu =1/\lambda.$ Then $\bar X \sim Gamma(shape = n, rate=n*\lambda),$ which has mean $E(\bar X) = \mu = 1/\lambda$, variance $V(\bar X) = \frac{\mu^2}{n} = \frac{1/\lambda^2}{n}$ and $SD(\bar X) = \frac{\mu}{\sqrt{n}} = \frac{1/\lambda}{\sqrt{n}}.$ [The reason for showing both $\mu$'s and $\lambda$'s is that some books define the exponential distribution using the mean and others using the rate.]

Then when using data $X_1, X_2, \dots, X_{10}$ to estimate the population mean $\mu,$ the standard error (standard deviation) of the mean is $\mu/\sqrt{n}.$

Thus you have (1) random sampling and (3) independence, but not (2) normality. However, the standard error is known.

For sufficiently large $n,$ by the Central Limit Theorem, $\bar X$ is $approximately$ normal. So for large $n$, one might use the $\bar X \pm 1.96\bar X/\sqrt{n}$ as an approximate 95% confidence interval for $\mu$.

A better 95% CI for $\mu,$ based on the gamma family of distributions, is valid for all $n$: Let $L$ and $U$ cut 2.5% from the lower and upper tails of $Gamma(n, n)$ (found using software). Then the confidence interval for $\mu$ is $(\bar X/U, \bar X/L).$ This method does not use the standard error.