4. Case studies

Calculations for the Bartlett test in {2} of Table 7 show that χ<sup>2</sup>

The ratio of the variation between stages to within features is v<sup>2</sup> = s3

This leads to an updated population variance is 1.525 less than s1

Variation sources SSD df s

Within stages 21.333 12 1.778

Variation sources SSD df s

Residual mean square 21.345 14 1.525

Subtotal 20.444 5

Total 41.778 17

Total 41.778 17

Table 9. ANOVA table—two-stage without interaction.

Table 8. Two-stage ANOVA table of Example 8.

An estimate of the population variance is s1

effects between stages are significantly discriminated.

2 /s1

between features within stages are almost the same.

Interaction.

Table 8 shows that v<sup>2</sup> = s2

70 Management of Information Systems

unbiased estimates of σ<sup>2</sup>

hypothesis that population variance is the same for all features is accepted at α = 5%.

part {3} of Table 7 is the calculation scheme for the terms in Table 8, where Subtotal equals Total minus Within stages or the sum of Between features within stages, Between stages, and

which by far exceeds the 95% percentile of Fisher distribution F0.95(1,12) = 4.75. That means the difference of the expected means between stages is different significantly. In other words, the

Similarly, in comparison of the variation within features and between features within stages,

difference between the expected means of features within stages is not significant or the effects

Beside the above effects, the interaction between stages and features is also a factor need to be considered. The ratio v<sup>2</sup> = 0.006/0.012 = 0.50 gives that such an interaction is not present in given dataset. Thus, both the lines labeled "Interaction" and "Within stages" give the same

The residual mean square is a sum of variations between the Interaction and Within stages.

obviously increases v<sup>2</sup> ratios. Table 9 analyzes the interaction without stage of Example 8.

Between stages 14.222 1 14.222 8.0 Between features within stages 6.210 2 3.105 1.747 Interaction 0.012 2 0.006 0.50

Between stages 14.222 1 14.222 9.328 Between features within stages 6.210 2 3.105 2.036

cal = 1.194 < χ<sup>2</sup>

<sup>2</sup> = 21.33/(2 <sup>3</sup> [3–1]) = 1.778, cf. Table 8. The

2 /s1

<sup>2</sup> = 3.105/1.778 = 1.747 < F0.95(2,12) = 3.89. This shows that the

, since a combination of these lines can improve the estimate of σ<sup>2</sup>

0.95(5) = 11.07, the

<sup>2</sup> = 14.222/1.778 = 8.0

<sup>2</sup> = 1.778 in Table 8, but

<sup>2</sup> v<sup>2</sup>

<sup>2</sup> v<sup>2</sup>

.

To evaluate the extent to which MIS is being used to attain achievements of long-term planning, short-term planning in the South-West Nigerian universities [14], all selected features are f1: Construction of building in the university, f2: Student enrolment projection, f3: Manpower projection, f4: Staff recruitment exercises, f5: Establishment of new faculties and department, f6: Designing university academic program, f7: Stock library with books and journals are considered in long-term evaluation. For short-term, f1: Promotion of Staff, f2: Staff Training and Development, f3: Appointment of Deans or Heads of Departments or Divisions, f4: Appointment of Committee Members, f5: Allocation of offices to staff, f6: Allocation of Residential Quarters, f7: Allocation of Lecture room/theaters, f8: Full-time equivalent or Teacher-Students Ratio, and f9: Maximum Teaching Load are considered.

In evaluation of the extent of our university for a 5-year strategy planning, the following features are used f1: Effectuation rights and obligations of students, f2: Promotion of international cooperations, f3: Library, equipment and material facilities, f4: Potential of Scientific R&D and transfer of technology, f5: Capacity of organization and management, f6: Design of university academic programs, f7: Promotion of academic operations, f8: Capacity of manpower projection, f9: Management of finance and resources. These basic features are factors to evaluate whether the university attains its goal and objectives. Each basic feature is evaluated in the scale of 100 but here it is illustrated in the one of 20 points.

Example 9: Let fi, i = 1, 2,…, 9 be features characterized as the extent of an MIS as above. ωij, j = 1, 2,…, 12 is a value that is evaluated as the ith feature by the jth evaluator in a shorten marking scheme of 20. Calculations for the single-stage ANOVA table are shown in Table 10.

