Question

A professional employee in a large corporation receives an average of μ = 38.8 e-mails per day. Most of these e-mails are from other employees in the company. Because of the large number of e-mails, employees find themselves distracted and are unable to concentrate when they return to their tasks. In an effort to reduce distraction caused by such interruptions, one company established a priority list that all employees were to use before sending an e-mail. One month after the new priority list was put into place, a random sample of 39 employees showed that they were receiving an average of x = 31 e-mails per day. The computer server through which the e-mails are routed showed that σ = 15.4. Has the new policy had any effect? Use a 5% level of significance to test the claim that there has been a change (either way) in the average number of e-mails received per day per employee. What are the null and alternate hypotheses?

Answer 1

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 38.8
Alternative Hypothesis: μ ≠ 38.8

Rejection Region
This is two tailed test, for α = 0.05
Critical value of z are -1.96 and 1.96.
Hence reject H0 if z < -1.96 or z > 1.96

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (31 - 38.8)/(15.4/sqrt(39))
z = -3.16

P-value Approach
P-value = 0.0016
As P-value < 0.05, reject the null hypothesis.

There is sufficient evidence to conclude that the there has been a change (either way) in the average number of e-mails received per day per employee