MAKING INFERENCES ABOUT A LOCATION

PARAMETER

One-Sample Sign Test

LEARNING OBJECTIVES

At the end of this topic, students should be ableto: 1.Test hypotheses using the one sample sign

test.2.Determine the test value using large sample

approximation .

HISTORY

• Reported by John Arbuthnot in 1710

• It is called the sign test because we may convert the data for analysis to a series of plus and minus signs.

INTRODUCTION

• One-sample sign test -- requires data converted to plus and minus signs to test a claim regarding the median

– Change all data to + (above H0 value) or – (below H0 value)

– Any values = to H0 , remove from sample size (n-1)

ASSUMPTIONS

a) The sample available for analysis is a random sample of independent measurements from a population with unknown median M.

b) The variable of interest is measured on at least an ordinal scale.

c) The variable of interest is continuous. The n sample measurements are designated by X1, X2, ….,Xn.

HYPOTHESES

A Two sided : H0:M=M0, H1:M≠M0

B One-sided : H0:M≤M0, H1:M>M0

C One-sided : H0:M≥M0, H1:M<M0

TEST STATISTIC

Do not reject H0

have equal number of + sign and – sign

Reject H0

have small number of either + / - signs

value of k=smaller number of +/- sign

DECISION RULE

1. For case A, reject H0 at the α level of significance if the probability is less than or equal to α/2.

2 For case B and C, reject H0 at the α level of significance if the probability is less than or equal to α.

STEPS TO SOLVE PROBLEMS

• State the hypotheses• Compute the test statistic• Find critical value• Make a decision, the null hypothesis will be

rejected if the test statistic is less than or equal to the critical value

• Make a conclusion

HOW TO USE THE TABLE?

• The probability can obtained from a table of binomial probabilities in Table A.1

P(K ≤ k| n,0.50)

Where k = r (test statistic)

n = the number of + or – signs

p = 0.50

EXAMPLE

In a study of myocardial transit times, Liedtke et al.* measured appearance transit times in a series of subjects with angiographically normal right coronary arteries. The median appearance time for this group was 3.50 seconds. Suppose that the another research team repeated the procedure on a sample of 11 patients with significantly occluded right coronary arteries and obtained the results shown below.

* – Liedtke, A. James, Harvey G. Kemp, David M. Borkenhagen, and Richard Gorlin,” Myocardial Transit Times From Intra-coronory Dye-Dilution Curves in Normal Subjects and Patients with Coronory Artery Disease,” Amer.J.Cardiol.,32 (1973),831 – 839.

Could the second team conclude, at the 0.05 level of significant , that the median appearance transit time in the population from which its sample was drawn is different from 3.50 seconds?

Appearance transit times for 11 patients with significantly occluded right coronary arteries

Subject 1 2 3 4 5 6 7 8 9 10 11

Transit time, sec

1.80 3.30 5.65 2.25 2.50 3.50 2.75 3.25 3.10 2.7 3.0

Sign - - + - - 0 - - - - -

Find the probability from the Table A.1

k= r =1, α=0.05, n=10, p=0.5

Since k ≤ 1, we take k= 0 and k=1,

Using the table, when k= 0, the value is 0.0010 and when k= 1, the value is 0.0098.

So, P(K≤1|10,0.5)= 0.011

Since this was a two-sided test, the probability is 2(0.011)=0.022

Make the decision

Since 0.022<0.05, the null hypothesis is rejected.

Conclusion

There is enough evidence to support the claim that the median appearance transit time in the population from which its sample was drawn is different from 3.50 seconds.

LARGE SAMPLE APPROXIMATION

• For sample size 12 or larger, we use the normal approximation to the binomial to find test value.

z = (K + 0.5) – 0.5n 0.5 √n

where, K = smaller number +/- sign n = sample size

EXERCISES

Case A1. An educational researcher believes that the

median number of faculty for proprietary (for-profit) colleges and universities is 150. The data provided list the number of faculty at a selected number of proprietary colleges and universities. At the 0.05 level of significance, is there sufficient to reject his claim?

372 37 133 92 179

111 119 342 140 243

165 142 126 140 109

95 136 64 75

191 137 61 108

83 171 100 96

136 122 225 138

149 133 127 318

The data provided list the number of faculty at a selected number of proprietary colleges and universities

Answer: Hypotheses :H0 = M = 150 (claim)H1 = M ≠ 150Conclusion:We reject H0

There is enough evidence to reject the claim

Case C2.Based on past experience, a manufacturer

claims that the median lifetime of a rubber washer is at least 8 years. A sample of 50 washers showed that 21 lasted more than 8 years. At α=0.05, is there enough evidence to reject the manufacturer’s claim?

Answer :Hypotheses: H0:M ≥ 8 (claim) H1: M < 8Conclusion Do not reject H0.

Not enough evidence to reject the claim

Case C3.Lenzer* reported the endurance scores of animal

during a 48-hour session of discrimination responding. The median score for animal with electrode implanted in the hypothalamus was 97.5. Suppose that the experiment was duplicate in another laboratory, except that electrode were implanted in the forebrain of 12 animals. Assume that investigators observed the endurance scores shown.

use the one sample sign test to see whether the investigators may conclude at the 0.05 level of significant that median endurance score of animals with electrodes implanted in the forebrain is less than 97.5.

Scores 83.0 89.1 97.7 84.4 97.8 94.5 88.3 97.5 83.7 94.6 85.5 82.6

*Lenzer, Irmingard I.and Carol A. White,” Statiation Effects in Continuous Reinforcement Sussecive Sensory Discrimination Situations,” Physiolog,. (1973) 77 - 82

Answer: Hypotheses: H0 : M ≥ 97.5H1 : M < 97.5 (claim)Conclusion :We do not reject H0

Do not have enough evidence to support the claim

Case B4. Iwamanto* found that the mean weight of a

sample of a particular species of adult female monkey from a certain locality was 8.34 kg. suppose that a sample of adult females of the same species from another locality yielded the weight shown below.

*Iwamanto,Mitsuo,“Morphological Studies of Macaca Fuscata: VI, Somatometry,” Primates 12 (1971) 151 - 174

Weight

8.30

9.50

9.60

8.75

8.40

9.10

9.25

9.80

10.05

8.15

10.00

9.60

9.80

9.20

9.30

Can we conclude that the median weight of the population from which this second sample was drawn is greater than 8.41 kg? use the one sample sign test and a 0.05 level of significance. What is P value for this testAnswer: HyphothesesH0 : M ≤ 8.41HI : M > 8.41 (claim)Conclusion Reject HO

Have enough evidence to support the claim