 This is the another example of the MANOVA. Now, in this example, what we will do is that we do not have raw data. We have the data in the form of matrices, the data in the form of some of the squares. So, how will we use this? This is the example number 2. The random sample of size 27, first for 38 and the 28. From 3 by variate normal population, the following mean vectors and the sum of square cross product given here. These are the mean vectors and these are the sum of squares. Now, you have a mean vector. Why do we call it bivariate? Because it is dimension 2. 3 bivariate normal population. Because it is dimension 2, that is why we call it bivariate normal population. Tests at 10% level that the mean vectors of 3 population are equal. Now, what we have to do is that these are the mean vectors of 3 population are equal. And what is the size? First, for 27, then 38 and the 28. Now, here is the solution. We have 6 steps. First is to construct the hypothesis. The hypothesis is the mean vector 1, mean vector 2 equals to mean vector 3. All 3 means are equal. And the alternative hypothesis, at least 1 pair mu i is not equals to mu t. All 3 means factors are not equal. The second step is the 10% level of significance and the third is the test statistic. This is the test statistic where Wilkes-Lamda equals to this one. The F statistic under H naught is distributed as F with 2 G minus 1 and 2 N minus D degree of freedom with N equals to N1 plus N2 plus N3. We will add up which is equals to N. Fourth step is the calculation. Calculation W. W which is equals to, we have already seen another formula. Use that formula because W is its simplest form. Because we solve it according to S because we have the sum of squares given. We are using the sum of squares. This is the general formula. We have said this formula further. This is the sum of squares A i. Now, sum of A i. What will happen to you? What is the first value of W? 4011.5. Sum of A i. First A is the value. Then second A is the value. Then this. So A1 plus A1 vector plus A2 vector plus A3 vector. After adding up, we have W value. Similarly, B between sum of squares. This is the form of matrices between sum of squares. That is why we are using it because the data matrices are given in this form. According to this, we will find the values between sum of squares. Then is the x bar. Small x bar vector. How will we find it? This is the x bar. You know how to find the combined mean. Similarly, we are finding this x bar mean vector. N1 x bar 1 plus N2 x bar 2 vector plus N3 x bar. 3 vector divided by total N which is equals to N1 plus N2 plus N3. So after calculation, the mean vector x bar which is equals to this one. Then further N1. First, how many values do you have? You don't have 27 values. 27 multiplied by this. And take its transpose. You have first, second, third quantities. Then further, what do we have to do? How to find B? We have to sum up. And after summing up, we will get the value of B. That is, you have to add them up. After adding them up, we will get the value of B between sum of squares. So total. How will we sum total? Within plus between. If we add them up, then you will get T. T is equal to W plus B. That is the value. Now we have to take out the modulus of within. We have to take out the modulus of total. Modulus. Because we have to take the ratio. It is equal to the ratio of two, which is the within over total. The modulus of them has come to you. So the value of W. Then W. Entered in F statistics. Finally, calculated F value has come. 192.0148. Fifth step is the critical region. There is the calculated value F which is greater than critical value. With these degree of freedom. This is the alpha by 2. 10% layer. We have here 0.05. 4178. 4178. So if you don't have 178, then check on infinity or interpolate it. 2.36. 2.36. And here you have calculated Ff192. 192 is greater than 2.36. So conclusion is the calculated value F falls in the critical region. So we will reject H0. It is concluded that the mean vectors in three populations are not equal. So we kept null hypothesis. All three means vector are equal. So our conclusion is that we have rejected the null hypothesis. So conclusion is. All three population means vector are not equal. This is the final conclusion.