The calculated value χ <sup>2</sup> cal = 9.432 in Table 10 does not exceed χ<sup>2</sup> 0.95(8) = 15.51, the hypothesis on equality of variances is accepted. The population variance is estimated as s<sup>2</sup> = 1185.58/ 99 = 11.976. The corresponding ANOVA table for this dataset is given in Table 11.

Here, as variance ratio v<sup>2</sup> = 8.907 far exceeds F0.95(8,99) = 2.06, it is unreasonable to assume that all the expected means of features are the same. This can also be seen from Table 10, where all sum of features from f1 to f4 are less than the ones of features from f5 to f9.

A more detailed examination revealed that the nine features can be partitioned into two groups, namely A = {f1, f2, f3, f4} with the first four features and B = {f5, f6, f7, f8, f9} with the remainders. Each group of features can be seen as a treatment and its observation sample includes all observations in the same group. Since, it would be reasonable to consider the variation between features into three portions between: the features from A, the features from B, and between group A and B. Calculations in this consideration are extracted from Table 10 and illustrated in Table 12.


Table 10. Calculations for single-stage ANOVA dataset.


Table 11. Single-stage ANOVA table of Example 9.

In comparison with the variance within features s<sup>2</sup> , the variance ratios v<sup>2</sup> = 23.243/ 11.976 = 1.941 < F0.95(3,99) = 2.66 and v<sup>2</sup> = 113.60/11.976 = 2.371 < F0.95(4,99) = 2.43 in Table 13 show that there is no essential difference between features in the same group. Since the third ratio v<sup>2</sup> = 183.33/11.976 = 15.309 is far greater than F0.95(1,99) = 3.9, the features in group A and B do have different expected mean.

Example 10: Assume that in an MIS, there are two stages that need ANOVA with the same set of features. In each stage, samples of evaluations in marking scheme of 20. Let ωνij be an integral value in marking scheme of 20 that evaluates the ith feature given by the jth evaluator


Table 12. Calculations for ANOVA between group A and B.


Table 13. ANOVA table of two groups A and B.

In comparison with the variance within features s<sup>2</sup>

Total 2038.917 107

{2} Bartlett df:8 Anova

Variation sources SSD df s

Within features 1185.583 99 11.976

B do have different expected mean.

Table 11. Single-stage ANOVA table of Example 9.

: 1.078 χ<sup>2</sup>

72 Management of Information Systems

Table 10. Calculations for single-stage ANOVA dataset.

Si2

Σlogsi 2

log.s<sup>2</sup>

11.976 = 1.941 < F0.95(3,99) = 2.66 and v<sup>2</sup> = 113.60/11.976 = 2.371 < F0.95(4,99) = 2.43 in Table 13 show that there is no essential difference between features in the same group. Since the third ratio v<sup>2</sup> = 183.33/11.976 = 15.309 is far greater than F0.95(1,99) = 3.9, the features in group A and

Between feature 853.333 8 106.667 8.907

/ni 630.75 320.33 432 833.3 1496.3 1323 1323 1776.3 2408.3 SSDi 122.25 233.67 124 204.7 135.67 69 55 83.667 157.67 logSSDi 2.087 2.369 2.09 2.311 2.133 1.839 1.740 1.923 2.20

/k: 1.036 c: 1.034 ΣSi: 1023 ΣSSDi: 1185.58

/fi: 10,543 ΣSi2

cal 9.432 ΣSi2

f1 f2 f3 f4 f5 f6 f7 f8 f9

ωi1 13 2 10 2 14 10 9 9 11 ωi2 1 1 3 13 11 10 10 10 14 ωi3 7 5 0 11 13 7 11 15 15 ωi4 9 15 2 17 10 7 11 13 15 ωi5 3 8 9 7 17 9 7 17 20 ωi6 6 2 5 5 14 10 11 15 17 ωi7 7 2 7 2 13 11 15 11 5 ωi8 10 3 10 10 11 13 13 11 15 ωi9 8 13 7 7 10 11 11 8 17 ωi10 11 6 9 8 9 9 10 14 15 ωi11 5 2 7 8 3 14 7 10 15 ωi12 7 3 3 10 9 15 11 13 11 {1} Si 87 62 72 100 134 126 126 146 170 SSi 753 554 556 1038 1632 1392 1378 1860 2566

