Algebra, two-sample t-statistic, variance

[_joey]

What have you tried so far? You'll probably want to use the result from part (i).

The question explicitly states to use the result from part (i). I am also thinking of applying this identity to the numerator of b : http://s8.postimage.org/88li4bjvn/pic3.png

[Dason]

I'm not sure you need that identity. I think you should apply the result of part (i) to the denominator. For the numerator just do something similar to what you did to show part (i) by breaking it up into two pieces - the parts where X_i = 0 and the parts where X_i = 1.

[_joey]

I'm not sure you need that identity. I think you should apply the result of part (i) to the denominator. For the numerator just do something similar to what you did to show part (i) by breaking it up into two pieces - the parts where X_i = 0 and the parts where X_i = 1.

I am stuck with the term in the numerator.

[Dason]

Note that if you break apart the sum in the numerator into two parts (one where X_i = 0 and one where X_i = 1) then the (X_i - Xbar) part is constant so you can pull it out of the sum. Also note that where the sum is only over the Y_i where X_i = 0. Then if you multiply both sides of that equality by you get the equality .

Play around with this for a while and see if you can simplify things.

[_joey]

Okay! I can see how to solve it now. Thanks again!

[matix1080]

Hi Danson, I was wondering if you could expand on how to break the sum apart in two peices... does the pooled variance formula come into play?

Hot ****. Was that so hard? Now I know that X is actually just a dummy coded indicator variable so that if a person is in the never-steroid group and if they're in the ever-steroid group . It could be flipped around but that is very important information.

So now we know that and we can break the sum apart into two pieces - one piece where all the X_i are 0 and one piece where they're all 1. Have you tried this yet?