Question

Day Labs	Agent	Transaction
24	1	1
24	1	2
21	1	3
20	1	4
21	1	5
25	1	6
21	1	7
27	1	8
23	1	9
21	1	10
24	1	11
26	1	12
23	1	13
24	1	14
23	1	15
23	1	16
23	1	17
25	1	18
22	1	19
25	1	20
18	2	1
20	2	2
20	2	3
24	2	4
22	2	5
29	2	6
23	2	7
24	2	8
28	2	9
19	2	10
24	2	11
25	2	12
21	2	13
20	2	14
24	2	15
22	2	16
19	2	17
26	2	18
22	2	19
21	2	20
10	3	1
11	3	2
8	3	3
12	3	4
12	3	5
10	3	6
14	3	7
9	3	8
8	3	9
11	3	10
16	3	11
12	3	12
18	3	13
14	3	14
13	3	15
11	3	16
14	2	17
9	3	18
11	3	19
12	3	20
15	4	1
13	4	2
18	4	3
16	4	4
12	4	5
19	4	6
10	4	7
18	4	8
11	4	9
17	4	10
15	4	11
12	4	12
13	4	13
13	4	14
14	4	15
17	4	16
16	4	17
17	4	18
14	4	19
16	4	20
33	5	1
22	5	2
28	5	3
35	5	4
29	5	5
28	5	6
30	5	7
31	5	8
29	5	9
28	5	10
33	5	11
30	5	12
32	5	13
33	5	14
29	5	15
35	5	16
32	5	17
26	5	18
30	5	19
29	5	20

Use data above please.

a) Prepare aligned box plots of the data. Do the factor level means appear to differ? Does the variability of the observations within each factor level appear to be approximately the same for all factor levels?

b) Obtain the fitted values.

c) Obtain the residuals. Do they sum to zero in accord with (16.21)?

d) Obtain the analysis of variance table.

e) Test whether or not the mean time lapse differs for the five agents; use α =10. Sate alternatives, decision rule, and conclusion .

f) What is the P-value of the test in part (e)? Explain how the same conclusion as in part (e can be reached by knowing the P-value f.

g) Based on the box plots obtained in part (a), does there appear to be much variation in te mean time lapse for the five agents? Is this variation necessarily the result of difterences in the efficiency of operations of the five agents? Discuss.

Answer 1

a)

Variability of the observations appears same for the transactions but different for agents

b) & c) Fitted values and residuals in the table below

Fits and Diagnostics for Unusual Observations

Obs	Day Labs	Fit	Resid	Std Resid
26	29.00	24.04	4.96	2.05	R
29	28.00	21.64	6.36	2.63	R
53	18.00	12.56	5.44	2.25	R
57	14.00	20.50	-6.50	-2.66	R
82	22.00	27.72	-5.72	-2.36	R

the sum of the residuals is not zero

d) Variance table as below

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Regression	23	4337.00	188.57	24.46	0.000
Agent	4	4210.01	1052.50	136.51	0.000
Transaction	19	141.76	7.46	0.97	0.507
Error	76	585.99	7.71
Lack-of-Fit	75	573.49	7.65	0.61	0.795
Pure Error	1	12.50	12.50
Total	99	4922.99

e) & f) Method - One way ANOVA analysis to check the means

Null hypothesis	All means are equal
Alternative hypothesis	Not all means are equal
Significance level	α = 0.1

Analysis of Variance

Source	DF	Adj SS	Adj MS	F-Value	P-Value
Agent	4	4195.2	1048.81	136.91	0.000
Error	95	727.7	7.66
Total	99	4923.0

Since the P-value is less than the level of signifance (o.1 or 10%), we can conclude that not all the means for the agents are equal

g) Yes there is a significant difference between the mean time lapse between the agents and due to this variation there is a difference in the efficiencies of the operation of five agents