Mathematical

Instrumentation

Use of Directional and Non Directional

DEFERENCE OF MEANS FOR DEFENDANT (CORRELATED) SAMPLES

One-Way Classification for a Balanced Array

One Way Classification for Unbalanced , Array

Two-Factor Analysis of VarianceDegrees of Freedom in a Two-Facto

r ANOVA

TABLE OF CONTENTS

USE OF DIRECTIONAL AND NON- DIRECTIONALThe choice of whether to use a non directional test depends upon several factors .If the nature of the study is such that there is very little basis for guessing the outcome of the study, then the non- directional test is generally chosen.In the course of comparing one group with another and some indicators are available to favor one over the other, then a directional test may be used.

STEPS IN HYPOTHESIS TESTING:

• The procedure in hypothesis testing can be summarize in the following steps;– Step 1; Formulate the null hypothesis (Ho) that

there is no significant difference between items being compared. State the alternative hypothesis(Ha) which is used in case Ho is rejected.

– Step 2; Set level of significant,– Step 3; Determine the test to be used. Used z-

test if population standard deviation is given, and t-test if the standard deviation is given from the sample.

Step 4; Determine the tabular value for the test. For a z-test, used the table of the critical values of z based on the area of the normal curve. For a t-test, one must first compare for the degrees of freedom, then look for the tabular value from the table of t-distribution. For a single sample, df= number of items -1 or in symbols. df= n-1

For two samples, the degrees of freedom is calculated as df= n₁+n₂-2Where n₁ refers to the number of items in the first sample and n₂ refers to the number of items in the second sample.

Step 5; Compute for Z or T as needed, using any of the following formula;

A. Z –TEST1. SAMPLE MEAN COMPARED POPULATION MEAN

or

XZ

N

X NZ

Where: = sample mean

= population mean

= population standard derivation

n = no. of items in the sample

2. Comparing two sample means

Where:

X

1

2

1

2

mean of the first sample

= mean of the second sample

n number of items in the first sample

n number of items in second sample

population standrad

X

X

deviation

1

1 2

2

or 1 1

X Xz

n n

1 2

2 21 2

1 2

X XZ

s s

n n

B. t-testThe t-test is a statistical tool we can use in analyzing small samples, usually less than thirty, without having to worry about distorted result.

1.Sample mean compared with population mean

or ( ) 1x nt

s

1

xt

s

n

Where;= sample mean= population mean

s = sample mean n= number of items in the sample

X

2. Comparing two samples for independent samples

or

Where;x₁= mean of the first samplex₂= mean of the second sampleS₁= Standard deviation of the firstS₂= standard deviation of the secondn₁= number of items in the first samplen₂= number of items in the second sample

1 2

2 2

1 1 2 2

1 2 1 2

1 1 1 1

2

X Xt

n S n S

n n n n

1 2

2 21 2

1 2

X Xt

s sn n

3. Difference of means for dependent (correlated) samples

or

Where; = average of the differences between the given items, the pre-test and the post-test

= Standard deviation of (x₁-x₂)n= Number of pairs in the sample

2( )2

1

dt

n d d

n

dt

n

d

1 21

( )n

n

x xd

n

Step 6; compare the computed value with its corresponding tabular

value then state your conclusion based on the following guidelines.

A.Reject ho, If the absolute computed value is equal to or greater than the absolute tabular value. There is a significant differences.

B. Accept ho,If the absolute computed value is less than the absolute tabular

value.

A. Sample mean compared with populationA researcher knows that the average height of Filipino women is 1.525 meters. A random sample of 26 women was Taken and was found to have a mean height of 1.56 meters, with a standard deviation of 0.10 meter. Is there a reason to believe that the 26 women in the sample are significantly taller than the others at 0.05 significant level?

• Step 1; Ho, the sample is not significantly taller than the other Filipino women.

Ha, the sample is significantly taller than the others

• Step 2; Set = 0.05• Step 3; The standard deviation given is based on the

sample so t-test must be used.

( =1.525) or Ho:x x

Application

( >1.525) or Ho:x x

• Step 4. Look for the degree of freedom. Since only the sample is given, df=n-1 df =26-1

df=25• Step 5. The given values in the problem are:

Step 6. The absolute value is greater than the absolute tabular value.

