Multiple Comparisons (Only Computing All Treatments With a Control)

8/3/2019 Multiple Comparisons (Only Computing All Treatments With a Control)

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MULTIPLE-

COMPARISON

PROCEDURE FOR USEWITH FRIEDMAN

TEST


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Learning outcome:

Students are able to calculate and

use multiple-comparison formulato solve problems.


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For treatment test (Friedman test) theirdata just allow them to conclude that not all

sampled populations or all treatment effects

are identical.

So we use multiple-comparison procedure

to see where the differences located.

( ONLY USE WHEN WE REJECT H0)


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FORMULA : | Rj

Rj

| z

Where;

Rj

and Rj

is the jth and jth treatment rank

total where z is a value from table A.2

that corresponding to


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WAY TO USE FRIEDMAN MULTIPLE-COMPARISON

STEP 1 : MAKE SURE THAT ITS REJECT H0, THEN WE CAN USE

MULTIPLE-COMPARISON

STEP 2 : CALCULATE TO FIND VALUE Z BY USING TABLE

A.2

STEP 3 : FIND VALUE Z

STEP 4 : COMPARED | Rj Rj | WITH Z

STEP 5 : ONLY | Rj

Rj

| Z WILL RESULT THE PAIR

DIFFERENCES.


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EXAMPLE

FROM PREVIOUS EXAMPLE

(GROUP 12 FRIEDMAN TEST)


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Example 1 (Page 265)

Hall et al. * compared three methods of determiningserum amylase values in patients with pancreatitis.

The result are shown in table 7.2. We wish to know

whether these data indicate a difference among the

three methods. Given

* Hall, F.F., T. W. Culp,T. Hayakawa, C. R. Ratliff, and N. C.Hightower,"An Improved Amylase Assay Using a New StarchDerivative,Amer. J. Clin. Pathol.,53 (1970),627-634

00.0=


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Serum amylase values (enzyme units per 100 ml of

serum) in patients with pancreatitis

Table 7.2

SpecimenMethods of determination

A B C

1 4000 3210 61202 1600 1040 2410

3 1600 647 2210

4 1200 570 2060

5 840 445 1400

6 352 156 249

7 224 155 224

8 200 99 208

9 184 70 227


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After convert the original observations to ranks, we

have

SpecimenMethods of determinationA B C

1 2 1 3

2 2 1 3

3 2 1 3

4 2 1 3

5 2 1 3

6 3 1 2

7 2.5 1 2.5

8 2 1 3

9 2 1 3

RA

= 19.5 R B

= 9 RC

= 25.5


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By equation, we have

0.00

1110.111

=

=

0 0

0

,0 0

00( )0 0

( )0

k

r j

j

k b

R b kbk k

=

= =

= ++

0 0 000

( . . ) ( )( )( )111111 1 1 1 1

( ) ( ) ( )0 0 0= + +


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DECISION

From table A.11,

Since then we reject

CONCLUSION

Enough evidence to support the claim that the three

methods does not all yield identical results.

0H

0

( . , )0000 .0000 =

0

( , )0 0k . ,000 0k = =

. .000 0000>


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Since we reject H0 , then we can find where the differences are located by

using Multiple-comparison Procedure.

From above we know that :

b = 9

k= 3

= 0.05

Ra = 19.5

Rb = 9Rc = 25.5

THEN, WE KNOW THE FORMULA IS : | Rj Rj| z

AND Z IS A VALUE FROM TABLE A.2 CORRESPONDING TO

FIRST, WE FIND

= 0.05 3(3-1) = 0.0083


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Then find z value:

Z = 0.5 0.00833

= 0.4916667 approximate to 0.4917 (table A.2)z = 2.39

Then we substitute value z into : z

We get z = 2.39

= 10.1399

Then : | Rj Rj| 10.1399


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Now we want to check where the differences occur.

So the three pairs of differences | Rj Rj| are

| Ra Rb| = | 19.5 9 | = 10.5

| Ra Rc| = | 19.5 25.5 | = 6

| Rb Rc| = | 9 25.5 | = 16.5

Thus, we can conclude that, pairs that yield different

result is pair Ra Rb and Rb Rc . Because it is

greater and equal to 10.1399.

EXERCISE


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Multiple Comparisons (Only Computing All Treatments With a Control)

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