00:01
We're told that in the past the average teaching evaluation score at a college has been 4 .1.
00:05
The administration believes the move to distance learning has had a negative impact on teaching effectiveness.
00:10
So they sampled 50 courses, course sections, and the sample showed an average teaching evaluation score of 4 .03, sample standard deviation.
00:23
And what we want to do is test whether the administration's belief is legitimate.
00:27
So the null hypothesis is, well before we do that, let's actually state the alternative.
00:31
Because this is the claim.
00:32
The administration's claim is that the mean has actually gone down.
00:37
Teaching effectiveness has gone down.
00:38
So that means the mean is less than 4.
00:41
Therefore, the alternative would be the complement of that, which would be the mean is greater than or equal to 4 .1.
00:47
You might even see the null as the mean is equal to 4 .1, but the main thing is you're testing for the alternative that it's actually significantly less than 4 .1.
00:56
And we have a large enough sample size.
00:58
So by the central limit theorem, which says if your sample size is greater than or equal to 30, the distribution about the sample mean would be approximately normal.
01:06
So we can use our one sample t -test.
01:11
And we should say we're going to test this hypothesis with a confidence interval.
01:29
And we'll also do the t -test approach too.
01:33
All right.
01:38
So let's go ahead and do the t -test first.
01:40
We're going to get a p -value.
01:43
Because we're going to test this set of hypotheses at the alpha 0 .05 level of significance, which is going to correspond with a 95 % confidence interval for mu.
01:57
So it's going to be, we're going to reach the same conclusion, albeit a different method.
02:04
So we'll do the hypothesis test first since we're already.
02:07
So our t -statistic is found by taking the sample mean, x -bar minus mu over s over root n.
02:15
Please note that whole term is in the denominator.
02:19
I've seen students mess up because they forget to include a parentheses around the standard deviation over root n term.
02:32
So just be aware of that.
02:35
And so we get our t -score to be negative 1 .414.
02:44
And the p -value associated with that is 0 .08.
02:52
And here, the way we do this is this t -disfunction within our spreadsheet.
02:58
You put in the absolute value of your test statistic, 49.
03:02
That is our degrees of freedom, which in this case is found by taking n minus 1.
03:08
And then since we're looking strictly less than, this is our number of tails.
03:19
And we're testing strictly less than.
03:21
So therefore, it is a one -tailed test...