Question

The mean work week for engineers in a start-up company is believed to be about 62 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she conclude at the 5% significance level that the mean work week is shorter than 62 hours?Data (length of mean work week): 70; 50; 55; 60; 65; 55; 55; 60; 50; 55.

Answer 1

From the data given using Minitab ,we get

sample mean is =57.5

sample standard deviation is s = 6.3465

sample size is n=10

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H_o: μ ≥ 62

H_a: μ < 62

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Test Statistics

The t-statistic is computed as follows:

Based on the information provided, the significance level is α=0.05 and degree of freedom df = n - 1 = 10 -1 = 9 , the critical value for a left-tailed test is t_c​ =−1.833

Since it is observed that t=−2.242< t_c​=−1.833, it is then concluded that the null hypothesis is rejected.

Hence, she can conclude at the 5% significance level that the mean work week is shorter than 62 hours.