Question

Before changes to its management staff, an automobile assembly line operation had a scheduled mean completion time of 13.7 minutes. The standard deviation of completion times was 1.3 minutes. An analyst at the company suspects that, under new management, the mean completion time, μ, is now less than 13.7 minutes. To test this claim, a random sample of 30 completion times under new management was taken by the analyst. The sample had a mean of 13.1 minutes. Assume that the population is normally distributed. Can we support, at the 0.01 level of significance, the claim that the population mean completion time under new management is less than 13.7 minutes? Assume that the population standard deviation of completion times has not changed under new management.

Perform a one-tailed test. Then complete the parts below.

(a) State the null hypothesis H0 and the alternative hypothesis H1.

H0: μ ≥ 13.7
H1: μ < 13.7

(b) Determine the type of test statistic to use.

▼(Choose one) Z-test

(c) Find the value of the test statistic. (Round to three or more decimal places.)

(d) Find the p-value. (Round to three or more decimal places.)

(e) Can we support the claim that the population mean completion time under new management is less than 13.7 minutes?

Yes
No

Answer 1

The null hypothesis is that the mean completion time is greater than or equal to 13.7 minutes. The alternative hypothesis is that the mean completion time is less than 13.7 minutes. Using a Z-test, the test statistic is calculated to be -2.747. The p-value is approximately 0.003, which is less than the significance level of 0.01. Therefore, we can support the claim that the population mean completion time under new management is less than 13.7 minutes.

Step-by-step explanation:

(a) Null hypothesis (H0): μ ≥ 13.7

Alternative hypothesis (H1): μ < 13.7

(b) The type of test statistic to use is a Z-test.

(c) The value of the test statistic is calculated using the formula:

Z = (Sample mean - Population mean) / (Population standard deviation / sqrt(sample size))

Z = (13.1 - 13.7) / (1.3 / sqrt(30)) = -2.747

(d) The p-value is the probability that the test statistic is as extreme as the observed value, assuming the null hypothesis is true. Since this is a left-tailed test, we need to find the probability of getting a Z-value less than -2.747. Using a Z-table or calculator, we can find that the p-value is approximately 0.003.

(e) Since the p-value (0.003) is less than the significance level of 0.01, we can reject the null hypothesis. Therefore, we have enough evidence to support the claim that the population mean completion time under new management is less than 13.7 minutes.