00:01
For this problem, we are looking for a 95 % confidence interval for difference in proportions.
00:08
The way we find this is we take our point estimate for the difference in proportions, which is the difference in sample proportions, plus or minus the z -score for a tail proportion of 0 .025, noting that this is the one tail proportion, 0 .025, multiply that by the square root of the pooled proportion times 1 minus the pooled proportion times 1 over n1 plus 1 over n2.
00:37
The first thing that i'll do is use a table of values i have up here to find that z -score.
00:45
So we want a one tail proportion of 0 .025, or 95 % confidence.
00:49
We can see that the z -score is 1 .960.
00:52
Then what i'll need to do is compute the pooled proportion, p -bar.
01:04
It's going to be equal to sample proportion 1 times sample size 1 plus sample proportion 2 times sample size 2 divided by n1 plus n2.
01:17
Plugging in our given values, we have that that's going to be 0 .14 times 100 plus 0 .22 times 100 divided by 100 plus 100 for a result of 0 .18.
01:30
Now we have everything that we need to compute that margin of error term, so i'll calculate that out by itself down here.
01:40
So the margin of error is going to be 1 .960 times the square root of 0 .18 times 1 minus 0 .18 divided by, or pardon me, i need to correct my formatting there, 0 .18 times 1 minus 0 .18 times 1 over sample size 1 was 100 and sample size 2 was 100 as well.
02:05
So 1 over 100 plus 1 over 100.
02:08
And i'll just label my calculations here...