chi-squared-distributionquantilesself-study
For chi squared distribution, how would I find quantiles for the following three cases:
A) $P(X^2> X^2_{\alpha})=0.01$ when $v = 21$
B) $P(X^2 < X^2_{\alpha})=0.95$ when $v =6$
C) $P(X^2_{\alpha} < X^2 <23.209) = 0.015$ when $v = 10$
Here $X^2_{\alpha}$ is the $\alpha$-quantile of the $\chi^2_{v}$ distribution.
I'm not surprised this is confusing.
Usually the subscript "$1-\alpha$" indicates the chi-square value is at the $1-\alpha$ quantile of the distribution.
So that would suggest it has an area $1-\alpha$ to its left and $\alpha$ to its right.
That is to say, your option 1.
But on the other hand "$\alpha$-quantile" would more usually suggest the lower tail, your option 2.
So which is it:
Well, the statistic should reject when correlations are large, and the formula will produce large values of the test statistic when correlations are large.
So when the population correlations are zero, the upper tail of the null distribution, of area $\alpha$ forms the rejection region, so you need $\alpha$ in the upper tail.
Which is your option 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
By hand, you need to refer to a tabulated distribution of the $\chi^2$, which should be found easily on the web (e.g., this one). Let x denotes the quantile of interest, and v the degrees of freedom of the chi-square distribution. You just have to know that the total area under the curve (i.e., the density) equals 1 (this will help you to work through the third case), and that such Tables generally gives P(X < x) = p, for a certain v. Knowing x (resp. p), you can find the approximated value of p (resp. x). In the aforementioned Table, the first cell reads: P(X < 1.32) = 0.25, for a 1-df chi-square.
If you have R, the
qchisq()function gives you the requested quantiles (look at the on-line help to be sure of what is returned, esp. thelower.tailargument). For the preceding example, we would useqchisq(0.25, 1, lower.tail=FALSE).It is always a good idea to draw the corresponding density curve, as illustrated below. Note that p3 is also 1-P(X < q2).