 Hi, this is Dr. Don. I have a problem out of Chapter 5 in McClave. This is 8.5.63. And we need to determine the sample sizes for a difference between two means. We're given this information. We want to estimate the difference between two population means to correct within 1.5 with a 95% confidence interval. You have to interpret that as 1.5%. And that is your margin of error. But we want the width if we're going to use stat crunch and I'll show you that. The second thing we're given is that the variances, the population variances are approximately equal to 22. So since we're given the population variances, that means we know the population standard deviations, which means we use the normal distribution, the z distribution in order to solve this, not the t distribution. So the first thing to do is get the square root of 22 and you can use your calculator there and that is 4.7. And I find it better to round to 10 on these questions. Then we'll call a question help and open stat crunch. Okay, I have stat crunch open. We need to go to stat, z stat to sample, and we look down for width sample size. So we bring that up and we need to put our parameters in. We have a 95% confidence level, so that's correct. Our standard deviations, the sigma squares were 22 and if you take the square root of that, that's 4.7 for both 4.7, stat crunch needs the width of the intervals. So we're given the margin of error there, which is half the width. So our width would be three. And then we click compute and we get sample size per group 76 for both of the groups and that's the size sample they wanted here. I hope this helps.