Before |
After |
170 |
185 |
175 |
180 |
165 |
180 |
155 |
172 |
160 |
175 |
175 |
190 |
170 |
195 |
163 |
175 |
182 |
190 |
175 |
190 |
(a)
H0: Null Hypothesis: 0 ( the work schedule was not effective in increasing the performance rating significantly )
HA:Alternative Hypothesis: <0 ( the work schedule was effective in increasing the performance rating significantly ) (Claim)
(b)
We use Dependent Samples t test (Paired t test)
(c)
From the given data, values of d = Before - After are got as follows:
d = Before - After =-15, - 5, - 15, - 17, - 15, - 15, -25, - 12, - 8, - 15
From d values the following statistics are calculated:
n = 10
= - 14.2
sd = 5.320
Take = 0.05
df = 10 - 1 = 9
From table, critical value of t = - 1.833
Test Statistic is given by:
Since calculated value of t = - 8.426 is less than critical value of t = - 1.833, the difference is significant. Reject null hypothesis.
Conclusion:
The data support the claim that the work schedule was
effective in increasing the performance rating
significantly.
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