Median Test

8/3/2019 Median Test

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MAKING INFERENCESABOUT THE

DIFFERENCE BETWEENTWO LOCATIONPARAMETERS BY

MEDIAN TEST


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Learning Outcome

Student should be able to make

inferences about the differencebetween two location by using

Median Test.


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MEDIAN TEST

The simplest and most widely proceduresfor testing the null hypothesis of twopopulations that have the same median.

Only two-sided alternatives will bediscussed.


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Assumptions for Median Test1) The data consist of two independent random

samples : X1, X2, X3, ..., X n, and Y1, Y2, Y3, ..., Y n2) The first sample is from a population withunknown median, MX and the second sample isfrom a population with unknown median, MY.

3) The measurement scale employed is at leastordinal.4) The variable of interest is continuous.5) The two populations have the same shape.

6) If the two populations have the same median,then for each population the probability p is thesame that an observed value will exceed thegrand median.


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Hypotheses for Median Test

H0 : MX = MYH1 : MX MY


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Step 1:State the hypothesis:H0: MX=MY

H1: MXMY

Step 2:Find the grand median from the sample values.

Step 3 :Classify each sample observation according to twocriteria:whether it belongs to sample X or sample Ywhether it is above or below the computed sample

median

Step 4 :Draw the contingency table.


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SAMPLE

RELATION ON

THE SAMPLE

X Y TOTAL

ABOVE A B A+B

BELOW C D C+D

n1 = A+C n2 = B+D N=n1+n2

Where,A : The number of observations from sample X falling above the medianB : The number of observations from sample Y falling above the medianC : The number of observations from sample X falling below the medianD : The number of observations from sample Y falling below the median

CONTINGENCY TABLE


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TABLE 3.2Stroke-index values, milliliters, for patients admitted to themyocardial-infarction research unit of a university hospital

Diagnosis

Anterior transmusal infarction and anteriornecrosis(X)

Interior transmuralinfarction and interiornecrosis (Y)

25 13 9 46 31 43

25 30 17 20 21 42

17 20 37 25 38 30

26 23 20 17 19 20

18 26 11 36 38 29

30 12 32 54 41 1324 20 16 8 68 32

21 37 31 26 28 30


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Step 1:

Classify the sample observation as sampleX and Y.Anterior transmural infarction and anterior necrosis assample X

Interior transmural infarction and interior necrosis assample Y

H0:MX = MY

H1:MX MY (claim)

Assumption:This data consists of two independent random samples Xand Y.


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Step 4: Draw the contingency table.

Relationship to

25.5

Anterior

transmural

infarction and

anterior

necrosis

Inferior

transmural

infarction and

interior

necrosis

Total

Above 12 12 24

Below 20 4 24

Total 32 16 48


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Step 6: Making decision.

We are using standard normal distributionand the critical value are 1.96. Since -2.45are in the critical region, so we reject the null

hypotheses.

By using P-value method we get:

2(0.5 - 0.4929) = 0.0142


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Step 7: Conclusion.

Since werejectthe null hypothesis, wehave enough evidence tosupportthe claimthat the two population medians are not

equal.


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THANKS FOR WATCHING!!

http://www.youtube.com/watch?v=1

pju_nbB0i8&feature=youtu.be
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Median Test

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