ii An important factor in selecting software for wordprocess

(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).

Solution

(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