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MULTIPLE COMPARISON TEST


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Learning Objective

Understand the definitions used

in multiple comparison test.

Determine which means differ,using the multiple comparison

test if the null hypothesis is

rejected in the Kruskal-Wallis

non-parametric ANOVA.


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Learning Outcome

Make inference to related sample using

multiple comparison test.


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When we use multiplecomparison test?

When a hypothesis-testing procedure

such as Kruskal-Wallis test leads us to

reject the null hypothesis. Thus to conclude that not all sampled

populations are identical.

This test use to answer the question whichpopulations are different from which

others.


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Example 6.2

Cawson et al.* reported the data shown in Table

6.10 on cortisol levels in three groups of patients

who were delivered between 38 and 42 weeks

gestation. Group 1 was studied before the onsetof labor at elective Caesarean section, group 2

was studied at emergency Caesarean section

during induced labor, and group 3 consisted of

patients in whom spontaneous labor occurred

and who were delivered either vaginally or by

Caesarean section.


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Example 6.2

We wish to know whether these data provide

sufficient evidence to indicate a difference in

median cortisol levels among the three

populations represented.

*Cawson, M. J., Anne B. M. Anderson, A. C. Turnbull, andL. Lampe, Cortisol, Cortisone, and 11-Deoxycortisol

Levels in Human Umbilical and Maternal Plasma in

Relation to the Onset of Labour, J. Obstet. Gynaecol. Br.

Commonw., 81(1974), 737-745.


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Table 6.10

Antecubital vein cortisol levels in three groups

of patients studied at time of delivery

Group 1 262 307 211 323 454 339 304 154 287 356

Group 2 465 501 455 355 468 362

Group 3 343 772 207 1048 838 687


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Kruskal-Wallis One Way Analysis of Variance

by Ranks

Hypotheses

Ho: The three populations represented by

the data are identical.

H1: The three populations do not have the

same median.


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Ranks corresponding to data of Table 6.10

Group Ranks Rank sums

1

2

3

4 7 3 8 14 9 61 5 12

16 18 15 11 17 13

10 20 2 22 21 19

R1= 69

R2= 90

R3= 94


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Test Statistic

H=


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N =22

R1= 69

R2= 90

R3= 94

H 222.2)222(22

99

2

99

22

22

)222(22

22 222 =+

++

+=


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Since the sample sizes all exceed 5, we use the

chi-square table to determine whether the

sample medians are significant different.

critical value:

k=3

degree of freedom = k-1, 3-1=2

by looking at table A.11, the critical value

is 9.210 for =0.01.


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H= 9.232

critical value= 9.210

H> critical value

9.232 > 9.210

Decision :

We reject Ho at the level of significance.

Conclusion :

There is enough evident to support the claim that is the three

populations represented by the data are not identical. We conclude

that the medians of the populations represented are not all equal

that is the median cortisol levels are not the same for all three typesofpatients.The P value is between 0.01 and 0.005.


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As we reject Ho, we must continue with multiple

comparison test to see which populations are different

from which others.

When we apply this multiple-comparison procedure, we

use what is known as an experimentwise error rate.

The experimentwise error rate, which represents a

conservative approach in making multiple comparisons,

holds the probability of making only correct decisions at 1-

when the null hypothesis of no difference among

populations is true.


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STEPS:

1) Obtain the mean of the ranks for each sample, and let Ri be the

mean of the ranks of the ith sample and Rj be the mean of therank of thejth sample.

2) Select an experimentwise error rate of, which we think of as an

overall level of significance. Our choice of is determined in part by2 k, the number of

samples involved, and is larger for largerk.


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If we have ksamples, there will be a total of

pairs of sample that can be compared apair of time.

We usually select a value of larger than2those customarily encountered in single-

comparison inference procedures-for

example, 0.15, 0.20,or 0.25 depending on the

size ofk.

a


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3) Find the value ofzin Table A.2 that has area to its

right.

4) Form the inequalityIfksamples are all of the different size

(6.7)

where Nis the number of observations in all samples

combined. The probability that Inequality 6.7 holds for all

pairs of means, when H0 is true, is at least 1- .2

)2( kka

++

jikka

ji

nn

NNZRR

22

99

)2(||

)2(2


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If the k samples are all of the same size

(6.8)

Any difference that is larger than the

right- hand side of Inequality 6.7 (or Inequality

6.8 if applicable) is declared significant at the

level.

2)2

(||)2(

2+

NNZRR

kk

aji


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Continue from Example 6.2

We will apply multiple comparison test to see

which populations are different from which

others.

1)Hypothesis

H0 : 1 = 2 H0 : 1 = 3 H0 : 2 = 3

H1 : 1 2 H1: 1 3 H1: 2 3


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To make all possible comparisons in order to

locate where the differences occur, let choose

an error rate of =0.15.

Samples involved are k=3,so that there will

be

So there are 3 comparisons to make.


