So we're looking at the reading of professors and we would be assuming that the mean rating for females is equal to the mean rating for males and alternately that they're different. We just want to find out if there's a difference between so we will be doing a two tailed test and I always like to draw a little picture. But remember we're assuming that the difference between these two is zero, and we can also write hypotheses that way, if we so choose so our smallest sample sizes, let's see. Our smallest sample size is 40. Therefore we will use 39 degrees of freedom and we will find our test statistics by taking the difference in our means. And so we're going to take that 3.79 minus the 4.1 And then we will be dividing it by the first standard deviation is 0.51 squared divided by that sample size. And then the second standard deviation is 0.53 square divided by its sample size, which is 53. And when I do that calculation, I find out that that test statistic comes out to be negative 2.25 And now we want to find, because we're doing a two tailed tests, we have this as one of our test statistics and the other one would be up here for our two tailed tests. And these two areas together give us our p values. So we want to find the probability of getting a test statistic less than or equal to negative 2.25 And then we want to double it for the other table, other tail. And we could look up the critical value for this as well. But I'm not going to set that up this way, I'm going to go through and use my T. C. D. F. So my T. C. D. F. I'm gonna have my left bound be like negative uh 10,000 and then my upper will be this negative 2.25 And I will have my degrees freedom being 39. Again that's like this isn't a conservative way. And then we're going to multiply that probability times two. And when I do that I find out that I get 0.4975 By just a tad bit. This is less than 5%. So we do have evidence to reject the not we have sufficient evidence to reject. And now and claim that the ratings are different mm for males and females. Now let's find that confidence interval. And because we did a two tailed test we would be finding them with a 5% significance level. We would be finding a 95% confidence interval and our confidence interval with 39 degrees of freedom. We don't have it in our table. So I'm going to actually use my inverse t to find that. And so I use my inverse T. I can put in the area and I'm gonna put in that lower areas point oh 25 since it's a 2.5% at the bottom tail, 2.5% of the top tail. And then I'll put in the degrees of freedom as 39. Again, this is a conservative approach and then we'll get that with that T value is so our T. Star value will be negative 2.227 So now when we get our confidence interval we need the difference. Let's get what that difference is. So that difference that 4.1 Well, I should say it the other way around the 3.79 minus the 4.1 That comes out to a difference of negative 0.22 it's a negative 0.22 plus or minus. And then we have our value for our T. Star value and then we'll use that big standard deviation and uh let me go back and find what those standard deviations are. So 0.51 point 53 0.51 squared and the 0.53 squared. And let's go back for the sample size is 40 and 53. Yeah. Right. Money. Mhm. Well, I never noticed before how this was 0.53 and that was 53 interesting. Not that makes any difference. So let's find the margin of error, first of all. Yeah, so 2.227 times the square root of 0.51 squared, divided by the 40 plus 400.53 squared divided by 53. And that margin of air comes out to be oh goodness! Isn't this interesting about 0.22? So when we find this interval this bottom while the negative and then minus is going to be minus 0.44 and then when we add them, that's going to be zero. So we're right at that cut off point that uh those values would uh we would only have negative values, we don't have zero. So this would be just at that cut off point to say that we would have evidence to reject the null but notice how close we were to 5% here and that's why this is such a close and we're here to so then it talks about the comparison between another example that did this and had smaller sample sizes. And yes, there is a difference. When we use larger sample sizes, we can make differences. The same differences be significant with larger sample sizes. If these were smaller sample sizes, we would not have been able to say and got the same meaning, same standard deviation, we would not have been able to have sufficient evidence to reject them all.