A professional employee in a large corporation receives an average of µ = 41.7 e-mails per...
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 (>, not =, <) ___
The test is a:
right-tailed
left-tailed
two-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:
Student's t
Normal
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:
Fail to reject
Reject the null hypothesis.
Homework Answers
Answer a)
Since level of significance is 5%, so α = 0.05
The following null and alternative hypotheses need to be tested:
Ho: μ = 41.7
Ha: μ not = 41.7
The test is a two-tailed test as here we are not concerned about whether there is increase or decrease in the average number of emails received per day per employee. Our aim is to test the claim that whether there has been a change (either way) in the average number of emails received per day per employee.
Answer b)
In this case, we will be using normal distribution as population standard deviation σ is known.
Test Statistics
The test statistic z = -2.000 (Rounded to 3 decimal places)
Answer c)
P-value corresponding to z = -2 is 0.0455 (Obtained using online p-value calculator for for two-tailed test)
Answer d)
Since p = 0.0455 < α = 0.05, we reject the null hypothesis.
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