00:01
All right, so we have 1 ,400 computer chips in a sample, and we want to test the company's claim that 60 % of the chips fail in the first hour of use, and we're claiming that less than 60 % fail.
00:22
So our null hypothesis is that p, which is the proportion that fail, is equal to 0 .6, and our alternative, which is what we're trying to, the test is that p is less than 0 .6.
00:37
So this is a one -tailed and specifically a left proportion z test.
00:51
So we need to find a critical z score for an alpha, and let me put that in the box here, an alpha of 0 .02, that's our 2 % significance level.
01:02
And that critical z score is negative 2 .05.
01:06
So what we want to do next is compute the test statistic, and that's given by the the z score formula in the box there.
01:14
And p had is the sample proportion.
01:16
So that's 0 .58, because we said 58 % failed.
01:20
And we're going to subtract the claimed population proportion of 0 .6, and then divide that by the population proportion, 0 .6 times 1 minus that value, which will be 0 .4, over our sample size, which is 1 ,400...