EPI809/Spring 2008 1 Chapter 10 Hypothesis testing: Categorical Data Analysis.

EPI809/Spring 2008EPI809/Spring 2008 11

Chapter 10Chapter 10

Hypothesis testing: Categorical Hypothesis testing: Categorical Data AnalysisData Analysis


Learning ObjectivesLearning Objectives

1.1. Comparison of binomial proportion using Z and Comparison of binomial proportion using Z and 22 Test. Test.

2.2. Explain Explain 22 Test for Independence of 2 variables Test for Independence of 2 variables

3.3. Explain The Fisher’s test for independenceExplain The Fisher’s test for independence

4.4. McNemar’s tests for correlated dataMcNemar’s tests for correlated data

5.5. Kappa StatisticKappa Statistic

6.6. Use of SAS Proc FREQ Use of SAS Proc FREQ


Data TypesData Types

Data

Quantitative Qualitative

Discrete Continuous

Data

Quantitative Qualitative

Discrete Continuous


Qualitative DataQualitative Data

1.1. Qualitative Random Variables Yield Qualitative Random Variables Yield Responses That Can Be Put In Categories. Responses That Can Be Put In Categories. Example: Gender (Male, Female)Example: Gender (Male, Female)

2.2. Measurement or Count Reflect # in CategoryMeasurement or Count Reflect # in Category3.3. Nominal (no order) or Ordinal Scale (order)Nominal (no order) or Ordinal Scale (order)

4.4. Data can be collected as continuous but Data can be collected as continuous but recoded to categorical data. Example recoded to categorical data. Example (Systolic Blood Pressure - Hypotension, (Systolic Blood Pressure - Hypotension, Normal tension, hypertension ) Normal tension, hypertension )


Hypothesis Tests Hypothesis Tests Qualitative Data Qualitative Data

QualitativeData

Z Test Z Test 2 Test

Proportion Independence1 pop.

2 Test

2 or morepop.

2 pop.

QualitativeData



2 Test

2 or morepop.

2 pop.


Z Test for Differences in Z Test for Differences in Two ProportionsTwo Proportions


Hypotheses for Hypotheses for TwoTwo Proportions Proportions



Research Questions

Hypothesis No DifferenceAny Difference

Pop 1 Pop 2Pop 1 < Pop 2

Pop 1 Pop 2Pop 1 > Pop 2

H0

Ha

Research Questions




H0

Ha



Research Questions




H0 p1 - p2 = 0

Ha p1 - p2 0

Research Questions




H0 p1 - p2 = 0

Ha p1 - p2 0



Research Questions




H0 p1 - p2 = 0 p1 - p2 0

Ha p1 - p2 0 p1 - p2 < 0

Research Questions




H0 p1 - p2 = 0 p1 - p2 0

Ha p1 - p2 0 p1 - p2 < 0



Research Questions




H0 p1 - p2 = 0 p1 - p2 0 p1 - p2

0

Ha p1 - p2 0 p1 - p2 < 0 p1 - p2 > 0

Research Questions




H0 p1 - p2 = 0 p1 - p2 0 p1 - p2

0

Ha p1 - p2 0 p1 - p2 < 0 p1 - p2 > 0



Research Questions




H0 p1 - p2 = 0 p1 - p2 0 p1 - p2

0

Ha p1 - p2 0 p1 - p2 < 0 p1 - p2 > 0

Research Questions




H0 p1 - p2 = 0 p1 - p2 0 p1 - p2

0

Ha p1 - p2 0 p1 - p2 < 0 p1 - p2 > 0


Z Test for Difference in Two Z Test for Difference in Two ProportionsProportions

1.1. AssumptionsAssumptions Populations Are IndependentPopulations Are Independent Populations Follow Binomial DistributionPopulations Follow Binomial Distribution Normal Approximation Can Be Used for Normal Approximation Can Be Used for

large samples large samples (All Expected Counts (All Expected Counts 5) 5)

2.2. Z-Test Statistic for Two ProportionsZ-Test Statistic for Two Proportions

21

21

21

2121 ˆ where11

ˆ1ˆ

ˆˆ

nn

XXp

nnpp

ppppZ

21

21

21

2121 ˆ where11

ˆ1ˆ

ˆˆ

nn

XXp

nnpp

ppppZ


Sample Distribution for Difference Sample Distribution for Difference Between Proportions Between Proportions

