A professional employee in a large corporation receives an average of μ = 39.8 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 38 employees showed that they were receiving an average of x = 33.1 e-mails per day. The computer server through which the e-mails are routed showed that σ = 16.2. Has the new policy had any effect? Use a 10% 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.

Question

Terrence Mack

Mathematics

26 September, 08:42

A professional employee in a large corporation receives an average of μ = 39.8 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 38 employees showed that they were receiving an average of x = 33.1 e-mails per day. The computer server through which the e-mails are routed showed that σ = 16.2. Has the new policy had any effect? Use a 10% 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.

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Kristina Armstrong · Answer 1

Step-by-step explanation:

Null hypothesis: u = 39.8

Alternative: u = / 39.8

Using a one sample z test: the formula is

z = x-u / (sd/√n)

Where x = 33.1 u = 39.8, sd = 16.2 and n = 38

Thus we have:

z = 33.1-39.8 / (16.2/√38)

z = - 6.7 / (16.2/6.1644)

z = - 6.7 / 2.6280

z = - 2.5495

To be able to arrive at a conclusion, we have to find the p value, the p value at a 0.1 significance level for a two tailed test is 0.0108. This is way less than 0.1 thus we will reject the null and conclude that there has been a change (either way) in the average number of e-mails received per day per employee. Yes, the new policy had an effect.