In: Statistics and Probability
A management consultant took a simple random sample of 10 government employees and another simple random sample of 10 private sector workers and measured the amount of time (in minutes) they spent in coffee breaks during the day. The results were as follows:
Government Employees |
Private Sector Workers |
23 |
25 |
18 |
19 |
34 |
18 |
31 |
22 |
28 |
28 |
33 |
25 |
25 |
21 |
27 |
21 |
32 |
20 |
21 |
16 |
Does that data provide sufficient evidence to conclude that government employees take longer coffee breaks on average? Use the p-value method with α = 0.05.
(Round answers to 4 decimal places if necessary)
Test statistic =
P-value =
Reject null hypothesis?
Run t-Test: Two-Sample Assuming Equal Variances in excel followed by below procedures.
1. Go to data tab --> data analysis --> choose t-Test: Two-Sample Assuming Equal Variances
Test statistic = 2.7664
P-value = 0.0127
Reject null hypothesis?
If the p-value is less than 0.05, we reject the null hypothesis that there's no difference between the means and conclude that a significant difference does exist