Question

A sprinkler manufacturer claims that the average activating temperatures is at least 132 degrees. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133 degrees. Assume the population standard deviation is 3.3 degrees. Find the standardized test statistic and the corresponding p-value.

z-test statistic = -1.71, p-value = 0.0432

z-test statistic = 1.71, p-value = 0.0432

z-test statistic = 1.71, p-value = 0.0865

z-test statistic = -1.71, p-value = 0.0865

A university claims that the mean time professors are in their offices for students is at least 6.5 hours each week. A random sample of nine professors finds that the mean time in their offices is 6.1 hours each week. With a sample standard deviation of 0.49 hours from a normally distributed data set, can the university’s claim be supported at α=0.05?

Yes, since the test statistic is not in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported

No, since the test statistic is in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported

No, since the test statistic is not in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported

Yes, since the test statistic is in the rejection region defined by the critical value, the null is not rejected. The claim is the null, so is supported

Answer 1

(A) given that population mean = 132, sample mean = 133, sample size n = 32 and population standard deviation = 3.3

z statistics =

setting the values, we get

using z distribution, check 1.7 in left most column and 0.01 in top most row, select the intersecting cell, we get

p value = 0.0432

Therefore, option B is correct answer

(B)

given that population mean = 6.5, sample mean = 6.1, sample size n =9 and population standard deviation = 0.49

degree of freedom = n-1 = 9-1 = 8

using t distribution table, t critical value = -1.86

t statistics =

setting the values, we get

t statistics is less than t critical, so rejecting the null hypothesis and supporting the claim.

No, since the test statistic is in the rejection region defined by the critical value, the null is rejected. The claim is the null, so is not supported