00:01
For this problem, in the first part, in order to determine if we can use a normal sampling distribution, we want to check if n times p is greater than 5, and if n times 1 minus p is greater than 5.
00:17
So, we have that n is equal to 180, and p, the null hypothesized proportion, is 0 .27.
00:32
And i'll note that n times 1 minus p greater than 5, we could write that equivalently as n times q is greater than 5.
00:40
Q here would be equal to 0 .73.
00:45
So 180 times 0 .27, 48 .6, that's definitely greater than 5.
00:53
And 180 times 0 .73, 131 .4, definitely greater than 5.
00:59
So we can use a normal sampling distribution.
01:03
For determining our hypotheses, but a label as the second step, since our claim has a equality in it, the claim is p is greater than or equal to 0 .27.
01:19
Since there is an equality in the claim, that means our null hypothesis will be the claim, p greater than or equal to 0 .27, and the alternative hypothesis will be the complement of the claim, p less than 0 .27.
01:34
And so, let's see here, we'd want to choose option d as the template and then fill in the blanks with 0 .27 for each.
01:44
For the third step, if a normal sampling distribution can be used, we want to identify the critical values.
01:52
So let's see here, the critical value, which i'll label as z star, or actually they call it z not here, so that's what i'll use.
01:59
The critical value, since this is a left -tailed test, is going to be the negative z statistic for a tail proportion of alpha, or 0 .04.
02:11
So i'll go to a critical value table here.
02:14
You know, we want 0 .04 in the tail, so we'll go as close to it as i can find...