Median Test
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Transcript of 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
http://www.youtube.com/watch?v=1pju_nbB0i8&feature=youtu.behttp://www.youtube.com/watch?v=1pju_nbB0i8&feature=youtu.behttp://www.youtube.com/watch?v=1pju_nbB0i8&feature=youtu.behttp://www.youtube.com/watch?v=1pju_nbB0i8&feature=youtu.be