00:01
For this problem, we are told that a sample, or for the first problem here, sample of 48 people was conducted to see how many cups of coffee per day people buy at starbucks.
00:09
Sample added a mean of 2 .16 cups.
00:12
It is known that the standard deviation is 2 .5 cups.
00:14
We're asked, what is the margin of error for a 95 % confidence interval.
00:18
So to begin, for finding our margin of error, we need to find our z value necessary.
00:24
That's going to be about negative 1 point, or that would be, pardon me.
00:30
We want the magnitude of this is what we care about.
00:33
So that's z of 0 .025 gives us a score of 1 .96.
00:39
And so our margin of error is equal to z of 0 .025 times the standard deviation over the square root of the sample size.
00:48
So we have 1 .96 times 2 .5 divided by the square root of 48.
00:57
So we get that the margin of error is about 0 .7.
01:02
Then for part 2, sample of 16 people.
01:06
So i'm just going to make the necessary changes here because it's almost the same setup for the problem.
01:14
So we just change the standard deviation to 2 .6 and the sample size.
01:20
Oh, pardon me, actually.
01:22
Biggest change is that we'd have to use a t score here rather than a z score.
01:29
So this would be for a student t distribution of a.
01:31
Of 15 degrees of freedom.
01:35
And so now the margin of error is going to be 2 .13 times 2 .6 over root 16.
01:41
We find the margin of error there is 1 .3845.
01:46
Then we're asked how large a sample is needed in order to have a margin of error of 3 % when n equals 384, x equals 48, and alpha equals 0 .01.
01:57
So to begin, we'll want to find what 3 % of 48, so 0 .03 times 48, it's 1 .44.
02:08
So margin of error is 1 .44.
02:11
It needs to be equal to z of 0 .005 times, oh, are we given a standard deviation? one second here.
02:30
Okay, i'm guessing that it's supposed to have a standard deviation of 2 .5 as a continuation from a previous problem.
02:37
So it would be z of 0 .005.
02:40
Times 2 .5 over the square root of 384...