data entry manager is looking to see how effective the employees are in entering
ID: 3252451 • Letter: D
Question
data entry manager is looking to see how effective the employees are in entering different applications for processing. Three different employees were reviewed for 3 days on applications processed. The manager is looking determine if there is a difference in production from each employee.
Day 1
Day 2
Day 3
Person 1
120
160
150
Person 2
95
135
100
Person 3
140
75
175
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
905.5556
2
452.7778
0.350538
0.717828
5.143253
Within Groups
7750
6
1291.667
Total
8655.556
8
Since the F value 0.350 which is lower than the F critical value of 5.14 and P is 0.717828 we can conclude there is not a large difference in production for each employee. The significance level is 0.05 and since the P-value is higher than the significance level the hypothesis will not be rejected.
Suppose we had rejected the null hypothesis. How would we know which employee's performance was better?
Day 1
Day 2
Day 3
Person 1
120
160
150
Person 2
95
135
100
Person 3
140
75
175
Explanation / Answer
In case if we would have been rejected the null hypothesis, the result would have been that atleast one of the mean is different from the others. To check which employees performance was better, we need to perform pariwise comparison test and with post hoc analysis using Tukey - Kramer or other procedures, we can check which employees performance was better.
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