You are trying to see if employees are more content working from home than going into the workplace. You ask a group of employees how content they are with their job while working from the office and then later you ask the same question (how content they are with their job) while they are working from home. The results are in the table below. What is your conclusion?
Feb |
June |
D |
D2 |
|
19 |
22 |
3 |
9 |
|
22 |
20 |
-2 |
4 |
|
14 |
18 |
4 |
16 |
|
15 |
18 |
3 |
9 |
|
18 |
23 |
5 |
25 |
|
15 |
14 |
-1 |
1 |
|
17 |
21 |
4 |
16 |
|
20 |
24 |
4 |
16 |
|
13 |
18 |
5 |
25 |
|
21 |
23 |
2 |
4 |
|
SUM |
174 |
201 |
27 |
125 |
St Dev |
3.0984 |
3.1073 |
2.4060 |
8.5016 |
June | Feb | Difference |
22 | 19 | 3 |
20 | 22 | -2 |
18 | 14 | 4 |
18 | 15 | 3 |
23 | 18 | 5 |
14 | 15 | -1 |
21 | 17 | 4 |
24 | 20 | 4 |
18 | 13 | 5 |
23 | 21 | 2 |
∑d = 27
∑d² = 125
n = 10
Mean, x̅d = Ʃd/n = 27/10 = 2.7
Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(125-(27)²/10)/(10-1)] = 2.4060
Null and Alternative hypothesis:
Ho : µd = 0 ; H1 : µd > 0
Test statistic:
t = (x̅d)/(sd/√n) = (2.7)/(2.406/√10) = 3.5487
df = n-1 = 9
p-value = T.DIST.RT(3.5487, 9) = 0.0031
Decision:
p-value < 0.05, Reject the null hypothesis
Conclusion:
There is enough evidence to conclude that employees are more content working from home than going into the workplace.
Get Answers For Free
Most questions answered within 1 hours.