Question

3.)

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. Are the data statistically significant at level α? Based on your answers, will you reject or fail to reject the null hypothesis?

Select one:

a. The P-value is greater than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.

b. The P-value is less than than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.

c. The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.

d. The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.

e. The P-value is less than than the level of significance and so the data are not statistically significant. Thus, we fail to reject the null hypothesis.

Answer 1

3)

Solution :

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :   = 39.8

H_a :    39.8

= 33.1

= 39.8

= 16.2

n = 38

Test statistic = z

= ( - ) / / n

= (33.1 - 39.8) / 16.2 / 38

= -2.55

Test statistic = -2.55

P(z < -2.55) = 0.0054

P-value = 2 * 0.0054 = 0.0108

= 0.10

P-value <

c. The P-value is less than than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.