00:01
Following is the solution to number 14 at one -way a nova test.
00:04
And this is about the mean sales prices for three cities.
00:09
And the null hypothesis here is that the mean sale prices are the same for these certain houses.
00:13
And then the alternative is that at least one of them is different.
00:16
The second step is to find the critical value.
00:19
And you need three pieces of information to find the critical value.
00:22
One is your alpha, your significance level.
00:24
And that significance level in this case, and that's usually given to you is 0 .10.
00:27
You also need the degrees of freedom for the numerator, which is, the number of categories, in this case the number of cities minus one.
00:33
So there were three cities that we looked at minus one is two.
00:36
So degrees of freedom for the numerator is two.
00:38
Degrees of freedom for the denominator is the total number of data values minus the number of categories.
00:43
So in this case, if you counted up those data values, there were 31 minus the three cities that we looked at.
00:48
So 31 minus three is 28.
00:50
So that's what we need.
00:51
So from there, you can use a table or you can use software.
00:54
I'm going to use software.
00:55
So i wrote a program and i called it inverse i'm not going to show you how to write this program.
00:59
You can youtube it if you wish, but it makes it easier for me.
01:05
So the area is the alpha value, so we'll put in point 10 for that.
01:09
Degrees of freedom for the numerator was 2, and then degrees of freedom for the denominator was 28.
01:14
And that's going to spit out my f star, my alpha value, or my critical value, which is about 2 .503, let's call it.
01:21
So 2 .503 is my f star.
01:24
So 2 .503.
01:29
So anything greater than 2 .503, we're going to reject h0.
01:34
Anything less than 2 .503, then we're going to fail to reject the null...