00:01
We are assuming that the sample size in the first group is 10.
00:04
The sample mean is equals to 3 .1.
00:08
In the second group, we have a sample size of 8.
00:13
A sample mean equals to 2 .7.
00:16
A sample standard deviation here equals to 0 .7, and for the first group, equals to 0 .5.
00:25
And we want to compute a 95 % confidence interval for the difference.
00:30
Of means.
00:32
So, so we are assuming here that the variance are equal.
00:38
In the question, they say that we should assume that the variance have equal, that the populations have equal variances.
00:48
And also, they also specify in the question that we are considered normal populations.
00:54
So consider that our sample size here is quite small, but we can assume and use the following formula here to compute the confidence interval, which is given by x -bore in the first group minus the x -bore in the second, plus and minus the t score, multiplied by the square root of a pool variance, which is given by the sample size for the first group, multiply by the variance in the sample plus n2 minus 1 the variance in the second group divided by n1 plus n2 minus 2.
01:40
And this we still need to multiply by 1 divided by n1 plus 1 divided by n2.
01:48
This t score here is the quantile in a t student distribution, which has a 2.
01:56
An area left to it, since we assume a 95 % confidence interval, equals to 0 .975.
02:05
And this t student here has the degrees of freedom equals to n1 plus n2 minus 2.
02:14
So, n1 here is equal to 10, n2 is equal to 8...