00:01
So we have data for the day shift and the afternoon shift, and the day shift a number of defects or problems are listed here, and then the 8, 9, 12, and 15.
00:15
And we're supposed to find out, do we have evidence that the afternoon shift has more defects than the day and using a 5 % significance level? and it seems kind of like a silly question because all the afternoons are smaller numbers.
00:36
So the answer is going to be no.
00:40
What will carry out the hypothesis test? i suspect that they may have written it the wrong way.
00:45
But we're going to have the difference be the afternoon score less the day score.
00:53
And so we have our mean listed.
00:59
As less than or equal to zero as our assumption and alternately that the difference is greater than zero.
01:09
And so our differences, and i'll put those in blue, that difference is negative two, that difference is negative three, that difference is negative three, and that difference is negative four.
01:20
So when i go through and find the mean difference for those values, and let me clear that out, we find that the mean difference is negative 3 for that sample, and i find that the standard deviation of those differences comes out as .816, and it would round to 5, and our sample size is only 4.
01:45
And so we would have to use a test statistic that is a t value, and it would only have three degrees of freedom.
01:52
And if we wanted to find our critical value, a critical value would have for our 5 % 3 % percent.
02:00
Significance level would have all 5 % in that upper tail since this is greater than.
02:06
And again, as i'm saying, this test doesn't really make sense.
02:09
And the value that has 5 % in the upper tail is a t value, that critical value, 2 .353.
02:19
And we would be rejecting our null if we were having our value higher, which we're actually not going to because we have this as a negative difference...