Example 10: Assume that in an MIS, there are two stages that need ANOVA with the same set of features. In each stage, samples of evaluations in marking scheme of 20. Let ωνij be an integral value in marking scheme of 20 that evaluates the ith feature given by the jth evaluator

, the variance ratios v<sup>2</sup> = 23.243/

/n <sup>S</sup><sup>2</sup>

/nm: 853.33

<sup>2</sup> v<sup>2</sup>

from the νth stage, ν = 1, 2, i = 1, 2,…, 7, j = 1, 2,…, 8. This dataset is in Table 14 including calculation for ANOVA.

Using Bartlett test in {3}, χ<sup>2</sup> cal = 9.507 not exceed χ<sup>2</sup> 0.95(15) = 22.36, so population variances are the same with the pool variance of s2 = 1185.58/96 = 2.105. Table 15 shows this ANOVA.

The ratio v<sup>2</sup> = s2 2 /s1 <sup>2</sup> = 3.932/2.063 = 1.906 is less than F0.95(6,98) = 2.15, the difference between the expected means within stages is not significant. Similarly, v<sup>2</sup> = s3 2 /s1 <sup>2</sup> = 37.723/ 2.063 = 18.29 > F0.95(1,98) = 3.96, the expected means between stages are discriminated.

Since v<sup>2</sup> = 0.057/0.339 = 0.167 less than the 95% percentile of Fisher distribution, any interaction does not exist. Thus, "Interaction" and "Within stages" variation sources are combined to s1 <sup>2</sup> = (202.125 + 0.339)/104 = 1.947 a better estimation for σ<sup>2</sup> than 2.063 in Table 15.

The case of m = 1 and k = 2 has been presented in the previous subsection with group A, B. In [15], ANOVA has been used to specify whether a statistical relationship exists between human development index and security index. The authors in [16] have used the ANOVA combined with regression analysis to assess and evaluate student MIS of a university.

In this subsection, the student test is presented in comparison with the effects of f from the two stages or treatments. Let {ωij} i = 1, 2 and j = 1, 2,…, ni be two observation samples of sizes ni drawn from the two treatments of the feature f. Using Eqs. (21)–(23), the means ϖ1, ϖ<sup>2</sup> and variances s1 2 , s2 <sup>2</sup> are calculated with df1 = n1–1, df2 = n2–1.

The equality of population variances is tested using Fisher distribution with v<sup>2</sup> = s1 2 /s2 2 . If <sup>v</sup><sup>2</sup> < Fα/2(df1, df2) or <sup>v</sup><sup>2</sup> > F1<sup>α</sup>/2(df1, df2), it is unreasonable to assert that the population


Table 14. Calculations for two-stage ANOVA dataset.

variances are equal. Otherwise, the pool variance of these treatments is s2 = (SSD1 + SSD2)/ (df1 + df2).


Table 15. Two-stage ANOVA table of Example 10.

The equality of the expected means from treatments is tested by the student distribution based on the difference ϖ<sup>0</sup> = ϖ<sup>1</sup> � ϖ2. If this hypothesis is correct, there are two cases:


The hypothesis that the two expected means of the feature f from the treatments are equal is rejected at a level of significance α when |tcal|>t1 � <sup>α</sup>/2(df). Otherwise, the confidence interval of the difference η between the two means is

$$
\mathfrak{a}\mathfrak{o} + \mathfrak{t}\_{a/2}(\mathrm{df})\mathfrak{s}\_{\mathfrak{o}} < \mathfrak{\eta} < \mathfrak{w}\mathfrak{\eta} + \mathfrak{t}\_{1-a/2}(\mathrm{df})\mathfrak{s}\_{\mathfrak{o}} \tag{31}
$$

where t1 � <sup>α</sup>/2(df) is the 100(1�α/2)% percentile of the student distribution, t1�<sup>α</sup>/2(df) = �tα/2(df).

