LEARNING OUTCOMES At the end this subtopic student should be able

to :Make inference about the equality of two

dispersion parameter by using Ansari- Bradley test.

ASSUMPTIONS

A. The data consist of two random samples and from populations 1 and 2, respectively.

B. The population distribution are continuous.C. The two samples are independent.D. The data are measured on at least an ordinal

scale.E. The two population are identical (including equal

medians) except a possible difference in dispersion.

HYPOTHESES

• CASE A. (Two-sided) H0 :σ1 = σ2 H1 :σ1 ≠ σ2

• CASE B. (One-sided) H0 :σ1 ≤ σ2 H1 :σ1 > σ2

• CASE C. (One-sided) H0 :σ1 ≥ σ2 H1 :σ1 < σ2

TEST STATISTICTo obtain the test of statistic,

Arrange the combined set of n1 + n2 = n’ measurements in order from smallest to largest.

Assign ranks to the ordered measurements as follows: the smallest measurement and the largest measurement are each given a rank of 1. The second smallest measurement and the second largest measurement are each given a rank of 2.

n1 + n2 is an even number the array of ranks will be 1,2,3....,(n’/2),(n’/2),...3,2,1.

 

If n1 + n2 is an odd number, the array of

ranks will be 1,2,3,...,(n’ – 1)/2, (n’ + 1)/2,(n’ –

1)/2,...3,2,1.

T = ∑ Ri, in other word T is the sum of ranks

assigned to the X values.

DECISION RULE.a. We reject H0 if T is either greater than or

equal to the larger critical value of x or less than the lower critical value of x. (CASE A)

b. We reject H0 if T is less than the critical value of x.(CASE B)

c. We reject H0 if T is greater than the critical value of x.(CASE C)

HYPOTHESIS REJECT H0

H0 :σ1 = σ2

H1 :σ1 ≠ σ2

H0 :σ1 ≤ σ2,

H1 :σ1 > σ2

H0 :σ1 ≥ σ2 H1 :σ1 < σ2

21 ,n/2,

2/1

or

nxT

xT

21 ,,1 nnxT

21 ,, nnxT

Example.Gordon* et al. reported cardiac-index values for two

groups of patients, as shown in table 1. All patients were

seen initially for severe aortic valvular disease requiring

prosthetic valve replacement. Data on the cardiac index

were obtained after operation. Group 1 consisted of

patients with normal prosthetic valve function. Group 2

consisted of patients with abnormal prosthetic valve

function. We wish to know whether the dispersion with

respect to the variable of interest is different in the two

populations represented by these samples. Let α = 0.05.

* Gordon, Richard F.,Moosa Najmi, Benadict Kingsley, Bernard L. Segal, and Joseph W.

Linhart, “Spectroanalytic Evaluation of Aortic Prosthetic Valve,” Chest, 66 (1974), 44-

49.

TABLE A.8 TABLE A.8

Table 1

Cardiac index, liters/minute/M2, in two groups of patients following prosthetic valve replacement

Group 1 (x) Group 2 (y)

3.84 3.97

2.60 2.50

1.19 2.70

2.00 3.36

- 2.30

1. Make hypotheses

H0 :σ1 = σ2 H1 :σ1 ≠ σ2 (claim)

SOLUTION: 

2. Test statistic.On combining the two samples and ranking, we have the results shown in table 2.

 Observation

1.19

2.00 2.30 2.50 2.60 2.70 3.36 3.84 3.97

Group x x y y x y y x y

Rank 1 2 3 4 5 4 3 2 1

3. MAKE A DECISION.We know that T = 1 + 2 + 5 + 2 = 10n1 = 4, n2 = 5 and α/2 = 0.025

According to the table A.8, the value 0f T for α/2 = 0.025 is 16 while the value of T for 1-α/2 = 0.975 is 8.

10 is not between 16 .4.

