00:01
We have a distribution of batteries that have a standard deviation of 1 .25 hours, and we have 10 batteries selected, and we have a mean of 40 .5 hours from that sample.
00:11
And we want to know if we have evidence to show that the mean is actually greater than 40 hours.
00:17
And so we're doing a z test, and the z test because this standard deviation is known.
00:24
So the z value associated with this is the difference between the sample amine and the population, mean divided by the standard air.
00:33
And this is the z value that we get, and we're doing an upper tail test.
00:38
So that's one way is to find the p value, which is what we do in part b.
00:42
But since our alpha level is 0 .05, i looked up what z value has 5 % in the upper tail.
00:50
And this is our critical value.
00:52
And we can see that this test statistic is smaller.
00:55
It is not in the reject region.
00:58
This is where we reject the null, and this is where we fail to reject.
01:04
And so we fail to reject the null, so we don't have evidence to say that the mean is higher than 40.
01:10
Now, part b wanted to know what the p value was, and truthfully, i usually find the p value, and know and even look up the critical value, but in any case, the p value is about 10%.
01:21
And our alpha level is 0 .05, and this is greater than that level, so we would fail to reject the null if we had used the p value.
01:30
Now, the next question asked you to find the probability of failing to reject the null at a 5 % significance level if we have the actual mean being 39 hours.
01:44
And this is really finding the likelihood of a type 2 error or a beta error.
01:49
And so we want to find, well, where would we reject the null, fail to reject the null? and we know that the z value from the previous question for where we have cutoff point is when the z value is 1 .45, 1 .645...