Sampling Distribution of the Sample Mean Concerning estimations/intervals:


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*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!
 A: 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.
