Question

A professional employee in a large corporation receives an average of µ = 41.7 e-mails per day. An anti-spam protection program was installed in the company's server and one month later a random sample of 45 employees showed that they were receiving an average of ?̅= 36.2 e-mails per day. Assume that ơ = 18.45. Use a 5% level of significance to test whether there has been a change (either way) in the average number of emails received per day per employee.

a.) What is α? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

α =

H0: μ =

H1: μ Select an answer > < not =  

The test is a Select an answer [left-tailed],[two-tailed],[right-tailed] test

b.) Identify the Sampling Distribution you will use. What is the value of the test statistic?

The best sampling distribution to use is the Select an answer[Student's t]or [normal] Distribution

The test statistic (z or t value) is =

c.) Find or estimate the P-value for the test.

The p-value is =

d.) Conclude the test.

Based on this we will Select an answer [Reject] or [Fail to reject the null hypothesis].

Answer 1

Solution :-

Given that ,

= 41.7

= 36.2

= 18.45

n = 45

This is the two tailed test .

The null and alternative hypothesis is ,

H0 :   = 41.7

Ha :    41.7

Test statistic = z

= ( - ) / / n

= (36.2 - 41.7 ) /18.45 / 45

= -2.00

The test statistic = -2.00

P - value = 2 * P ( Z < -2.00 )

= 2 * 0.0228

= 0.0456

P-value = 0.0456

= 0.05

0.0456 < 0.05  

P-value <

Reject the null hypothesis .

There is sufficient evidence to the test claim .