00:01
So in this question, we are supposing that the proportion of students, first year students, that bought or purchased used textbooks in the past year was 9 .1 divided by 102, which in this case is 0 .8922.
00:24
And if you compute the proportion for the second year students, we are going to have 95 purchase like in the last year used textbooks.
00:38
So basically here we have that this proportion, 069 .3.
00:44
Now this question is asking us to construct what is the 90 % confidence interval for the difference of proportions.
00:55
So we have a formula to compute this, which is basically given by p -hat minus p -1 -hat, which is the first year, in relation to the second year, plus and minus the one value, which is the z score, which is based on this confidence level, multiplied by this square root, p -1 -hat, 1 -1 -hat, divided by the sample size.
01:26
For the first year, p2 hat, 1 minus p2 hat, and this is multiplying, and 2.
01:36
So in our case, using the information that we have, we should plug like first the value of p1 hat, 8922, minus 06934, and like i said, this score depends on this confidence level.
01:53
But for the proportions, we usually use the standard normal distribution to compute this z value.
02:01
So basically, this zed value is here, and the area left to this number is 0 .95, because this value here is 90%, which means that we have 10 % left out of 100.
02:14
So this area, here is always what is left, divided by 2 when we are computing a confidence interval.
02:21
So this is 5%.
02:23
So because this is 5%, this means that this area here is 0 .95.
02:28
So if you use a z -tab, you're going to get that this value is 645.
02:34
Now we can plug this here, and we should repeat like the values that we have, b1, hat.
02:44
So 0 .89 -2, multiply by 1 minus 0 .89 .02, divided by the sample size 4 .2, divided by the sample size, for the first group, which is 102, then do the same thing for the second group, which is this second year student.
03:04
So 069 -34 divided by the sample size.
03:12
One, in this case, is 135.
03:17
So basically, this part here, and this value here is the same.
03:21
So 0 .6934 because it is hard to see.
03:26
But it's the same here, like just 1 minus the p2 hat.
03:31
Now if you compute this port 1 .645 multiplied by this square root, you're going to get 0 .08 254.
03:41
And if you compute this difference, you're going to get 0 .1988, which means that the lower bound of our interval is the difference between these two numbers, which is 0 .116 to 6.
03:58
And the upper bound is the sum of these two numbers, which is 0 .28 .13...