Question

The manager of OfficeSpaceInc. wishes to determine whether employees with an office spend more or less “down time” on the computer than employees working in a cubicle. She tracks the monitors of 7 cubicle-workers (denoted as X) and 3 office-workers for 8 hours each. She finds that the cubicle-workers spend an average of 38 minutes of “down time,” and the office-workers spend an average of 49 minutes of “down time,” with a pooled sample standard deviation of 7 minutes. Suppose the manager determines a level of significance for the test equal to 0.01.1.1

Write down the null and alternative hypotheses. Is this a left-, right-, or two-tailed test?

Calculate the t-score of the samples.

What is the p-value of this sample?

In two sentences, what should the manager conclude from this hypothesis test? Then, in one sentence, explain how you came to this decision

Answer 1

Ho: u1 = u2

Ha: u1 =/ u2

Here, u1 and u2 is the downtime on the computer for employees with an office and working in a cubicle respectively. This is a two-tailed test. Let's perform the independent samples t-test.

Given information:

n1 = 3, n2 = 7

s_p = 7

X1 = 49

X2 = 38

The formula for the t-statistic is:

t = (49 - 38) / √(49*(1/7 + 1/3))

t = 11/4.83 = 2.277

df = n1 + n2 - 2 = 3+7-2 = 8

At df = 8 and t-statistic = 2.277, p-value = 0.0103

As the p-value < 0.05, we can reject the null hypothesis.

The manager should conclude that he has strong statistical evidence to say that the average downtime for employees with an office is significantly higher than the average downtime for employees working in a cubicle. The conclusion was made after a significant p-value was found and as the t-statistic is positive, this tells us that u1 - u2 > 0 and the difference is positive.