The Ho is rejected (The Samples is significantly taller than the others.)

X=1.56 meters = 1.525 meters

26 s = 0.10

X- 0.035 5t= t= =

s 1 0.10n-1

1.56-1.525t= t= 1.75

0.10 0.353

t= 0.10

25

n

B. Comparing Two Sample Means or Independent Samples

A teacher wishes to test whether or not the modular method of teaching is more effective than the traditional method. She picks two classes of approximately equal intelligence ( verified through an administered IQ test). She gathers a sample of 18 students to whom she uses the modular method and another sample of 14 students to whom she uses the traditional method. After the experiment, an objective test revealed that the first sample got a mean score of 28.6 with a standard deviation of 5.9, while a second group get a mean score of 21.7 with a standard deviation of 4.6. Based in the result of the administered test, can we say that the modular method is more effective than the traditional method.Step 1; Ho; The modular method is as effective as the traditional method.

Ha; The modular method is more effective than the traditional method.

1 2;Ho x x

1 2Ha x x

Step 2; Set ∞= 0.05

Step 3; Since the standard deviations or the samples are given, the t-test is used.

NOTE; x₁= modular method x₂= traditional method

Step 4; Look for the degrees of freedom. For two samples

df= n₁ +n₂ -2Z = 18 + 14 -2 =30

Step 5. The given values in the problem are;

X₁= 28.6 S₁= 5.9 n₁= 18 X₂= 21.7 S₂= 4.6 n₂= 14

1 2

2

1 1 2 2

1 2 1 2

2

1 1 1 12

28.6 21.7

18 1 5.9 14 1 4.6 1 118 14 2 18 14

6.9

17 34.81 21.160.06 0.07

32 2

6.9t =

(28.895)(0.13)

6.9

3.7566.9

t= 1.94

t=3.56

X Xt

n s n s

n n n n

t

t

t

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Step 6; The computed value is in the rejection region. It is greater than the tabular value 1.697. the Ho is rejected. (the modular Method of teaching is more effective than the traditional method.

Diff erence of Means for Dependent (Correlated)

Samples

A group of employees was given in attitude test on E-VAT. Then, they were shown a film favorable to implementation of E-VAT. The test was then re-administered. Is there a significant deference bet. Their attitude before and after they have seen the film? Use ∞= .05. the data are shown below.

x₁ x₂15 25

17 19

19 23

23 27

23 30

21 20

19 23

18 21

9 17

8 17

0 1 21.1 :H

1: 1 21.2 H

1: 1 21.2 H

1.3 interval measurement

1.3 interval measurement

1.5 =.05 n=10

x₁ X₂ D²

15 25 10 100

17 19 2 4

19 23 4 16

23 27 4 16

23 30 7 49

21 20 -1 1

19 23 4 16

18 21 3 9

9 17 8 64

8 17 9 81

2 2 ( )D X X

1 172X 2 222X 50D 2356D

22

1

Dt

N D D

N

2

50

10(356) (50)10 1

50

3560 25009

t

504.607

10.853

1.7 the tabular value of t at .05 with df=9 is 1.833n (table 2) Since the computed value, 4.607 is greater than the tabular value, then the null hypothesis is rejected.1.8 there is a significant deference bet. The attitudes of the employees before and after they have seen the film.

