00:01
So we're going to be assuming that the mean is five or five.
00:05
Let's see what unit that was.
00:08
5 inches, 0 .5 inches.
00:10
I'm sorry, 0 .5 inches.
00:14
And alternately, we'll assume that it is not equal to.
00:20
It didn't give any information there about direction.
00:22
So we're just going to assume that it is a two -tail test.
00:27
And so you had four different settings.
00:29
When n was 13, and if the 10, t value were 1 .6 and the alpha level is equal to 0 .05, we want to know what decision we would make.
00:41
And it says what conclusion would be appropriate in each situation, whether we're going to reject the null or not.
00:47
So let's draw, and then while we're at it, why don't we put down what this next one is as well? because that's dealing with the same sample size, with the t value being negative 1 .6, and the alpha level being still 0 .05.
00:59
So let's draw our picture here and we'll make our decision.
01:02
So if we have an alpha level of 0 .05, that means for a two -tailed test, we'll split that alpha to each tail.
01:12
And we have degrees of freedom for both of these problems being 12.
01:17
And so we want to find what the area is in the upper tail.
01:22
And using our chart for 0 .025, we would find that the critical t -vest, here would be positive 2 .179 and down here is going to be a negative 2 .179 and we would reject the null if we're in either end and we can see that in this case for a t -value being 1 .6 that's here here we would fail to reject the null so we would say we have no evidence to say that the mean is different from 0 .5.
02:04
We would say it means probably 0 .5.
02:07
And likewise if we have a negative 1 .6 for a test statistic, we would again fail to reject the null and say that the mean is most likely this distribution is centered at 0 .5.
02:26
Now let's look at our other two scenarios...