For instance, from Table 12, the variances in groups sA <sup>2</sup> = 69.729/47 = 1.484 and sB <sup>2</sup> = 113.60/ 59 = 1.925 give v<sup>2</sup> = sA 2 /sB <sup>2</sup> = 1.30 less than t0.995(106) = 2.35. It is accepted the variances in group A and B are equal. The pool variance is estimated by s<sup>2</sup> = (SSD1 + SSD2)/(df1 + df2) = 113.60/ 106 = 1.729. Also, Table 12 gives so <sup>2</sup> = s<sup>2</sup> .[1/47 + 1/59] = 0.06611 and tcal = (11.7–6.6875)/so = 19.68, this so far exceeds t0.995(106) = 2.606. The student test for these two treatment shows the expected mean of group B so far exceeds the one of A. The 99.5% confidence interval of the difference between these expected means is 11.7 � 6.6875 � 2.606 � √0.06611 or (4.342, 5.683).

Similarly, Table 15 shows that there is no difference in evaluating features by evaluators within stages in Example 10. It is reasonable to group features in each stage to each other and using the method of comparison between two treatments of a feature as above.

#### 5. Conclusion

variances are equal. Otherwise, the pool variance of these treatments is s2 = (SSD1 + SSD2)/

: 0.314 S<sup>2</sup>

cal: 9.507 SS1 + SS2 – S<sup>2</sup>

f1 f2 f3 f4 f5 f6 f7 ω1i1 16 15 17 14 14 13 14 k = 2 ω1i2 13 14 11 12 10 13 12 m = 7 ω1i3 14 16 15 14 15 14 16 n = 8

{1} S1i 103 114 107 104 104 103 107 742 SS1i 1347 1634 1459 1370 1368 1339 1447 9964

{2} S2i 112 123 118 113 112 113 116 807 SS2i 1576 1907 1760 1607 1584 1601 1688 11,723

/ni 1568 1891 1741 1596 1568 1596 1682 11641.9 SSD2i 8 15.88 19.5 10.88 16 4.875 6 81.125 logSSD2i 0.903 1.201 1.29 1.036 1.204 0.688 0.778 7.101 (S1i+S2i)2 46,225 56,169 50,625 47,089 46,656 46,656 4973 343,149 {3} Bartlett Anova S=S1 + S2 1549

/(mn): 21460.9

/(kmn): 21423.2 37.723

/(kmn): 23.589

/(kmn): 263.77

/ni 1326 1625 1431 1352 1352 1326 1431.1 9843 SSD1i 20.88 9.5 27.88 18 16 12.88 15.88 121 logSSD1i 1.32 0.978 1.445 1.255 1.204 1.11 1.201 8.512 f1 f2 f3 f4 f5 f6 f7

ω1i4 12 14 13 12 14 12 15 ω1i5 13 15 14 10 13 14 13 ω1i6 10 12 11 13 12 10 12 ω1i7 12 14 13 14 13 13 12 ω1i8 13 14 13 15 13 14 13

ω2i1 16 17 18 14 15 14 15 ω2i2 14 15 13 12 11 13 14 ω2i3 15 18 15 14 16 15 16 ω2i4 13 14 14 15 14 13 14 ω2i5 14 16 15 13 15 14 13 ω2i6 13 14 13 15 14 14 15 ω2i7 14 15 16 14 13 15 14 ω2i8 13 14 14 16 14 15 15

<sup>Σ</sup>logSSDi./(km)log(n1): 0.270 S<sup>2</sup>

<sup>Σ</sup>logSSDi: 15.61 c: 1.051 <sup>Σ</sup>(S1i+S2i)2 /(km)<sup>S</sup><sup>2</sup>

χ2

ΣSSDi: 202.1 logs2

Table 14. Calculations for two-stage ANOVA dataset.

(df1 + df2).

S1i 2

74 Management of Information Systems

S2i 2

> It is dealt with this chapter the useful methods for choosing important features and supporting decisions of a given decision information system, presented in Section 2. The methods of ANOVA are introduced in Section 3 to evaluate features from the extent of an MIS.

The demonstrations of using such methods, through examples and case studies in Section 4 at our Faculty of Information System—University of Information Technology, showed that the efficiency of the proposed methods. The illustrated calculating schemes allow designing and coding computer programs for solving the above problems automatically.
