 Hi, this is Dr. Dine. I have a problem out of Chapter 10 on Kaya Square. This problem is a goodness of fit test and we know that because we're asked to test the claim of this social service organization that the distribution of educational attainment of mothers is uniformly distributed. So we need to test to see if the distribution of the data that we get matches that uniform distribution. And our alpha, our significance level is .05. They first want us to state the null and the alternative and in the Kaya Square goodness of fit, the claim is that the distribution is uniform and that is the null. The null is always that the distribution observed does meet the expected distribution. So the alternative is the complement which is that it's not uniform. And again the null is the claim. We need to find the Kaya Square critical value, the rejection region. We need to find the test statistic, the standardized test statistic if you will, and then make a decision. So we're going to do that using stat crunch. Remember if you're in my stat lab you can click on the little rectangle there and open up this data directly in stat crunch and I'm going to do that now. Okay I have stat crunch open and we have the data, the response categories, there's three of them, and then the frequencies in each of those categories. So we're going to get the critical value first and we do that by going to stat calculators Chi Square and it brings up the calculator. We need to enter the degrees of freedom and that's just a number of categories minus one. So we've got three K is equal to three so three minus one would be two. We need to enter our alpha which is point zero five and remember for our goodness of fit test the distribution we're interested is always a right tail so we make it point to the right and there's our rejection region beyond five point nine nine one which I think is the answer they wanted. So that's the first part. Then we're going to do the Chi Square test itself. As usual we go back to stat. Now we look down and if you look down toward the bottom you'll see there is a goodness of fit test there and we want the Chi Square test and it brings up this dialog box and we need to get the observed data. Now I'm going to click on that and that's our frequency column. The expected we don't have a column with the expected frequencies but we know that all cells are equal proportion which is a uniform distribution and I'm going to go ahead and get the expected values because I think they're always interested and we just click compute and we get our results here. We get our standardized test statistic of five point six seven eight. We got a p-value of point oh five eight which is greater than our significance level point oh five so that would tell us to re excuse me fail to reject the null and we get our expected distribution which is uniform which is what we wanted. So that tells us what we need. Now if we look over here with our critical area critical value in rejection region five point nine nine is the start of that. The five point six eight is close but it's not quite in there. It's just outside of the rejection region which matches with our p-value which is just over point oh five. So if we go back into the problem we see that we got the answers correctly five point nine nine one for the critical value. The rejection region is to the right of that critical value. The standardized test statistic is five point six eight eight and then we need to make a decision which was fail to reject and we always need to include the significance level. In this case it's at the five percent significance level. There's not enough evidence to reject the claim. Remember the null was the claim. So you need to remember that when you look at your conclusion. So there's not enough evidence to reject the claim that the distribution of educational attainment responses is uniform. So I hope this helps.