# Is it true that if a population is normally distributed, then the sample variance for any sample size is also normally distributed, so that we may use an interval of the form [(sample variance) pm (critical value)(standard deviation of sample variance)] to estimate the population variance?

Confidence Intervals Calculation and Interpretation
Could someone verify if my reasoning is correct?
1. Is it true that if a population is normally distributed, then the sample variance for any sample size is also normally distributed, so that we may use an interval of the form [(sample variance) $±$ (critical value)(standard deviation of sample variance)] to estimate the population variance?I understand that the form statistic $±$ (critical value) ∗ (standard dev. of statistic) is applicable to mean and proportion, but is it applicable to any statistic? I believe there is a more complex formula for variance, as it follows a chi-squared distribution and is not always normally distributed, even when the population is.
2. A 90% confidence interval for the height, in meters, of adults in Switzerland is 1.78 $±$ 0.2. Are we 90% confident that the sample mean of the next sample of adults taken in Switzerland will be between 1.58 and 1.98 meters tall?
Would this also be false? I understand that a particular confidence interval of 90% calculated from an experiment does not mean that there is a 90% probability of a sample mean from a repeat of the experiment falling within this interval.
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Layton Leach
Step 1
The sample variance of a random sample ${S}^{2}$ of n observations from a normal population is not normal. Its distribution can be described as follows: $\left(n-1\right){S}^{2}/{\sigma }^{2}\sim Chisq\left(df=n-1\right),$, where $\sigma$ is the population standard deviation. This information allows one to find a confidence interval for ${\sigma }^{2},$, using the chi-squared distribution.
For such a normal sample, a confidence interval for the population mean $\mu$ is $\overline{X}±{t}^{\ast }S/\sqrt{n}$ if the population standard deviation σ is unknown and thus must be estimated by S. Here t∗ is a number that cuts 2.5% probability from the upper tail of Student's t distribution with $n-1$ degrees of freedom.
In this formula $S/\sqrt{n}$ is called the '(estimated) standard error of the mean'. That terminology arises because $SD\left(\overline{X}\right)=\sigma /\sqrt{n}.$ which is estimated by $S/\sqrt{n}.$
Step 2
If the population SD $\sigma$ is known, then a 95% CI for $\mu$ is given by $\overline{X}±1.96\sigma /\sqrt{n},$ where 1.96 cuts 2.5% probability from the upper tail of a standard normal distribution.
Unfortunately, the meaning of the word confidence in 'confidence interval' has many different interpretations. What is considered correct in one text is scorned as misleading in another. So I am not going to venture a True/False answer to the question about statures of Swiss adults.