1 1 2 21 2 1 2

1 2

0 1 21 2

1 2

1 2

1 1 N ;

1 1N 0; :

,

p p p pp p p p

n n

pq under H p pn n

x xp

n n

2 21 2

1 2 1 21 2

~ N ;X Xn n


Z Test for Two Proportions Z Test for Two Proportions Thinking Challenge Thinking Challenge

You’re an epidemiologist for the US You’re an epidemiologist for the US Department of Health and Human Department of Health and Human Services. You’re studying the Services. You’re studying the prevalence of disease X in two prevalence of disease X in two states (MA and CA). In states (MA and CA). In MAMA, , 7474 of of 15001500 people surveyed were people surveyed were diseased and in diseased and in CACA, , 129 129 of of 15001500 were diseased. At were diseased. At .05.05 level, does level, does MAMA have a have a lowerlower prevalence rate? prevalence rate?

MA

CA


Z Test for Two Proportions Z Test for Two Proportions Solution*Solution*


Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:


HH00::

HHaa::

= =

nnMAMA = = nnCACA ==

Critical Value(s):Critical Value(s):



Decision:Decision:



HH00: : ppMAMA - - ppCACA = 0 = 0

HHaa: : ppMAMA - - ppCACA < 0 < 0

==

nnMAMA = = nnCACA ==




Decision:Decision:





== .05 .05

nnMAMA = = 1500 1500 nnCACA = = 15001500




Decision:Decision:





== .05 .05

nnMAMA == 1500 1500 nnCACA == 1500 1500


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject



00.4

15001

15001

0677.10677.

00860.0493.

0677.15001500

12974ˆ

0860.1500

129ˆ0493.

1500

74ˆ

Z

nn

XXp

n

Xp

n

Xp

CAMA

CAMA

CA

CACA

MA

MAMA

00.4

15001

15001

0677.10677.

00860.0493.

0677.15001500

12974ˆ

0860.1500

129ˆ0493.

1500

74ˆ

Z

nn

XXp

n

Xp

n

Xp

CAMA

CAMA

CA

CACA

MA

MAMA


Z = -4.00Z = -4.00




== .05 .05

nnMAMA = = 1500 1500 nnCACA == 1500 1500



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject


Z = -4.00Z = -4.00




= = .05.05

nnMAMA = = 1500 1500 nnCACA == 1500 1500



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject Reject at Reject at = .05 = .05


Z = -4.00Z = -4.00




== .05 .05

nnMAMA == 1500 1500 nnCACA == 1500 1500



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject Reject at Reject at = .05 = .05

There is evidence MA There is evidence MA is less than CAis less than CA


22 Test of Independence Test of Independence Between 2 Categorical Between 2 Categorical

VariablesVariables


Hypothesis Tests Hypothesis Tests Qualitative Data Qualitative Data

QualitativeData



2 Test

2 or morepop.

2 pop.

QualitativeData



2 Test

2 or morepop.

2 pop.


22 Test of Independence Test of Independence

1.1.Shows If a Relationship Exists Between 2 Shows If a Relationship Exists Between 2 Qualitative Variables, but does Qualitative Variables, but does NotNot Show Show CausalityCausality

2.2.AssumptionsAssumptionsMultinomial ExperimentMultinomial Experiment

All Expected Counts All Expected Counts 5 5

3.3.Uses Two-Way Contingency TableUses Two-Way Contingency Table


22 Test of Independence Test of Independence Contingency Table Contingency Table

1.1. Shows # Observations From 1 Shows # Observations From 1 Sample Jointly in 2 Qualitative VariablesSample Jointly in 2 Qualitative Variables


Residence Disease Status

Urban Rural Total

Disease 63 49 112 No disease 15 33 48 Total 78 82 160

Residence Disease Status

Urban Rural Total

Disease 63 49 112 No disease 15 33 48 Total 78 82 160

22 Test of Independence Test of Independence Contingency Table Contingency Table

1.1.Shows # Observations From 1 Sample Shows # Observations From 1 Sample Jointly in 2 Qualitative VariablesJointly in 2 Qualitative Variables

Levels of variable 2Levels of variable 2

Levels of variable 1Levels of variable 1


22 Test of Independence Test of Independence Hypotheses & StatisticHypotheses & Statistic

1.1.HypothesesHypotheses HH00: Variables Are Independent : Variables Are Independent

