In this problem were presented with the distributions of grades for a course given by two different professors, and we want to investigate whether one class had higher marks. So for party were asked what type of test this would be. So we can consider this to be two populations of student grades, one for each professor, and we want to see if these populations have different distributions. So this is a test of homogeneity for Part B were asked to write and no hypothesis. So the no hypothesis would be that there is no difference in the grade distribution for either professor and for Part C were asked to find the expected counts for each cell and explain why a Chi chi squared test is not appropriate. So these were the counter data in the table I provided. So the first thing to do is to give the row. Can't the row sums and the column sums. So with some of the rose we get 12 23 22 11 and for and then for the columns. This adds up to 40 and this is 32 which means there was a total of 72 grades, so if we wanted to take to calculate the calculate the expected number of counts for this great right here and assuming the no hypothesis that there's no difference between the distributions, we only have to look at the total. Ah, the total really counts to determine the probability. So if we want to calculate the expected count for this cell right here, we take the number of students who are who studied under that professor. So that's 40 and we multiplied by the probability of getting an A and the probability of getting an A is based on the total values, because we're assuming there's no difference between proper A and property, so we would take 40 and multiply it by 12 divided by 72. So remember, 40 is the total number of students under Professor Alfa, and 12/72 represents the probability of a student from either profit alphas class or profit is class getting in a and this comes out to about 6.67 So let's do one more. Let's let's look at the cell right here. So what are the expected counts for that? Total number of students under profit data is 32 and the probability of them getting a mark of a B is 23 divided by a 72. So let's write that out. 32 students times the probability they will get a B and that comes out to vote 10.22 So then you could do that for every cell. And I have done that using software, and these are my results here. And so if you look it so if we want to consider the conditions necessary for using Chi squared tests, one is the expected cell frequency has to be at least five for every cell, and we can see in this table that he expected Cell frequencies for many of the cells are less than five where at least for several of the cells. So therefore, that condition is met and therefore ah chi squared tests are not appropriate for this data.