 Hi Eileen. I have this problem that you're asking about and I know that you are primarily interested in how to find the Rejection region using stat crunch and to do that. We need to review just a little bit. We've got this Problem and we're told that the engineer has recorded the braking distances for two types of tires now that tells me off the top. This is a two sample test and we're also told that each end, each sample has 35 tires because n is greater than 30 and We also know that these are randomly selected and the samples are independent. Those two things tell me that we can use the z test to solve this as opposed to the t-test. We're given the Type A, the mean of the sample x-bar 1 and type B, the mean of the sample x-bar 2 and we're also given the sample standard deviations s1 and s2. Normally when you use the standard normal solution to find z you have to have the population standard deviation, but again because we've got more than an n of 30 we can use the sample standard deviations. Okay, the key to this, to finding the rejection region, first of all recognizing alpha in this case, it's 0.1 and the question is what are the null and the alternative? So let's look at those. The null in these difference of mean cases, and when we're comparing means we always subtract the 2 and we set the null to be that there is no difference, the mu 1 minus mu 2 is equal to 0. The alternative is the complement mu 1 minus mu 2 is not equal to 0 and because the inequality is in the alternative, the alternative is the claim. In other words the claim is the engineer says that the two types of tires have different means. Because the alternative contains the inequality, this is a two-tailed test. That means we have to put alpha over 2 on each end of the standard normal curve to find the rejection area. So let's look at stat crunch. Okay, I've opened up stat crunch and to find the rejection region, we just go to stat, calculators, normal, and we get our little sketch of our standard normal situation and we don't have to change the mean and standard deviation. What we have to do is put in the alpha, or in this case because it's a two-tailed test, we put in half of alpha, which would be 0.05 and I'm going to first calculate the left tail compute, and we see that's minus 1,6.45 rounding off. Now we know by symmetry, the upper tail, the right tail, would be just the positive version of that minus 1.645, but we can just check. I'm going to put in 0.05 again and click compute and there we see the upper rejection region. Again, 1.645. Anything, a z greater than 1.645 would mean the test is significant and reject the null. Or again, we go less than, we see our left rejection region. So a z that's either in the negative, the lower rejection region, a z that is smaller than minus 1.645 or a z that is greater than plus 1.645. Okay, just for fun, let's go ahead and run the hypothesis test. This time we go to z-stat, two samples with summary, and we bring up our dialogue box there. I think our first mean was 43 and again I'm just putting numbers in, I think our approximate 4.8, standard deviation 35 for the size, the second was 45 standard deviation 4.4 and the size was 35. We've got the hypothesis test already selected. The difference mu1 minus mu2 is zero and the alternative is not equal to zero. We click compute and we get our results here. We get our standardized z statistic of minus 1.87. That is on the lower side and minus 1.87 is smaller than minus 1.645. So it definitely falls in the lower rejection region. But we also get a p-value of 0.06, 0.07 rounded. That is smaller than the alpha of 0.10. So that also tells us to reject the null. And therefore the claim would be supported, I think.