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) $\pm$ (critical value)(standard deviation of sample variance)] to estimate the population variance?

I understand that the form statistic $\pm$ (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.

  1. A 90% confidence interval for the height, in meters, of adults in Switzerland is 1.78 $\pm$ 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.


  • 1
    $\begingroup$ Answer to your deleted question: You know the expected number which is 5. Therefore you "estimate" the standard deviation by using the known proportion: $\sigma_{ p}=\sqrt{0.5^2/10}=0.1581$ If you would not know the true proportion then you would estimate the standard deviation of the sample mean by using the result of the sample: $\hat \sigma_{ p}=\sqrt{0.4\cdot 0.6/9}=0.1633$ Here you have to divide by n-1. By doing that the estimator is unbiased. $\endgroup$ – callculus Mar 19 '16 at 19:16

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: $(n-1)S^2/\sigma^2 \sim Chisq(df=n-1),$ 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 $\bar X \pm t^*S/\sqrt{n}$ if the population standard deviation $\sigma$ 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 $SD(\bar X) = \sigma/\sqrt{n}.$ which is estimated by $S/\sqrt{n}.$

If the population SD $\sigma$ is known, then a 95% CI for $\mu$ is given by $\bar X \pm 1.96\sigma/\sqrt{n},$ 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.


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