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Let $X_1, ..., X_n$ be a random sample from a normal distribution with an unknown $\mu$ and unknown $\sigma$. The sample mean of this sample is $\bar{X}$. The following graph shows the density curve of the sampling distribution of the sample mean: enter image description here

The yellow shaded region represents where 95% of random sample means would fall and has an interval of $(\mu - L, \mu + L)$. Since there is a 95% chance that any random sample mean would fall in that interval, we can also say with 95% confidence that a given sample mean would contain the population mean within its confidence interval (CI). Finding the CI then for a sample mean simply requires centering the interval around the sample mean: $$(\mu - L + (\bar{X} - \mu), \mu + L + (\bar{X} - \mu)) = (\bar{X} - L, \bar{X} + L)$$

where $L = t_{\alpha/2} * \frac{s}{\sqrt(n)}$.

Unfortunately, this approach seems to fall apart when I try to apply it to finding the confidence interval for the sample variance. The sample variance has a sampling distribution of $\chi^2$ with $n - 1$ degrees of freedom. We can convert a sample variance to its corresponding $\chi^2$ value with the following formula:

$$\chi^2 = \frac{(n - 1)s^2}{\sigma^2}$$

Here is a plot of a chi-square distribution with $df = 4$. It can represent the distribution of the "chi-transformed" sample variances.

enter image description here

The red line is the mean of the distribution, the blue lines represent the $\chi^2$ values below which and above which lie 5% of the data. And the dotted green line represents the "chi-transformed" value of our theoretical sample variance.

I can no longer apply my "shift the interval" strategy because the distribution is skewed and also cannot go below 0. How do I intuit (mathematically and visually) the 90% confidence interval of the sample variance?

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  • $\begingroup$ What does the green line represent? $S^2$ or $ (n-1)S^2$? Not clear. Specify your parameters and the corresponding numbers. $\endgroup$ Commented Feb 20, 2022 at 0:50
  • $\begingroup$ @callculus42 Green line represents $\frac{s^2(n-1)}{\sigma^2}$ where s is the variance of a sample with 5 observations. $\endgroup$
    – Nova
    Commented Feb 20, 2022 at 3:07
  • $\begingroup$ For a normal model, the mean $\mu$ is a location parameter; it is estimated by $\bar X.$ which has a symmetrical distribution. The SD $\sigma$ is a scale parameter; it is estimated by $S,$ which has a right-skewed distribution. You may have to adjust your 'intuition' when moving from a CI for $\mu$ to a CI for $\sigma.$ $\endgroup$
    – BruceET
    Commented Feb 20, 2022 at 21:33

1 Answer 1

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General construction of confidence intervals

In general, let $W_n (\theta) $ denote a pivotal scalar statistic (dependent upon parameter $\theta$ and sample size $n$) with invertible CDF $F$ and $q$ quantile $w_q=F^{-1}(q)$. This notation means $w_q$ has left-tail area $q$ (in contrast to your notation, which uses the right-tail area).

Then the way we construct a $1-\alpha$ coverage CI for $\theta$ is by choosing any $q_1,q_2\in (0,1): q_2-q_1=1-\alpha$ and using the fact that

$$w_{q_1}\leq W_n(\theta)\leq w_{q_2}\quad (1)$$

occurs with probability $1-\alpha$. That is, $P(W_n(\theta)\in [w_{q_1},w_{q_2}])=1-\alpha.$ Here is an illustration:

enter image description here

The actual CI bounds for $\theta$ are then given by $W_n^{-1}(w_{q_1}),W_n^{-1}(w_{q_2}).$

Example: confidence interval for $\mu$ (iid normal data)

Let's take your example of finding the CI for the population mean $\mu$. If $\frac{\bar X-\mu}{S/\sqrt n}$ has a $t({\small \text{df}=n-1})$ distribution, then by $(1)$ we may construct a $1-\alpha$ coverage CI for $\mu$ $\forall q_1\in (0,\alpha)$:

$$t_{q_1}({\small \text{df}=n-1})\leq \frac{\bar X-\mu}{S/\sqrt n}\leq t_{q_1+1-\alpha}({\small \text{df}=n-1})\\ \implies \bar X-t_{q_1+1-\alpha}({\small \text{df}=n-1})\frac{S}{\sqrt n}\leq \mu\leq \bar X-t_{q_1}({\small \text{df}=n-1})\frac{S}{\sqrt n}.$$

Now recall the symmetry of the $t$ distribution: $t_{\alpha/2}({\small \text{df}})=-t_{1-\alpha/2}({\small \text{df}}) $. In the special case where you choose $q_1=\alpha/2$, then this symmetry implies a confidence interval of

$$\bar X-t_{1-\alpha/2}({\small \text{df}=n-1})\frac{S}{\sqrt n}\leq \mu\leq \bar X+t_{1-\alpha/2}({\small \text{df}=n-1})\frac{S}{\sqrt n},$$

which is exactly what you obtained (except for the notational difference mentioned before).

Example: confidence interval for $\sigma^2$ (iid normal data)

We can do the exact same procedure for finding the CI of the population variance $\sigma^2$. If $\frac{(n-1)S^2}{\sigma^2}$ has a $\chi^2({\small \text{df}=n-1})$ distribution, then by $(1)$ we may construct a $1-\alpha$ coverage CI for $\sigma^2$ $\forall q_1\in (0,\alpha)$:

$$\chi^2_{q_1}({\small \text{df}=n-1})\leq \frac{(n-1)S^2}{\sigma^2}\leq \chi^2_{q_1+1-\alpha}({\small \text{df}=n-1})\\ \implies \frac{(n-1)S^2}{\chi^2_{q_1+1-\alpha}({\small \text{df}=n-1})}\leq \sigma^2\leq\frac{(n-1)S^2}{\chi^2_{q_1}({\small \text{df}=n-1})}.$$

You could choose $q_1=\alpha/2$, but there is no symmetric simplification as in the case of the $t$ distribution.

Final thoughts

Finally, I would like to point out that it may seem arbitrary that we chose $q_1=\alpha/2$ in the above (since any $q_1\in (0,\alpha)$ can result in $1-\alpha$ coverage). I asked about this previously here, and you may find that post to be of interest.

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  • $\begingroup$ Thank you for your answer! I had a hard time following the "general construction of confidence intervals" initial section. Is there a resource you recommend that goes into the intuition behind the general construction in a step-by-step fashion? $\endgroup$
    – Nova
    Commented Feb 22, 2022 at 19:11
  • $\begingroup$ @Nova No prob. I know I use some terminology/notation in the "general construction" section but it was hard to avoid that because, well, it's general. Did you follow the two examples though? If you did, then you got the point. $\endgroup$ Commented Feb 22, 2022 at 19:17
  • $\begingroup$ @Nova As far as resources, it really depends on your background. You could try wiki for a start. $\endgroup$ Commented Feb 22, 2022 at 19:17
  • $\begingroup$ Yeah, the examples are great because I see how you are rearranging terms to get the statistic of interest in the middle of the inequalities. I guess what I am missing is what does $t_{q_1}$ and $t_{q_1 + 1-\alpha}$ terms mean. Are those the values on the t-distribution that correspond to certain areas/probabilities (like 0.05/0.95)? $\endgroup$
    – Nova
    Commented Feb 22, 2022 at 19:22
  • $\begingroup$ @Nova Right. I added a pic...lmk if that makes sense $\endgroup$ Commented Feb 22, 2022 at 19:31

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