5. The decision is do not reject H0.

6. Do not have enough evidence to support the claim that the two population dispersion parameter maybe equal.

 

21

2

xT

xT

Exercise 1 Reimer* et al. studied the effect of propranolol on the severity of myocardial

necrosis following 40 minutes of temporary coronary-artery occlusion in dogs.

One group of dogs was untreated, and a second group received propranolol 10

minutes before the occlusion. Following this procedure Reimer et al. recorded

the relative area of necrosis (percentage of fibers involved) in the posterior

papillary muscle of each dog’s heart. Table below gives partial results. Can we

conclude on the basis of these data that dispersion with regard to relative

necrosis differs in the two population represented? Let α = 0.05

* Reimer , Keith A., Margaret M. Rassmussen, and Robert B. Jennings, “ Reduction By Propranolol

of Myocardial Necrosis Following Temporary Coronary Artery Occlusion in Dogs, “Circ.Rest.,

33(1973), 353-363.

TABLE A.8

Untreated (x)

44.4 81.0

23.6 62.1

39.1 25.5 44.2 43.3 39.8 51.3

Prapanolol treated, 50mg/kg,iv (y)

0 4.5 5.6 6.1 22.6 30.8 13.4 1.3 45.0 30.3

Observation

0 1.3 4.5 5.6 6.1 13.4

22.6 23.6 25.5 30.3 30.8

Group y y y y y y y x x y y

Rank 1 2 3 4 5 6 7 8 9 10 10

Observation

39.1

39.8 43.3 44.2

44.4 45.0 61.3 62.1 81.0

Group x x x x x y x x x

Rank 9 8 7 6 5 4 3 2 1

3. Make a decision.We know that T = ∑ = 7+ 8 + 9 +8+

7+6+5+3+2+1 = 58n1 = 10, n2 = 10 and α/2 = 0.025

According to the table A.8, the value of T for α/2 = 0.025 is 69. where in the table the closest (0.0192)The value of T for 1-α/2 = 0.975 is 43. Where in the table the closest (0.9718)

iR

4.

21

2

xT

xT

4558

6958

5. DecisionDo not reject

6. Not enough evidence to support the claim. We conclude that the two population dispersion parameters may be equal.

0H

Exercise 2 Boullin* and O’Brien studied uptake and loss of 14C-

dopamine by platelets in five autistic children and

five normal controls. Part of their results are shown

in table below. Can we concluded from these data

that the two population represented differ with

respect to dispersion of uptake values? Let α =

0.05.

* Boullin, David J.,and Robert A.O’Brien, “Uptake and loss of 14C-dopamine By Platelets From Children With Infantile Autism,” J. Autism Child . Shizophrenia, 2 ( 1972), 67-74. TABLE A.

8

Table 1Austistic children (x) 433 347 328 607 478

Controls (y) 428 372 434 425 336

1. Make hypothesesH0 :σ1 = σ2 H1 :σ1 ≠ σ2 (claim)

2. Test statistic.On combining the two samples and ranking, we have the results shown in table 2.

Table 2

Observation

328

336 347 372 425 428 433 434 478

607

Group x y x y y y x y x x

Rank 1 2 3 4 5 5 4 3 2 1

3. Make a decision.We know that T = ∑ = 1 + 3 + 2 +4+

1 = 11n1 = 5, n2 = 5 and α/2 = 0.025

According to the table A.8, the value of T for α/2 = 0.025 is 20. where in the table the closest (0.0238)The value of T for 1-α/2 = 0.975 is 11 .where in the table the closest (0.9286)

4.

iR

21

2

XT

XT

5. DecisionDo not reject We conclude that the two population dispersion parameters may be equal.

0H

EXAMPLE

EXERCISE 1

EXERCISE 2

LARGE - SAMPLE APPROXIMATIONWhen the sample sizes exceed those found in Table

A.8, we may compute

(3.9)

If is even, and

(3.10)

If is odd. We compute for significance with appropriate values of the standard normal distribution.

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