 Hi, this is Dr. Don. I have a problem out of Chapter 8 in Larson. This is Section 8.3 on dependent samples, otherwise known as paired samples. In this case, we have a company who wants to know whether or not its ratings have changed from one year to the next. And the question is, is there enough evidence to conclude that the ratings have changed? The way we should interpret that is that the claim is that the ratings have changed, which is the first question here. That means this is a, that's a not equal for the mean difference, mu1 minus mu2, and that means that our hypotheses would be set up this way. The null is the mean difference is zero. The alternative is the mean difference is not equal to zero. So let's jump into this. I'm going to click on the rectangle, open this data in stat crunch, bring this up and expand things a little bit here. What we're going to do first of all is run stat, t stats, paired samples. My sample one is last year, sample two is this year. We want to save the differences in order to find the S of D and D bar easily. This is a hypothesis test for the mean difference equal to zero, and it's a not equal for the alternative. So I'm going to click compute. And we get our answers here. We get our standardized test statistic 1.106. And we get a p value of 0.305, which means if our alpha is 0.05, that this test is not statistically significant, which means we fail to reject the null hypothesis. But here we wanted to use the rejection area method. So we'll find the critical values here and look down alpha is 0.05. So I'm going to go back to stat, calculators, get my t, and we need to input the degrees of freedom, which it tells us right there, seven, which is just the number of pairs minus one. And we've got an alpha of 0.05. But this is a two tail. So we put 0.025. And that gives us a lower critical value of minus 2.365 and an upper by symmetry of plus 2.365. And again, because our test statistic is 1.1, which is somewhere in here, it's definitely not in either one of the rejection zones. Therefore, again, it is failed to reject the null hypothesis fail to support the claim that the mean differences are different. Okay, so we want to get the D bar, the output from the t test will give us that, which is this mean here, the mean of the differences is 0.625. It does not give us S of D, the standard deviations of the differences. And that's the reason that I save the differences. So here we call when we ran the test, I checked that little box there to have us save the differences. There's my differences. So I go back to stat, summary stats, columns, and I just want to click on my differences. And I want to get the mean, hold down the control key, standard deviation, and click compute. And here we get again, the mean, which is D bar is 0.625. And the standard deviation S of D, which is just the standard deviation of these differences, is 1.598. So quickly done using stat crunch.