00:01
For this problem, we're given some sample data and we want to find the minimum sample size needed for a 95 % confidence interval.
00:12
So the data we're given is that we want our sample mean to be within 5 of the population mean, so that means we have a margin of error less than or equal to 5.
00:25
And we also know that the population standard deviation is known, it's 44.
00:32
Okay, so let's recall how we actually construct a confidence interval and use that information to describe how we'll find our bound for the sample size.
00:44
So to find the confidence interval, you take whatever your sample mean is, which i'll call x -bar, and you add or subtract the margin of error term, which is given as follows.
00:54
It's a z -score that corresponds with this level of confidence, 95%, times the population standard deviation sigma, divided by the square root of n, your sample size.
01:05
So given this bound for the margin of error, and as i said, this term here is itself the margin of error, we can rearrange and get an inequality for n.
01:15
But first we're going to need this z -score, so let me briefly explain how we get that.
01:21
So imagine you have a standard normal distribution, the mean is 0, the standard deviation is 1, then this z -score is such that the area between minus z sub c and plus z sub c is equal to 95 % of the area under the curve.
01:39
So that's 0 .95 from here to here.
01:43
And due to the symmetry of the situation, that means the remaining area is equally distributed in these two tails, which means 0 .025 area in each tail, which then tells you that the area to the left of the z -score must be 0 .975, which i'm now going to write down as a cumulative probability.
02:08
So now that we have this cumulative probability, there are many methods to turn this into the z -score, tables, programs, calculators, whatever you like.
02:19
I'm going to use the inverse normal function on the ti -84, it's called invnorm, and it takes in this cumulative probability of 0 .975 plus the mean and standard deviation of your distribution, and we're using a standard normal, so it's 0 and 1 respectively.
02:41
So i'm going to put this into my calculator and see what i get.
02:46
And it looks like i get a z -score of 1 .960 to three decimal places.
02:53
So we'll need that later, let me just circle it...