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From table A.2, we find the z with 0.025 of

the area to its right to be 1.96.

The mean of the rank for the threesamples:

222.2)2(2

22.2

)2(==

kka

2.222

222

==R

99.992

99

2==R

222

222

==R


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TABLE A.2


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2)Test to compare group

Compare groups 1 and 2

8.1 > 6.57

|222.2||| = ji RR

2.2=

++=

2

2

99

2

99

)222(2222.2

22.2=

+

+

jikk

a nn

NNZ

22

99

)2(

)2(2

C 1 d 3


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Compare group 1 and 3

8.77 > 6.57

+

+=

2

2

99

2

99

)222(2222.2

|22.222.2||| = ji RR

22.2=

22.2=

++

ji

kk

a nn

NNZ

22

99

)2(

)2(

2

C 2 d 3


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Compare group 2 and 3

0.67 < 7.35

|22.2222||| = ji RR

22.2=

22.2=

++

jikka

nn

NNZ

22

99

)2(

)2(2

+

+=

2

2

2

2

99

)222(2222.2


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3) Decision

a) 8.10 > 6.57, reject the H0. This

comparison is significant.

b) 8.77 > 6.57, reject the H0. Thiscomparison is significant.

c) 0.67 < 7.35 , do not reject the H0. Thiscomparison is not significant.


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4) Conclusion :

a)There is enough evident to support the claimthat there are different between median group 1

and 2, we conclude that subjects for elective

Caesarean section before the onset of labortend to have lower cortisol levels that women

experiencing induced labor at emergency

Caesarean section.

b)There is enough evident to support the claim that there


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b)There is enough evident to support the claim that there

are different between median group 1 and 3. We

conclude that the women in population 1 also tend to

have lower cortisol values than the women inpopulation 3.

c)There is not enough evident to support the claim that

there are different between median group 2 and 3. We

cannot conclude that the populations represented by

samples 2 and 3 differ with respect to their median

cortisol levels.


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Exercise 6.27, page 2541)Four tour guides are employed at a state-operated

tourist attraction of historical interest. Each guideconducts tours of exactly 20 persons each during

peak season. Each tour is timed. A researcher with

the state department of tourism conducted a study

compared the tour guides with respect to efficiency.

For each guide the researcher selected a simple

random sample of 6 tours conducted during the

month of June, and the required for each wasrecorded. The results are shown in Table 6.32.


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Time required to conduct tours by fourtour guides

Tour guide_________________________________A B C

D______________________

37 36 38 30

31 33 38 34

33 38 39 31

30 36 36 30

39 37 38 35

31 39 37 34


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Can the researcher conclude on the basis of these

data that the four tour guides differ with respect to

the average time it takes them to conduct a tour? Let

=0.05, use the Kruskal-Wallis test to determine the

P value and make all possible comparisons.


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Exercise 6.12, page 249

2) Ali and Sweeney* determined the protoporphyrin levels in15 normal,

healthy laboratory workers and in 26 patients admitted with acute

alcoholism. Their results appear in Table 6.12. Can we conclude from

these data that the normal workers and the two groups of alcoholics differ

with respect to average protoporphyrin level? Let =0.15,

determine the P value and make all possible comparisons.

*Ali, M. A. M., and G. Sweney, Erythrocyte Corproporphyrin andPhotoporphyrin in Ethanol-Induced Sideroblastic Erythropoiesis, Blood,

43(1974), 291-295.

Protoporphyrin level milligrams/100 ml RBC in


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Protoporphyrin level, milligrams/100 ml RBC, inthree groups of subjects

Normal 22 27 47 30 38 78 28 58 72 56 30 39 53 50 36

Alcoholicswith ringsideroblastsin bone

marrow

78 172 286 82 453 513 174 915 84 153 780

Alcoholicswithoutringsideroblasts

in bonemarrow

37 28 38 45 47 29 34 20 68 12 37 8 76 148 11


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Exercise 6.13, page 249

3) Torre et al.* recorded the changes in rat cerebral and extracerebral (platelet)

serotomin (5-HT) after intraperitoneal administration of LSD-25 and 1-

methyl-d-lysergic acid butanolamide (UML). They also took measurements

on 11 controls. The results are shown in Table 6.13. Do these data provide

sufficient evidence to indicate a difference among the three groups? Let

=0.20, determine the P value and make all possible comparisons.

*Torre, Michele, Filippo Bogetto, and Eugenio Torre , Effect of LSD-25 and 1-Methyl-d-Lysergic

Acid Butanolamide on Rat Brain and Platelet Serotonin Levels, Psychopharmacologia,36(1974), 117-122.

Brain serotonin (5-HT) nanograms per gram in


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Brain serotonin (5-HT), nanograms per gram, inthree groups of mice

Controls 340 340 356 386 386 402 402417 433 495 557

LSD, 0.5 mg/kg 294 325 325 340 356 371 385402

UML, 0.5 mg/kg 263 309 340 356 371 371 402417


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