00:01
Here's the solution to number 16, and we're basically just going to repeat number 15 with this information, but this time the sample size changed.
00:07
So a lot of this is basically the exact same logic.
00:14
So to find alpha, that's the probability of a type 1 error, which means we're going to reject h .0 whenever h .0 is actually true.
00:20
So that means we're going to find the probability that x bar is less than or equal to 191, or the x bar is greater and equal.
00:30
To 209, given that, mu in fact, equals 200.
00:35
So basically the same thing.
00:38
We're just going to do, so mu is 200.
00:41
It's supposed to be in the middle.
00:43
191, and then here's 209.
00:47
So we're going to shade to the left and to the right.
00:50
So we're finding those probabilities.
00:51
So we're going to do the normal cdf.
00:53
I'm just going to do cdf.
00:54
So normal cdf, where the lower bound is negative infinity, the upper bound is 191.
01:00
The mean is 200, and the standard deviation this time is the 15 divided by a square root of 25.
01:09
So that actually is a smaller standard deviation is 3.
01:12
And we're going to add that to the cdf of 209 being the lower bound, infinity being the upper bound.
01:19
The mean is 200, and the standard deviation is also 3.
01:23
15 divided by a square root of 25, so 15 over 5 is 3...