df.10 .05 .025 .01 .005 .0005

Level of Significance for a non-directional test

.20 .10 .05 .02 .01 .001

1 3.078 6.314 12.706 31.821 63.657 636.619

2 1.886 2.92 4.303 6.965 9.925 31.599

3 1.636 2.353 3.182 4.541 5.841 12.924

4 1.533 2.132 2.776 3.747 4.604 8.610

5 1.476 2.015 2.571 3.365 4.032 6.689

6 1.440 1.943 2.447 3.143 3.707 5.959

7 1.415 1.895 2.365 2.998 3.499 5.408

8 1.397 1.860 2.306 2.896 3.355 5.041

STUDENTS DISTRIBUTIONLevel of Significance for One-Tailed test

13 1.350 1.771 2.160 2.650 3.012 4.221

14 1.345 1.761 2.145 2.624 2.977 4.140

15 1.341 1.753 2.131 2.602 2.947 4.073

16 1.337 1.746 2.120 2.583 2.921 4.015

17 1.333 1.74 2.110 2.567 2.898 3.965

18 1.330 1.734 2.101 2.552 2.878 3.992

19 1.328 1.729 2.093 2.539 2.861 3.883

20 1.325 1.725 2.086 2.528 2.845 3.850

21 1.323 1.721 2.080 2.518 2.831 3.819

22 1.321 1.717 2.074 2.508 2.819 3.792

23 1.319 1.714 2.069 2.500 2.807 3.768

24 1.318 1.711 2.064 2.492 2.797 3.745

25 1.316 1.708 2.060 2.485 2.787 3.725

26 1.315 1.706 2.056 2.479 2.779 3.707

27 1.314 1.703 2.052 2.473 2.771 3.690

28 1.313 1.701 2.048 2.467 2.763 3.674

29 1.311 1.699 2.045 2.462 2.756 3.659

30 1.310 1.697 2.042 2.457 2.750 3.646

40 1.303 1.684 2.021 2.423 2.704 3.551

60 1.296 1.671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.373

infinite 1.282 1.645 1.960 2.326 2.576 3.291

One-Way Classification for a

Balanced Array

Let us illustrate how to apply the formulas for this type of problem with the following example:

Example: The manager of a training output for sales representative would like to known if there might be a significant differences among three method of instruction (Method A, B and C) that are employed for trainees. A different method is used for each of three separated group of trainees. The same topic is discussed for all groups and the same set of examination questions are given at the end of training seminar. A sample of examination scores of six trainees from each of group is taken. Based on the level of significance of 1%, test to determine if there is a significant differences in the scores of group (that is, if there is a significant differences in the method of instruction employed).

Trainee Number

Per group

Method A

Method B

Method C

1 82 80 92

2 93 86 96

3 90 78 85

4 88 82 90

5 85 76 89

6 92 84 94

Solution:Note the differences in grade

within samples and between samples. W hat we would like to determine, however, is whether such differences are significant.

The first step is to solve for the squared values of each of the grades in the three column. Reproducing the entries of the three X columns and get the squared values that now form three additional derived columns, we have:

82 80 92 6724 6400 8464

93 86 96 8649 7396 9216

90 78 85 8100 6084 7225

88 82 90 7744 6724 8100

85 76 89 7225 5776 7921

92 84 94 8464 7056 8836

2( )AXAX BX CX2( )BX

530AX 486BX 2( ) 94,762CX 546CX

2( )CX

2( ) 46906AX 2( ) 39,436BX

The next step is to get the sums of all six columns. These are shown above. The entries in the last three columns (the squared columns) are arrived at by squaring the entries in the X columns, for examples: First entries for

: 6724 :6400 :8464

Before we solve for SSt and SSc, let us identify the following:r = number of rows = number of trainees per samples = 6

2( )AX2( )BX

2( )CX

C = number of columns = number of samples = 3 (the last three columns are not include)Degree of freedom: dft = rc-1 dfc = c-1 dfw = c(r-1) = 6(3)-1 = 3-1 = 3(6-1)

=18-1 = 2 =3(5) =17 = 15

2 2 2 2

Terms for the formula of SSt:

( ) ( ) ( ) ( )

= 46,906+39,436+49,762

= 136,104

A B CX X X X

( )

=530+486+546

=1,562

A B CX X X X

2 2 2 2

2 2 2

Term for formula of SSc :

( ) ( ) ( ) ( )

(530) (486) (546)

= 280,900+236,196+298,116

=815,212

A B CX X X X

22

2

2

2

Substituting these values into the formulas for SSt,

SSc, and SSw we have:

( )( )

(1562) 136,104

6(3)

= 136,104-135,546.89

= 557.11

( )1( )

1

6

XSSt X

rc

XSSc X

r rc

815,212 135,546.89

= 321.78

SSw = SSt-SSc

= 557.11-321.78

= 235.33

Substituting the computed values of SSc and SSw into

the formulas for MSSt and MSSw, we have:

