# Understanding the role of the chi-squared distribution in the confidence interval for the variance

In my textbook they have this inequality:

$$\chi_{1-\frac{\alpha}{2}}^2 < \frac{(n-1)s^2}{\sigma^2} < \chi_{\frac{\alpha}{2}}^2$$

which later becomes this statement:

$$\frac{(n-1)s^2}{\chi_{\frac{\alpha}{2}}^2 } < \sigma^2 < \frac{ (n-1)s^2}{ \chi_{1-\frac{\alpha}{2}}^2}$$

Now I know the whole idea is to find the confidence interval for $\sigma^2$ the variance, but I was wondering if the distribution for the variance is normal. I also don't understand why the chi square is squared.

When I look at the picture in the book that shows a right skewed graph with the chi squares labeled (i.e. $\chi_{0.95}^2 = 4.575$ and $\chi_{0.05}^2 = 19.675$), I get the impression that I'm looking at something similar to $z$ scores. What are these chi squares? Do they represent the number of standard deviations away from the mean?

I have also asked this question on math.stackexchange section but I haven't heard back from anyone.

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Variance is not normally distributed, because variance is the average of the squared deviations of each datum from the mean of the distribution. If all data points in your dataset are identical, then the deviations would each be zero, and so would the squared deviations and their average. Thus, $0$ is the lowest variance possible. On the other hand, the normal distribution ranges from $-\infty$ to $\infty$. Therefore, variance cannot be normally distributed.
The chi-squared distribution is related to $z$-scores. A $z$-score is a quantile of the standard normal distribution. That is, it is the value of a data point from a normal distribution with mean $0$ and variance $1$ (e.g., if the distribution was standardized first). The distribution of $z$-scores that have been squared is $\chi^2_\text{df=1}$. To understand this connection more fully, let's examine the formula for the variance:
$$s^2 = \frac{\sum_{i=1}^N(x_i-\bar x)^2}{N-1}$$ If you were to multiply both sides by $(N-1)$ (as in the numerator of the middle of your top set of inequalities), then you simply have a sum of squares. The sum of squared deviations is distributed as chi-squared. In other words, the squaring already exists in $s^2(N-1)$, and so you need a distribution that accounts for that. (To answer one of your specific questions at this point, it is not the number of standard deviations of something from your mean.)
Now if you want a two-sided $1-\alpha$ confidence interval for anything (including this as a special case), you find the quantiles that correspond to the $\alpha/2$ percentile and the $1-\alpha/2$ percentile. In this case, you do that for the appropriate chi-squared distribution, which is chi-squared (since these are sums of squared deviations as noted above) with $\text{df} = N-1$. This value is then scaled as described in your second set of inequalities. (As to how we got from the first set to the second set, it is just algebra.)
Therefore, to answer your first question, no..the sample variance is NOT normally distributed. As for your second qestion about why its chi-SQUARED...I have no idea, but if a random standard normal variable is represented by X, then $X^2$ is distributed chi-squared with 1 degree of freedom. For me at least, I alwasy think that the chi-squared symbol looks like $X^2$, so that's my only input for this question.