SOLUTION: The Human Resources Director for a large company is studying absenteeism among hourly workers. A sample of 120 day shift employees showed 15 were absent more than five days last ye

Click here to see ALL problems on Probability-and-statistics

Question 327054: The Human Resources Director for a large company is studying absenteeism among hourly workers. A sample of 120 day shift employees showed 15 were absent more than five days last year. A sample of 80 afternoon employees showed 18 to be absent five or more times. At the 0.01 significance level can we conclude that there is a higher proportion of absenteeism among afternoon employees?
a. State the null and alternate hypotheses.
b. State the decision rule.
c. Compute the value of the test statistic.
d. Compute the p-value.
e. What is your decision regarding the null hypothesis?
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The Human Resources Director for a large company is studying absenteeism among hourly workers.
---------------------
A sample of 120 day shift employees showed 15 were absent more than five days last year.
-------------
A sample of 80 afternoon employees showed 18 to be absent five or more times.
-------------
At the 0.01 significance level can we conclude that there is a higher proportion of absenteeism among afternoon employees?
-------------
a. State the null and alternate hypotheses.
Ho: p(aft)-p(day) = 0
Ha: P(aft)-p(day) > 0
-----------------------------
b. State the decision rule.
reject Ho if ts > 2.3263
==============================
c. Compute the value of the test statistic.
I ran a 2-PropZtest and got:
z = 1.8665
---------------------------
d. Compute the p-value.
P(z > 1.8665) = 0.0310
===========================
e. What is your decision regarding the null hypothesis?
Since the p-value is greater than 1%, fail to reject Ho.
===============================
Cheers,
Stan H.
==================