SSt

MSScdfc

321.78

2

= 160.89

SSw

MSSwdfwc

235. 33

15

= 15,69

Substituting the computed values of MSSc and MSSw

into the formula of Fc, we have:

MSScFc

MSSw

160.89

15.69

=10.25

Let us now look for the tabular value of the F statistic (that is, Ft) by referring to the table on F Distribution in Appendix 4a. To identify our tabular value we need: Two degree of freedom– A numerator value corresponding to dfc. – A denominator value corresponding to dfw.The level of significance.For our problem, these value are:

dfc = 2dfw = 15

Level of significant = 1%Base on these value, the tabular value in the table is:

Ft =6.36 (Note: These are two F table s. One for 1%,second for 5% level of significance.)

Going through the four-step solution for test of Hypothesis, we have:

1. 8Ho : There is no significant difference in the scores of three groups of trainees.

2. This translates to: There is no significant differences in the three method of instruction employed.

3. Ha : There is a significance differences in the scores of the three group of trainees.

4. The computed value for F statistic : Fc = 10.25 5. The tabular value is : Ft =6.36 6. Conclusion

Since the computed value of F (that is, Fc =10.25) is greater than the tabular value (that is, Ft = 6.36), we reject Ho(and accept Ha)

Therefore: there is a significant difference in the grades of three group of trainees

This means further that: There is a significant differences in the three method of instruction employed.

What, then, is the implication of this conclusion? Since there is a significant differences in the methods, one has to determine which method is the best and to employ that method. Through several applications, one is able to determine which method will yield consistently high grades for trainees.Suppose the conclusion is that three is no significant differences in the grades? Then the three method may be considered equally effective and any one of the method may be used for training purposes.

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One Way Classification for Unbalanced , Array

In an unbalanced array, the samples do not have the same array of observations the formulas are slightly different from those used in a balanced array. Let us summarized for an unbalanced array. These are:

Note: the formulas of the other term remain the same. These are:

2

2( )X

SSt Xn

22 ( )( ) XX

SScr n

• .

1dft n

1dfc c

dfw n c

SScMSSc

dfc

SSwMSSw

dfw

MSScFc

MSSw

The slight differences compared to the formulas of a balanced array are:

• The denominator for the second term of SSt and SSc is n instead of rc.

• In the first term SSc, r appears inside the brackets instead of outside. This means r is distributed into each of the entries.

• dft is n-1 instead of r c-1• dft is n-c instead of c(r-1)

2( )X

Note: the formulas of the other terms remain the same. These are:

SSw MSSc Fcdfc MSSw

The values of all the terms will again be solved by following the same sequence as in the case of a Balanced array.

Let us now apply the formulas for an Unbalanced array in the following example. The example is essentially the same as in the preceding example of a Balanced array except that we introduce changes in the number of observations for examples B and C, that is:

• Sample A : the same 6 observations• Sample B : the first 5 observations only• Sample C : the first 4 observations only

This means that only 5 and 4 trainees respectively are taken for samples B and C

Thus the column entries are:

82 80 92 6724 6400 886493 86 96 8649 7396 921690 78 85 8100 6084 722588 82 90 7744 6724 810085 76 __ 7225 5776 ____92 __ __ 8464 ____ ______ __ __ ____ ____ ____530 402 363 46,906 32,380

33,005

AX BX CX 2( )AX 2( )BX 2( )CX

Identifying the values of terms in the formulas, we have:

n = the number of observations in the three samples

= observations in samples A, B, and C together

= 6+5+4

=15

Degrees of Freedom

dft = n-1 dfc = c-1 dfw = n-c

= 15-1 = 3-1 = 15-3

= 14 = 2 = 12

Terms in the formula SSt

=

= 46,909 + 32,380 + 33,005

= 112,297

= = 530 + 402 + 363 = 1295

2( )X 2 2 2( ) ( ) ( )A B CX X X

( )X A B CX X X

Terms in the formula SSc

.

