Can someone explain why when variance is considered equal we can use a pooled estimator and the test statistic is based off of n1+n2-2 df but if variance is unequal we have to use modified degrees of freedom to find the test statistic?
Anyone?? Bueller??? Bueller???
Without going into the math (and proving it can get ugly), think of it this way - if the variances are equal, you can estimate the standard error as if the two samples were really one big sample, so it’s essentially of size (N1 + N2) -less the two degrees of freedom for the two samples.
However, if the variances are unequal, your degrees of freedom will be based on a somewhat complicated weighted average of the two sample sizes (and variances).
Luckily, from what I understand, they typically don’t make you know DF - but they do want you to know that at you can either pool or weight variances, and you’ll have to test for equality of variances (and for the purposes of the exam, that test is an F-test).
Cool thanks that was what I was looking for. Also my basic Algebra skills thank you for not going g into the math.