A company decided to test the hypothesis that the average time a company's employees are spending to check their private e-mails at work is more than 6 minutes.A random sample of 40 employees were selected and they averaged 6.6 minutes.It is believed that the population standard deviation is 1.7 minutes.The α is set to 0.05.The critical value for this hypothesis test would be ______.
A) 6.0
B) 7.4
C) 6.4
D) 7.9

Question

Quizplus · Accepted Answer

The hypothesis test uses a t-distribution since the population standard deviation is unknown and the sample size is less than 30. The null hypothesis is that the average time spent checking private emails is 6 minutes or less (H0: μ ≤ 6), while the alternative hypothesis is that it is more than 6 minutes (Ha: μ > 6). The significance level α is 0.05, so the critical value can be found using a t-distribution with 39 degrees of freedom (40 employees minus 1). Using a t-table or a calculator, the critical value is 1.684. The test statistic can be calculated using the formula:

t = (x̄ - μ) / (s / √n)

where x̄ is the sample mean (6.6 minutes), μ is the hypothesized population mean (6 minutes), s is the sample standard deviation (1.7 minutes), and n is the sample size (40 employees). Plugging in the values, we get:

t = (6.6 - 6) / (1.7 / √40) ≈ 2.16

The calculated t-statistic is greater than the critical value of 1.684, so we reject the null hypothesis and conclude that there is evidence to support the claim that the average time spent checking private emails is more than 6 minutes. Therefore, the critical value for this hypothesis test would be the t-value at 0.05 level of significance with 39 degrees of freedom, which is approximately 1.684. The closest option is C) 6.4.

A Company Decided to Test the Hypothesis That the Average