00:01
So i see that you need help with this question and it says, assuming that the population has been approximate normal distribution, if the mean sample size, n is 12, has a sample mean of 34, and a sample standard deviation of 8, find the margin of error at 95 % confidence interval.
00:19
Round the answer to two decimal places.
00:21
So the first thing that you need to do is you need to find your t -score.
00:26
So you're going to do 1 minus 0 .95, which equals 0 .05.
00:32
You're going to take 1 minus 0 .05, divide that by 2, and that's going to equal 0 .975.
00:40
Then you're going to take t, 0 .975.
00:44
You're going to multiply that by your degrees of freedom, which is 11, because 12 minus 1 is 11, and that's going to equal 2 .2.
00:52
Then you're going to take your mean of 34.
00:56
You're going to do plus or minus your 2 .2 t -score times your standard deviation of 8 divided by your square root of 12...