00:01
So we have in our first sample that n sub 1 is 75 binomial trials, and r sub 1, the number of successes, is 45.
00:10
In the second independent, we have n sub 2 is equal to 100, and the number of successes is 65.
00:17
And at a 5 % significance level, we want to test the claim that the probabilities of the two for successes is different.
00:27
So we would be in let's see where you want to leave your hypotheses.
00:33
C is to give your hypotheses.
00:36
We are going to assume that the proportion between these two that they are equal and alternately if we want to show that they are not equal and then it asks us in a compute the pooled probability.
00:54
So that pooled value, the pooled value is to take the total number of successes, which is that 45 plus 65, and then 75 plus 100.
01:08
So that comes up to be 110 over 175.
01:13
And 110 divided by 175 comes comes out to be about approximately 0 .6286, about 63%.
01:23
Now on part b, it says check the assumptions, what distribution.
01:27
We will use a standard normal.
01:31
So a standard normal distribution.
01:37
We are going to assume that the two proportions are equal or that the difference is zero.
01:45
And our standard deviation where the difference distribution for the p -hat 1 minus the p -hat 2 is going to be at 0 and the standard deviation of the p -hat 1 minus the p -hat 2 is going to end up being the pooled p, 1 minus the pooled p, and then 1 over the first sample size and 1 over the second sample size...