2( )

2

The first term in the formula of SSc is

This means that we should first determine the value

for each sample after which we get the sum

of these computed results for the three samples

X

r

X

r

22(530) 280,900

Sample A = 46,816.676 6

AX

r

22(402) 161,604

Sample B: = 32,320.805 5

BX

r

22(363) 131,769

Sample C: = 32,942.254 4

CX

r

Therefore, the first term of the formula for SSc is:

Substituting these computed values into the formulas of SSt and SSc, we have:

2 2 2 2

6 5 4

A B CX X X X

r

2 2 2530 402 363

6 5 4

46,816.67 32,320.80 32,942.25 112,079.72

2

2

t X

XSS

n

1,677,025112,297

15

112,297 111,801.67

495.33

22XX

cnr

SS

112,079.72 111,801.66

278.06

Solving now for SSw, MSSc, MSSw, and Fc, we have:

WSS SSt SSc 495.33 278.06 217.27

278.06139.03

2

SScMSSc

dfc

217.2718.10

12

SSwMSSw

dfw

139.037.68

18.10

MSScFc

MSSw

Let us now identify the tabular value, Ft, based on the following information:

Degree of Freedom numerator = 2Degree of Freedom, denominator = 12Level of Significance = 1%

From the table of F Distribution in Appendix 4a, Ft=6.93. Going through the four-step solution, we have :1. Ho: There is no significant difference in the

grades of the three groups of trainees.This means that: there is no significant difference in the three methods of instruction.Ha: There is significant difference in the grades of the three groups of trainees.

3. The tabular value is : Ft = 6.934. Conclusion:– Since the computed value (that is, Fc

= 7.68) is greater than the tabular value (that is, Ft = 6.93), we reject Ho (and accepts Ha)

– Therefore: There is a significant difference in the grades of the three groups of trainees.

– Therefore: There is a significant difference in the three methods of instruction employed.

2. The computed value is : Fc= 7.68

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Table 3. CRITICAL VALUES OF FFractions for .05 (lightface) and .01 (boldface)

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Two-Factor Analysis of VarianceIn the contingency Table 15.1 a schematic representation of a

two-factor ANOVA is presented. In this table we have a design with treatment groups of size 10(N) and 2 factors: a column factor, C and a row factor R the column factor has 4 levels and the row factor 3. In the table 15.1 there are four columns, c=4, and three rows, r=3. This causes the table to contain 12 cells. Each of these cells correspond to a treatment group and each contains a sample that has been uniquely treated the ten individuals in the upper-left hand cell received the treatment associated with the level 1 of factor R and level 1 of factor C.

The effects of the treatment combination r₁ and c₁ will be shown when the size of the mean of this cell is compared with the means of the 11 other treatments. The effects of an entire factor, called factor main effects, are calculated through a study of the marginal means. That is, the differences between the marginal means

1, 2, 3, 4 C C C CX X and X

Table 15.1 MODEL OF TWO-FACTOR ANOVA DESIGN

Factor C: ColumnLevelC1 Level C2 Level C3 Level C4

Level 1 r₁ Nr₁=40 Xr₁

Level 2 r₂ Nr₂=40 Xr₂

Level 3 r3 Nr₃=40Xr₃

Factor R: Nt=120Rows Xt

N11 =10X11

N12=10X12

N13=10X13

N14=10X14

N21=10X21

N22=10X22

N23=10X23

N24=10X24

N31=10X31

N32=10X32

N33=10X33

N34=10X34

Show the main effects of fator C across all levels of factor C. Similarly, differences between the marginal row means when analyzed, show the main effects of factor R.

Partitioning the Sum-of-SquaresThe usual formula for the total sum-of-squares is again used here:

The within sum-of-squares is obtained by the following formula:

Where is the number of cell groups and N is the number in each cell.

