How do I do an f-test in jmp?
As stated previously:
Trex,You started along the right path, but there are a few things that are incorrect in your last post.1. You said 1.14 is the ratio of the square of two variances. I think I know what you meant to say, but want to make sure. The statistic to use is the ratio of the square of the std dev's. In other words, the ratio of the variances, not the ratio of the square of the variances.2. Your script was almost correct. If the ratio of variances is 1.14, and the degrees of freedom are 581 and 568 (the df are n1-1 and n2-1), then the p-value for a two-sided test is found by 2*(1-F Distribution(1.14, 581, 568)) = 0.1168Alternatively, you can compare 1.14 to an F quantile. If you use an alpha level of 0.05, then the quantile is found by F Quantile(.975,581,568) = 1.18. The 0.975 = 1-(0.05/2).Since the p-value>.05 and 1.14 < 1.18, we can conclude equal variances at the 0.05 level.If you have the raw data, you can feed it into Fit Y by X and do the F-test.
There's an "F Distribution" function, along with an assortment of other functions for calculating quantiles from statistics and vice versa, all of which are described in the online help under "Probability Functions". If you run the following lines of script you'll get a couple of the familiar results from performing tests with the F, t and Normal distributions falling out, showing how several of them work:
F95 = F Quantile(0.95, 1, 10000); show(char(F95, 6, 3)); t95 = sqrt(F95); show(char(t95, 6, 3)); show(char(F Distribution(F95, 1, 10000), 6, 3)); show(char(t distribution(t95, 10000), 6, 3)); show(char(Normal distribution(t95), 6, 3));
The 1 and 10000 are my numerator and denominator degrees of freedom for the F-test; an F-test on (1 and N) df is equivalent to a t-test on the square root of the F statistic on N df (but note that the F-test is one-tailed, whereas the t-test is two-tailed - and only one tail is reported in the result), and if N is large enough, the t-test approximates to a Normal test.