00:01
So we have a online course for psychology and the general population of students get a mean on the final exam with a standard deviation of 12.
00:10
And we want to know if we have evidence that it is effective to give the class online.
00:17
So i don't see any evidence of a direction.
00:20
So we will assume that the mean for the online class is the same as the general population and alternately it is different.
00:29
The mean does not equal 71.
00:32
And this is a z -test.
00:34
So we can find that z -value by taking the sample mean which was that 76 minus the population mean we're assuming and then that's population standard deviation divided by the square root of 36.
00:50
And so we can see that this is going to end up being, this is 6.
00:55
So 12 divided by 6 is standard error is 2.
00:57
And so we're going to have 5 divided by 2 is going to end up being a 2 .5.
01:04
So the z -value for this is going to end up being the area greater than or equal to 2 .5 and then double it.
01:15
Two times that probability and let me just quickly look up that 2 .5.
01:20
Area actually below negative 2 .5 is 2 times 0 .0062...