1, 2, 3,r r r

22

t

tt t N

SS

2

2 k

w t NSS

SSь is not computed directly in this type of analysis; instead, it is partitioned into the following: the main effect of C(SSс), the main effect of R (SSr) and the interaction effect (SSс•г). Thus:

The computational formulas for these follow:

Where and are the sums of the X values in each column and row respectively, and Nс and Nг are the sample sizes of each column and row.

c r

.b c r c rSS SS SS SS

2 2c t

c tc N N

SS

22tr

r tt N N

SS

The SSc r is found by subtraction:

After all the sums of squares have been determined, the usual ANOVA table is set up and each SS is divided by its appropriate degrees of freedom to obtain the corresponding mean square (MS) or variance. The test of the main effects of factors C and R and the interaction effect are then evaluated by making an F test for each by dividing each MS by the MSw. This latter mean square is sometimes referred to as the error term.

c r t w rc rSS SS SS SS SS

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Degrees of Freedom in a Two-Factor ANOVA

Degrees of freedom for this type of problem are determined as follows:

Where k is the number of samples and N is the size of each sample, assuming N’s are equal.

1

1

1 1

col c

row r

c r c r

df

df

df

1

1t t

w k

df N

df

:bdf

TABLE 15.2 Scores of 80 individuals of 80th sexes on an attitude scale after the administration of

four different films

FILM1 2 3 4

MALES10 14 13 168 12 12 12.6 8 10 104 4 9 94 4 9 94 4 7 72 3 4 72 2 4 62 2 4 51 1 2 5

∑=43 54 74 86∑=257

FEMALES

14 14 10 18 12 13 10 16 12 12 10 15 10 11 9 14 8 10 9 13 6 9 7 10 4 7 7 10 2 6 6 10 2 3 5 10 1 2 4 9 ∑=71 87 77 125

∑=360∑c=114 141 151 211

∑c=617

An Example

In table 15.2 are the responses of 40 female and 40 male university students to a attitudinal scale. Each group of 40 was randomly divided into two groups of 10 each then each of these groups of 10 was shown one of four different films on a controversial subject. Later the attitudinal scale on this subject was administered to each individual. For the study, alpha was set at .05.

First the scores in each of the columns are summed, as shown in the table, then the two rows are summed. Both of these sums come to 617. The is obtained by squaring each of the separate scores and summing the squares. The squares and sum can most readily be obtained by the student using a pocket calculator The resulting is 6159. Then we proceed to obtain the various sums-of-squares:

2

2 2 2 2 210 8 6 ....9

(1) 2

2t

t tt

SS N

2617

615980

380,6896159

80

6159 4758.6

1400.4

(3)

22w t

NSS

2 2 2 2 2 2 2 243 54 74 86 71 87 77 125

106159 51,801

615910

6159 5180.1 978.9 2 2

c t

c tc

N NSS

22 2 2 2 617114 141 151 211

20 80

100,199 380,689

20 80

5009.95 4758.61 251.34

(2)

(5)

22tr

r tr

N NSS

22 2 617257 360

40 80

195,649 380,689

40 80

4891.22 4758.61 132.61c r t w c rSS SS SS SS SS

1400.4 978.9 251.34 132.61 37.6

(4)

The number of degrees of freedom for the various parts of this problem are:

Then the data entered into ANOVA table:

1 80 1 79t tdf N ( 1) 8(10 1) 72w kdf N 1 4 1 3c cdf 1 2 1 1r rdf

1 1 4 1 2 1 3 1 3c r c rdf

Source of Variation SSc df MS F P

Between

Columns 251.34 3 83.78 6.16 <.05

Rows 132.61 1 132.61 9.75 <.05

Interaction 37.6 3 12.53 .92 >.05

Within 978.9 72 13.6

Total 1400.4 79

The following decision model has been built for these data

Not

NotInteraction effects are not present.Interaction effects are presents.

0 1 2 3: c c cH

1 :H 0H0 1 2 3: r r rH

1 :H 0H0 :H

1 :H: .01

c : 0 DR Reject if 5.49;otherwise, do not reject. obsH F r : 0 DR Reject if 5.49;otherwise, do not reject. obsH F c r : 0 DR Reject if 4.11;otherwise, do not reject. obsH F

In this problem for columns, teachers, exceeds the .01 table value of for This leads to the rejection of and to the conclusion that the main effect factor, teacher, is significant. On the other hand, for texts is less than the 0.01 table value ofSo in this case is accepted and we conclude that the interaction effect is also significant at the .01 level. The presence of significant has to be acknowledged when taking about the significant main effect for teacher. From the data in Table 15.3, it appears that students of teacher 2 using text 1 did better with those with the other combination of factors. Methods exist for carrying out these post hoc comparisons (such as a test for simple effects and individual comparisons); however, these are too complicated for an elementary text.

