00:01
So for this problem, since we are testing to see if the mean value is different from nine ounces, we'd have that our null hypothesis is that the mean value is equal to nine ounces, and our alternate hypothesis is that the mean value simply does not equal nine ounces.
00:19
So this is going to be a two -tailed test, and we're told that we have alpha equals 0 .01 as our level of significance.
00:28
So i'll start off by finding our rejection region.
00:33
The rejection region is going to be all z scores, or actually i'll put it this way, r is going to be the set of z scores, such that the magnitude of the z score is greater than z for a proportion of 1 minus alpha over 2 in the tail.
00:53
So that would be 0 .99, or 0 .005 in the tail.
01:00
So if we want a...
01:02
So i'm just going to use my software here for finding that z value.
01:06
So i'd put in normal distribution.
01:09
0 .005.
01:10
I'll get a negative value, but we really just care about what the proportion in the tail is...