(ii) An important factor in selecting software for word-processing and database
ID: 3020371 • Letter: #
Question
(ii) An important factor in selecting software for word-processing and database management systems is the time required to learn how to use the system. To evaluate three file management systems, a firm designed a test involving five word-processing operators. Since operator variability was believed to be a significant factor, each of the five operators was trained on each of the three file management systems. The data obtained follow. SYSTEM A B C 1 16 16 24 2 19 17 22 OPERATOR 3 14 13 19 4 13 12 18 5 18 17 22 (a) Compute all that apply: SST-------------, SSTR--------------, SSBL-------------, SSE-------------, MSTR--------------, MSBL-------------, MSE---------------. (b) Using = .05, test to see whether there is any difference in the mean training time (in hours) for the three systems (give full ANOVA table values).
Explanation / Answer
(ii) An important factor in selecting software for word-processing and database management systems is the time required to learn how to use the system. To evaluate three file management systems, a firm designed a test involving five word-processing operators. Since operator variability was believed to be a significant factor, each of the five operators was trained on each of the three file management systems
operator
A
B
C
Total
1
16
16
24
56
2
19
17
22
58
3
14
13
19
46
4
13
12
18
43
5
18
17
22
57
Total
80
75
105
260
SST =162+162+….222 - 2602/15 =4682-2602/15 =175.3333
SSTR = 802/5 +752/5 +1052/5 -2602/15 =103.3333
SSBL=562/3+582/3+462/3+432/3+572/3-2602/15 =64.6667
SSE=175.3333-103.3333-64.6667=7.3333
MSTR=103.3333/2=51.6667
MSBL=64.6667/4=16.1667
MSE=7.3333/8=0.9167
ANOVA table
Source
SS
df
MS
F
Treatments
103.3333
2
51.6667
56.36
Blocks
64.6667
4
16.1667
17.64
Error
7.3333
8
0.9167
Total
175.3333
14
Test: to see whether there is any difference in the mean training time (in hours) for the three systems
calculated F=56.36
table value at 5%, F( 2,8) =4.46
Calculated F=56.36 > table value 4.46
The null hypothesis is rejected.
We conclude that there is difference in the mean training time (in hours) for the three systems.
operator
A
B
C
Total
1
16
16
24
56
2
19
17
22
58
3
14
13
19
46
4
13
12
18
43
5
18
17
22
57
Total
80
75
105
260
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