obsF5.49F of 2.27.df

0HobsF

for of 2.27. F df

0H

Another Application of the Two-Factor Design

In some cases there are two treatments or experimental variations and only one observation of each combination of factors. We shall illustrate this type of design using the data in Table 15.4 where the ratings of 6 trainees by 4 supervisors are presented. The methodology presented below is similar so that the given before, except that the sum of squares related to interaction and to error effect are not separated or partitioned.

Note that in Table 15.4 various squaring and summing have carried out. This illustration is one of the simpler two-way ANOVA applications in that although there are two variables, rater and rate, there is only one observation recorded in each cell or for each combination of the two variables. Such a model is referred to as a random model, since both trainees and supervisors are assumed to be random samples from normal populations.

Table 15.4The ratings of Six Trainees By Four

SupervisorsRater Squares of

RatingsTrainee A B C D A B C D

1 10 6 8 7 31 100 36 64 49

2 4 5 3 4 16 16 25 9 16

3 8 4 7 4 23 64 16 49 16

4 3 4 2 2 11 9 16 4 4

5 6 6 6 7 27 36 64 36 49

6 9 7 8 7 31 81 49 64 49

40 34 34 31 139 306 206 226 183 ∑=921

ofRows

. The analysis of variance proceeds as follows:

1) The total sum-of-squares (SSt) is found in the

usual manner. 2

2

2139 921

24 921 805

116

tSS

(2) The sum of the squares for rows, trainees(SS?) is

next obtained. 22 2 2 2 2 2 13931 16 23 11 27 31

4 243557

8054

889.25 805

84.25

rSS

(3) The sum of the squares for columns, Raters (SSc), is then calculated.

22 2 2 2 13940 34 34 31

6 24cSS

4873805

6

812.16 805

7.16

(4) After the SS for both rows and columns have been obtained. This leaves the residual or error sum squares, SSc. The error term is a combination of interaction and error effects.

116 84.25 7.14

116 91.41

24.59

e t r cSS SS SS SS

(5) Finally a table set up:

Source of Variation

df SS MS F P

Rows(trainees) 5 84.25 16.85 10.27 <0.01

Columns(raters)3 3 7.16 16.85 1.46 >0.01

Error 15 24.59 1.64

Total 23 116.00

The first F in this table is obtained by dividing the variance for rows by the error variance This obtained F of 10.27 is highly significant for 5.15 degrees of freedom, and is far beyond the 1 percent point. This indicates that the supervisors did discriminate among trainees. The second F is obtained by dividing the variance for columns, Raters, by the error variance . With 3.15 degrees of freedom the obtained F of 1.46 is not significant, indicating that the supervisors did not differ significantly among themselves in their ratings of the trainees. In other words, their ratings were reliable.

/r eMS MS

/c eMS MS

Data from this technique are often used in determining the reliability of raters. Reliability is discussed in some detail in Chapter 18. By the use of the following formula, the reliability of the ratings of a single rater is obtained:

where k is the number of columns and the other terms are as used above:

1

r e

r e

MS MSr

MS k MS

16.85 1.64

16.85 4 1 1.64r

15.2

16.85 4.92

15.2121.77

.70

The reliability of all four raters taken together is obtained by:

r e

ttr

MS MSr

MS

16.85 1.64

16.85

15.21

16.85 .90

In this chapter we have described only a few of the simple uses of the analysis of variance. To go further is beyond the scope of this book. The interested student will find in more advanced texts the large number of experimental designs that can be analyzed by the application of the analysis of variance.

In this chapter we have described only a few of the simple uses of the analysis of variance. To go further is beyond the scope of this book. The interested student will find in more advanced texts the large number of experimental designs that can be analyzed by the application of the analysis of variance.

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Prepared by;

Wily A. AgloboJohn Leonard Calinog

Fevilyn DimasuayKaren Joy D. ElchicoMary Grace Maxilum