00:01
For this problem, we have salaries of 40 college graduates.
00:05
So let me use n for the size of the sample here.
00:09
And they had a mean salary of $69 ,600.
00:20
So i'll call that x -bar.
00:24
And the standard deviation for salaries of college graduates is given as $16 ,568.
00:33
And using this information, we want to construct a 95 % confidence interval for the true population mean, the average salary of a college graduate.
00:45
So let's recall how we do that.
00:47
You start with your sample mean, and then you add or subtract the margin of error to get the endpoints of your confidence interval.
00:54
And that margin of error consists of a z -score associated to the level of confidence you want, times sigma over root n.
01:01
So we have sigma and n.
01:03
Let's just get this z -score.
01:06
So to get the z -score, start with a standard normal distribution, which looks something like this.
01:12
Mean 0, standard deviation 1.
01:15
Then this z -score is such that the area under the curve between minus z sub c and plus z sub c is equal to 95 % of the area under the curve, since that's the level of confidence we're working at.
01:28
So this area i'm shading in red here is 0 .95, as the total area under the curve is 1.
01:35
That means the remaining 0 .05 area is equally distributed in these two tails, because the curve is symmetric...