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 p-value for this hypothesis test would be ______.
A) 0.0644
B) 0.0129
C) 0.0268
D) 0.0395

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test since we don't know the population standard deviation and the sample size is less than 30.

Null hypothesis: H0: μ ≤ 6 (the average time spent checking private e-mails is less than or equal to 6 minutes)

Alternative hypothesis: H1: μ > 6 (the average time spent checking private e-mails is more than 6 minutes)

The test statistic can be calculated as:

t=(x̄-μ)/(s/√n)=(6.6-6)/(1.7/√40)=2.35

Degrees of freedom= n-1 = 40-1 = 39

Using a t-distribution table or calculator with 39 degrees of freedom and α=0.05, the critical value is 1.684.

The p-value is the probability of obtaining a test statistic as extreme as t=2.35 or more extreme, assuming the null hypothesis is true.

The p-value can be calculated as P(T>2.35)=0.0129

Since the p-value (0.0129) is less than the level of significance (0.05), we reject the null hypothesis. Therefore, we can conclude that the average time employees spend checking their private emails is more than 6 minutes.

A Company Decided to Test the Hypothesis That the Average