When constructing, for example, a $90\%$ confidence interval for the population variance using the chi-squared distribution, we have:
\begin{align}
& P\left(a<\frac{(n-1)S^2}{\sigma^2}<b\right) \\
= {} & P\left(\frac{\sum(\bar{X}-X_i)^2}{b}<\sigma^2<\frac{\sum(\bar{X}-X_i)^2}{a}\right)=0.9, \\ & \text{ where } \frac{(n-1)S^2}{\sigma^2}\sim\chi_{n-1}^2.
\end{align}
In my course, we then find $a, b$ such that $$P(\chi_{n-1}^{2}<a)=P(\chi_{n-1}^{2}>b)=0.05.$$
My question is, given that the chi-squared distribution is asymmetric for small $n,$ why do we pick $0.05$ for both sides? Surely we’d have a shorter confidence interval if we weighted more of the $0.1$ probability to one of the sides, instead of splitting equally?
Best Answer
Because the chi-squared distribution is skewed, the sample variance is not generally at the center of a 95% CI for the variance (for normal data).
You are correct to say that you can often get a narrower interval by taking something like probability 2% from one tail and 3% from the other, than by taking 2.5% from each tail.
For practical purposes, the narrowest 95% interval may put almost all of the 5% probability in one tail, thus becoming nearly a one-sided interval. This may or may not be useful.
Thus, it has become more or less standard to use probability-symmetric intervals in general practice. If you are not showing a probability-symmetric interval, it is a good idea to report that you are not, and to explain why.
Example: Consider a normal sample of size $n=20$ with variance $\sigma^2 = 25.$
Seven 2-sided 95% CIs for $\sigma^2$ and their widths:
Note: The relevant one-sided 95% CI would give the upper bound $46.97848.$ Depending on the application, that might be exactly what you want.