00:01
All right, so we're given the following hypotheses that mu is 100 for the null and mu is not 100 for the alternative.
00:07
Excuse me, that the mu is not 100 for the alternative.
00:11
And we're given six samples here and we're assuming it's a normal population.
00:16
And we're going to know at the 0 .05 significance level, if we can conclude, the mean is different from 100.
00:23
So let's state the decision rule.
00:26
So we need the critical value.
00:27
So it's a two -tailed test.
00:30
So we're going to go to our t -table and look up five degrees of freedom.
00:39
So five degrees of freedom at the 05.
00:45
Where we go? 0 .05 here.
00:49
There we go.
00:49
Two -tailed test.
00:50
0 .05.
00:52
You could also get there by going 0 .05 for one -tailed, divide by 2 .025 for two -tale tests.
00:59
That's how we can get it.
01:00
So there's our value.
01:03
2 .571.
01:04
We also say it could be the negative of it, or sorry, we're looking for, is it greater than this value of 2 .5 or less than the negative 2 .5? and let's compute our statistic.
01:16
Oh, and it's already done for us...