Maybe it might help if I illustrate *why* it's not just a shift.

First, let's start by looking at the test statistic:

t = (xbar1 - xbar2)/(Sp*sqrt(1/n1 + 1/n2)) hopefully the terminology is clear enough here. I'm going to make a slight simplification and just define

SE = (Sp*sqrt(1/n1 + 1/n2))

so that our original reduces to

t = (xbar1 - xbar2)/SE

Great. Now under the null hypothesis that the two means are equal this has a central T distribution with v=n1+n2-2 degrees of freedom right? The main thing to note is that the numerator here (xbar1 - xbar2) has expected value of 0 since under the null hypothesis the means are equal. There are lots of other properties but this is the important one for our purposes.

Now let's consider what happens under the alternative hypothesis

t = (xbar1 - xbar2)/SE

Well the numerator xbar1-xbar2 doesn't have an expected value of 0 anymore. And I'm sure this whole time you've been saying "yes of course and that's why we apply the shift!". But I'm going to show why that isn't quite right. But let's try that approach and see what happens - can we get the expected value of the numerator to be 0? Sort of! We can subtract the expected and then add the expected value right? Let's give it a try

t = ((xbar1 - xbar2) - (mu1-mu2) + (mu1-mu2))/SE

= ((xbar1 - xbar2) - (mu1-mu2))/SE + (mu1-mu2)/SE

Nice. Now we have this broken into two pieces and the first one ((xbar1 - xbar2) - (mu1-mu2))/SE does have an expected value of 0 in the numerator and hey it *does* have a central T distribution. So you might be thinking - yes this is all proving my point because now we have a central T and then we're just adding something to it - that's basically a shift right? Wrong.

Let's look at what we're adding here:

(mu1 - mu2)/SE

The big hiccup here is the SE in the denominator. If that was a constant then this whole term would be constant and the idea of just shifting the T distribution would work perfectly. Unfortunately for us in this situation SE contains the pooled estimate of the standard deviation. Which means that whole term is a random variable itself. And that whole term *isn't* a central T distribution either. If SE was constant (like it is in the case for a Z-test) then this would all work and that's how we go about calculating power for a Z-test. It's just that the math ends up being trickier for this T-test situation.

Fortunately for us though we do have an answer for the overall question: Under the alternative our t-statistic follows a non-central T distribution. Proving that piece is left as an exercise for the reader but hopefully this provided some insight into why we can't just shift the central T to do power calculations here.

First, let's start by looking at the test statistic:

t = (xbar1 - xbar2)/(Sp*sqrt(1/n1 + 1/n2)) hopefully the terminology is clear enough here. I'm going to make a slight simplification and just define

SE = (Sp*sqrt(1/n1 + 1/n2))

so that our original reduces to

t = (xbar1 - xbar2)/SE

Great. Now under the null hypothesis that the two means are equal this has a central T distribution with v=n1+n2-2 degrees of freedom right? The main thing to note is that the numerator here (xbar1 - xbar2) has expected value of 0 since under the null hypothesis the means are equal. There are lots of other properties but this is the important one for our purposes.

Now let's consider what happens under the alternative hypothesis

t = (xbar1 - xbar2)/SE

Well the numerator xbar1-xbar2 doesn't have an expected value of 0 anymore. And I'm sure this whole time you've been saying "yes of course and that's why we apply the shift!". But I'm going to show why that isn't quite right. But let's try that approach and see what happens - can we get the expected value of the numerator to be 0? Sort of! We can subtract the expected and then add the expected value right? Let's give it a try

t = ((xbar1 - xbar2) - (mu1-mu2) + (mu1-mu2))/SE

= ((xbar1 - xbar2) - (mu1-mu2))/SE + (mu1-mu2)/SE

Nice. Now we have this broken into two pieces and the first one ((xbar1 - xbar2) - (mu1-mu2))/SE does have an expected value of 0 in the numerator and hey it *does* have a central T distribution. So you might be thinking - yes this is all proving my point because now we have a central T and then we're just adding something to it - that's basically a shift right? Wrong.

Let's look at what we're adding here:

(mu1 - mu2)/SE

The big hiccup here is the SE in the denominator. If that was a constant then this whole term would be constant and the idea of just shifting the T distribution would work perfectly. Unfortunately for us in this situation SE contains the pooled estimate of the standard deviation. Which means that whole term is a random variable itself. And that whole term *isn't* a central T distribution either. If SE was constant (like it is in the case for a Z-test) then this would all work and that's how we go about calculating power for a Z-test. It's just that the math ends up being trickier for this T-test situation.

Fortunately for us though we do have an answer for the overall question: Under the alternative our t-statistic follows a non-central T distribution. Proving that piece is left as an exercise for the reader but hopefully this provided some insight into why we can't just shift the central T to do power calculations here.