00:01
All right, so here we are going to do a z test because we know our population standard deviation is 600.
00:10
And we are going to test if the average is equal to 25 ,000 hours of use.
00:24
And i guess we're going to go with a two -teled test based off of the way it's worded.
00:29
I don't see any clear direction, so we're going to go with two -tell test.
00:36
So i'm going to enter in my data into my l1 in my calculator.
00:40
So 24807, 24 -968, 24502, 24502, 24 -361, 24 -302, 24 -361, 23 -304.
01:03
24 -852, 257, 24 -4 -147, 24 -416, 25 -47 -25 -478, 25 -0 -7 -3.
01:18
And i'm going to go to my test menu and do a z -test, and my data is typed in, 25 -1 -2 -3, 25 ,000 for my null -mean, my standard deviation, 600, l1, and then a two -tale test.
01:32
And what that gives us is a z score of negative 1 .629627.
01:42
And this is a, again, i was doing just a one sample z test.
01:48
So what that looks like in r, so your r code...