00:01
Okay, today we're going to be going over question 19, so it tells you the bonferoni adjustment, which is a multiple comparison correction used when several dependent are independent or being performed simultaneously.
01:18
So your h -0 is going to be that your mue are equal, and then your h -a, or your alternative, is that.
01:34
They are not equal.
01:46
So you have your alpha yet to find, which is going to be alpha ew, which is your 0 .05.
01:55
There are six samples.
01:57
So it's going to be 0 .08.
02:00
You need to know your degrees of freedom, which are going to be 24 because you have 24 values minus four categories.
02:06
So your degrees of freedom is 20.
02:09
To find out lsd is equal to t of a over 2 over your mean square of 1 over n1 or n1 or n5 plus 1 over n2 or nj.
02:32
So then you've calculated the values, you get 2 .086 is equal to your mean square, which is 0 .995, 1 over 6.
02:49
So plus one over six.
02:52
You have your six here because there are six samples for each machine...