[SOUND] To know whether or not students will perform better at a standardized test after completing a preparatory class, a group of 65 students are given a test before and then again after having gone through the training. So we want to test this claim at 5% level of significance. This is an example of a pair test, because the same people are being given a pre-test, then they go through a training, and then they're given a post-test. We're not comparing two different groups, one has been trained and the other one has not been trained. These are the same individuals. This makes it a pair test, and pair tests, by and large, are stronger tests when you're looking for differences. So when you are doing this, the hypothesis is just only about the mean difference, nothing else. Now in this case, they're saying that the difference would be better. So what they're saying is that alternative is that the average difference, after going to test, through the test is going to be greater than zero, which means that null is that, that difference is either zero or less. So we're going to test this at 5% level of significance. So let's see how would do this differently using Excel data analysis. So here we are and these are the test scores when the students went through pre-training and after training. So each student has a score that is before they went through the training, and after they have gone through training. So we're going to test these values. The ones in column B and column C. So we go to data. We go to data analysis just like before. However this time we going to scroll down and look for the t-Test Paired Two Samples for Means. So not the one that has been using before. So this is the one that we will use, the paired two sample for means, so click OK. The rest of it is going to look very similar to what we were doing before. Input for Variable 1 is in column B, this is before. And this is the after test scores, Variable 2. I'm going to mention that I have a label and then here I'm going to put a 0. Now I could have also given you a value here to test. I could have said the test scores goes up at least by ten points and that's what you would have put here. Alpha is 0.05. I'm going to Output Range here, click here and then I just I'm going to put it somewhere here. So here's our output. And one of things that you notice right away is that right up here it says that this is a t-Test for paired two sample. And of course in a paired two sample, we have to have exactly the same sample size that's a given, because the same people are being tested in pre-test and after test. So now we can say whether or not we have the same test scores or not. So in our problem we had said that it would be greater than zero, which means this was a one-tail test. So we're going to focus on the one-tail test and that is right here. And this value is extremely small. Definitely smaller than .05. So we will reject the null hypothesis, which means what? Our null hypothesis was that the difference would be less than or 0. We will accept the fact that based on our data, the training seems to increase the test scores for these students. Once again, we can use the information that we have here to even give what is the confidence intervals for he differences in the mean. So looking at the equation for the confidence interval, first of all, I need to find out what is the difference that I see in the means right now. And that's relatively simple. I'm going to find the difference in the mean by taking Post-Training minus Pre-Training, again it doesn't really make a difference but I like to talk about how much we expect the test score to go up. It would be easier to communicate here. Then we have the t-value. Now, we did a one-tail test and if I want to use that one-tail test, and for that I used the t-value that I see here, this is going to result in the 90% confidence interval. Let me again remind you why that is. For the one-tail test, the entire 5% is on one tail, and this is a value of 1.669, which is what you see right here. And we will have, the same on the other side because, again, confidence interval is going to have a margin of error that's going to be added and subtracted from it. So we can make a mistake on either end and there's a 5% chance on either side. So if these are 5%, the confidence interval that you're coming up with is going to be a 90% confidence interval. So this t-value, this t of alpha over two is going to be for 90% confidence interval. Now then the equation asks for standard deviation of the differences, and that part we cannot get from the output directly. I'm going to have to calculate this and that's what I'm going to do. So let's go back to our data. Which is right here. I'm going to create a column here called differences and the differences are just going to be the simply for each student. What was the difference between their score before and after? So, it's the difference between Pre-Training minus Post-Training. Again, it will not make a difference which one you subtract from which. Because we are going to use this column for finding out the standard deviation of the differences, so it doesn't make a difference. So then, now that you have it for one, just put your cursor in the corner, then you see the plus sign, double-click, and the entire thing will be picked up. To find the standard deviation for the values that you see in this column, column D is what we want, standard deviation of the differences. Then it's simply =STDEV.S, which we have been using all along of the values you see from here all the way to the 65th student, close the parentheses and press Return. So the standard deviation of differences here is 111.58. Now that I have this I can calculate the margin of error. Margin of error is simply is equal to t-value times, as you can see in the formula that you have, standard deviation of differences divided by square root of sample size and in this case you can pick either one of the ones that says 65. That's all we need and once I write all of it out I will find out that the margin of error is 23.1. So now I can come up with a lower bound and upper bound of the concentrical. So the lower bound is the mean difference minus margin of error and the upper bound is the mean difference plus the margin of error. But you see here is your 90% confidence interval of the differences. So what do you expect? You're 90% confident that the average score for a student will go up after going to this training. The average score for the student will go up by 24 points up to 70 points. The true difference is somewhere between 24 and 70, so at the minimum, a student will get a 24 point boost. Which is not bad, and at the optimistic side it's 70. But remember, any value between 24 and 70 is a plausible value.
