chi-squared-distributionhypothesis testingintuition
I understand that the chi squared distribution is the square of a normal distribution. If X is a normal variable, what are we squaring, the values of X or the corresponding probabilities?
I am unable to see the intuitive explanation as to how by squaring a normal variable, I can use the resultant distribution for hypothesis testing and confidence interval?
[I'll assume from the discussion in your question that you're happy to accept as fact that if $Z_i, i=1,2,\ldots,k$ are independent identically distributed $N(0,1)$ random variables then $\sum_{i=1}^{k}Z_i^2\sim \chi^2_k$.]
Formally, the result you need follows from Cochran's theorem. (Though it can be shown in other ways)
Less formally, consider that if we knew the population mean, and estimated the variance about it (rather than about the sample mean): $s_0^2 = \frac{1}{n} \sum_{i=1}^{n}(X_i-\mu)^2$, then $s_0^2/\sigma^2 = \frac{1}{n} \sum_{i=1}^{n}\left(\frac{X_i-\mu}{\sigma}\right)^2=\frac{1}{n} \sum_{i=1}^{n}Z_i^2$, ($Z_i=(X_i-\mu)/\sigma$) which will be $\frac{1}{n}$ times a $\chi^2_n$ random variable.
The fact that the sample mean is used, instead of the population mean ($Z_i^*=(X_i-\bar{X})/\sigma$) makes the sum of squares of deviations smaller, but in just such a way that $\sum_{i=1}^{n}(Z_i^*)^2\,\sim\chi^2_{n-1}$ (about which, see Cochran's theorem). That is, rather than $ns_0^2/\sigma^2\sim \chi^2_n$ we now have $(n-1)s^2/\sigma^2\sim\chi^2_{n-1}$.
Following the logic of the case with 1 df, in the case of 2 dfs we can also be expressed the threshold as the 97.5 percentile of a N(0, ~1.25) Normal distribution, which is about the square-root of 6.0.
In the 1 df case it doesn't just match at the 97.5 percentile -- all of the percentiles correspond in a similar way.
While there's definitely a connection in the 1 df case, this is not really relevant for other degrees of freedom. How did you arrive at 1.25 (aside from matching its 97.5 quantile to $\sqrt{6}\,$)?
You could as easily say "well, see it's connected to a logistic distribution" and then compute the corresponding scale parameter for a logistic by matching two quantiles in a similar fashion. That doesn't imply they're related in any meaningful sense.
For the chi-squared(1), the two corresponding curves lie one atop the other, rather than just crossing somewhere.
The sum of two chi-squared(1) variates is exponential with mean 2. You can match any individual quantile to some normal distribution but you can't choose a single standard deviation that will make all the quantiles correspond at the same time like you can with 1.df
e.g. The 75th percentile of a chi-squared(2) is about $1.67^2$; to get the 87.5 percentile of a zero-mean normal to be 1.67 you need $\sigma$ to be about $1.448$. Would you want to use a different $\sigma$ for each quantile? I don't see much understanding to be gleaned from that activity.
Best Answer
Let's say you have random numbers $x_i$. When you estimate the variance of the series, you have to calculate sums like $\sum_i x_i^2$. If your numbers are from normal distribution, then the sum is from $\chi^2$ distribution. If you need to know the confidence intervals of your variance estimate, then you can use $\chi^2$ distribution to get them.
Often your numbers are not from normal distribution. However, due to central limit theorem (CLT) the sums of non-normal random variables are still normal in some cases. So, if you look at the mean of the sample, it's from normal distribution $\bar x=\frac{1}{n}\sum_i x_i$. Hence, if you want to know the variance of the sample mean, you'll get to calculate the sums like $\sum_k \mu_k^2$, where $k$ is a sample. Again, you can use $\chi^2$ to get confidence intervals of the variance of the sample mean.