00:01
Okay, we're going to compute the statistical power of a situation where we're testing a null hypothesis of mu equals 2 .4 versus an alternative of mu not equal to 2 .4.
00:13
So the first thing we need to do is find basically the critical value at which we would reject the null hypothesis.
00:21
So if this is the distribution under our null hypothesis, there's some value.
00:26
And since this is a two -tail test, there's some value here.
00:30
And here, okay, which is the same x bar away, okay? and we are assuming that the mean that we're using is 2 .4 and that the population standard deviation that we're using is 11 .4.
00:52
Okay, so we need to find the x bar here and let's see what is our level of significance.
01:01
So our level of significance is alpha equals 0 .05, which means that we need this area here to be 0 .025, and this area here to be 0 .025, so that the two tail areas add up to 0 .5.
01:18
So we're going to find this x bar that's right here.
01:23
And the way that we're going to do that is we're going to look up on our z table, the value, such that 0 .025.
01:33
That proportion of the curve is underneath our values.
01:37
So i look up that value on z table, so i'm finding the z value that makes that happen.
01:44
And i look it up and i find that at that point, z is equal to 1 .96.
01:53
Okay? and also negative 1 .96 here.
01:58
So now i'm going to find the value that makes that happen going backwards from the z test statistic formula.
02:04
Which is x bar minus mu from the null hypothesis over sigma over the square root of n.
02:10
So that'll be 1 .96 equals x bar, which i don't know, minus 2 .4 over 11 .4 over the square root of n, which was they did a random sample of 195 days.
02:32
Okay.
02:32
And then using some algebraic manipulation, i will just need to multiply both sides by this denominator.
02:41
So i'm going to do 1 .96 times in parentheses 11 .4 divided by the square root of 195...