Assume the population has a normal distribution and the...

Assume the population has a normal distribution and the number of observations in a random sample is greater than fifty. If a z-test is used to test a null hypothesis about the population mean versus the alternative that the true mean is greater than that specified by the null hypothesis, what is the critical value for a one-tailed test? Assume the type one-error rate is 0.05. A) +2.052 B) +1.645 C) +2.576 D) +2.33

Transcript

00:01 In the asked question, we assume the population has a normal distribution and the number of observations in a random sample is greater than 50.

00:09 If a z test is used to test a null hypothesis about the population mean versus the alternative that the true mean is greater than that's satisfied by the null hypothesis.

00:22 It's required to find the critical value for a one -tail test.

00:27 Assume the type 1 error.

00:30 Type 1 error is 0 .05 this is the same as alpha to find this critical value let's visualize first the standard normal distribution it's a bell -shaped distribution like that centered at 0 and symmetric about it for a one -tail test it might be right tail or left tail we can decide from the alternative hypothesis hypothesis says that the true mean is greater than that satisfied by the null hypothesis is greater than new node as long as it's greater than means its right tail test we need z critical here to give us type 1 error or alpha to be the specified value 0 .05 how we can find these z critical we can use the standard normal table the standard normal table the standard table gives us the area to the left of a z score and we can find the z score by finding the area to its left first here if we can find this area we can enter the table by this area to find the corresponding z grade how we can find it it's very easy because we know the total area under the distribution is 1 this makes the dotted area here to be 0 .95 now we can enter the table by this value by this area to find z credit we search for an area of 0 .95 we can find it in this row and it's approximately between this value and this value then we can take the average z score between 1 .6 5 and 1 .66 it's going to be about 1 .6 5 sorry, this is wrong.

03:05 This must be 4 and 5.

03:08 This is wrong...

Question

Please give Ace some feedback