A new operator was recently assigned to a crew of workers who perform a certain
ID: 3436210 • Letter: A
Question
A new operator was recently assigned to a crew of workers who perform a certain job. From the records of the number of units of work completed by each worker each day last month, a sample of size five was randomly selected for each of the two experienced workers and the new worker. At the ? = .05 level of significance, does the evidence provide sufficient reason to reject the claim that there is no difference in the amount of work done by the three workers?
(a) Find the test statistic. (Give your answer correct to two decimal places.)
(ii) Find the p-value. (Give your answer bounds exactly.)
? < p < ?
(b) State the appropriate conclusion.
Reject the null hypothesis, there is not significant evidence of a difference in units of work between the workers.
Reject the null hypothesis, there is significant evidence of a difference in units of work between the workers.
Fail to reject the null hypothesis, there is significant evidence of a difference in units of work between the workers.
Fail to reject the null hypothesis, there is not significant evidence of a difference in units of work between the workers.
Workers New A B Units of work (replicates) 9 13 9 11 13 9 11 13 11 9 13 11 10 12 9Explanation / Answer
One factor ANOVA
Mean
n
Std. Dev
10.0
5
1.00
New
12.8
5
0.45
A
9.8
5
1.10
B
10.9
15
1.64
Total
ANOVA table
Source
SS
df
MS
F
p-value
Treatment
28.13
2
14.067
17.58
.0003
Error
9.60
12
0.800
Total
37.73
14
F= 17.58
(ii) Find the p-value. (Give your answer bounds exactly.)
P=0.0003
0.001 < p < 0.0001
(b) State the appropriate conclusion.
Reject the null hypothesis, there is not significant evidence of a difference in units of work between the workers.
Reject the null hypothesis, there is significant evidence of a difference in units of work between the workers.
Fail to reject the null hypothesis, there is significant evidence of a difference in units of work between the workers.
Fail to reject the null hypothesis, there is not significant evidence of a difference in units of work between the workers.
One factor ANOVA
Mean
n
Std. Dev
10.0
5
1.00
New
12.8
5
0.45
A
9.8
5
1.10
B
10.9
15
1.64
Total
ANOVA table
Source
SS
df
MS
F
p-value
Treatment
28.13
2
14.067
17.58
.0003
Error
9.60
12
0.800
Total
37.73
14
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