HHaa: Variables Are Related (Dependent): Variables Are Related (Dependent)



1.1.HypothesesHypothesesHH00: Variables Are Independent : Variables Are Independent


2.2.Test StatisticTest Statistic Observed countObserved count

Expected Expected countcount 2

2

n E n

E n

ij ij

ij

c hc hall cells

2

2

n E n

E n

ij ij

ij

c hc hall cells



1.1.HypothesesHypothesesHH00: Variables Are Independent : Variables Are Independent


2.2.Test StatisticTest Statistic

Degrees of Freedom: (Degrees of Freedom: (rr - 1)( - 1)(cc - 1) - 1)RowsRows Columns Columns

Observed countObserved count

Expected Expected countcount 2

2

n E n

E n

ij ij

ij

c hc hall cells

2

2

n E n

E n

ij ij

ij

c hc hall cells


22 Test of Independence Test of Independence Expected CountsExpected Counts

1.1.Statistical Independence Means Joint Statistical Independence Means Joint Probability Equals Product of Marginal Probability Equals Product of Marginal ProbabilitiesProbabilities

2.2.Compute Marginal Probabilities & Multiply Compute Marginal Probabilities & Multiply for Joint Probabilityfor Joint Probability

3.3.Expected Count Is Sample Size Times Expected Count Is Sample Size Times Joint ProbabilityJoint Probability


Expected Count ExampleExpected Count Example


Residence Disease Urban Rural

Status Obs. Obs. Total

Disease 63 49 112

No Disease 15 33 48

Total 78 82 160



Disease 63 49 112

No Disease 15 33 48

Total 78 82 160


112 112 160160

Marginal probability = Marginal probability =




Disease 63 49 112

No Disease 15 33 48

Total 78 82 160



Disease 63 49 112

No Disease 15 33 48

Total 78 82 160


112 112 160160

78 78 160160






Disease 63 49 112

No Disease 15 33 48

Total 78 82 160



Disease 63 49 112

No Disease 15 33 48

Total 78 82 160


112 112 160160

78 78 160160



Joint probability = Joint probability = 112 112 160160

78 78 160160




Disease 63 49 112

No Disease 15 33 48

Total 78 82 160



Disease 63 49 112

No Disease 15 33 48

Total 78 82 160


112 112 160160

78 78 160160



Joint probability = Joint probability = 112 112 160160

78 78 160160

Expected count = 160· Expected count = 160· 112 112 160160

78 78 160160

= 54.6 = 54.6


Expected Count CalculationExpected Count Calculation



Expected count = Row total Column total

Sample sizea fa f


Sample sizea fa f



Status Obs. Exp. Obs. Exp. Total

Disease 63 54.6 49 57.4 112

No Disease 15 23.4 33 24.6 48

Total 78 78 82 82 160


Status Obs. Exp. Obs. Exp. Total

Disease 63 54.6 49 57.4 112

No Disease 15 23.4 33 24.6 48

Total 78 78 82 82 160


112x82 112x82 160160

48x78 48x78 160160

48x82 48x82 160160

112x78 112x78 160160


Sample sizea fa f


Sample sizea fa f


HIV STDs Hx No Yes Total

No 84 32 116 Yes 48 122 170 Total 132 154 286

HIV STDs Hx No Yes Total

No 84 32 116 Yes 48 122 170 Total 132 154 286

You randomly sample You randomly sample 286286 sexually active sexually active individuals and collect information on their HIV individuals and collect information on their HIV status and History of STDs. At the status and History of STDs. At the .05.05 level, is level, is there evidence of a there evidence of a relationshiprelationship??

22 Test of Independence Test of Independence Example on HIVExample on HIV


22 Test of Independence Test of Independence SolutionSolution



HH00: :

HHaa: :

= =

df = df =



Decision:Decision:


20

Reject

20

Reject



HH00: : No Relationship No Relationship

HHaa: : Relationship Relationship

= =

df = df =



Decision:Decision:


20

Reject

20

Reject





= = .05.05

df = df = (2 - 1)(2 - 1) = 1 (2 - 1)(2 - 1) = 1



Decision:Decision:


20

Reject

20

Reject





= = .05.05

df = df = (2 - 1)(2 - 1) = 1 (2 - 1)(2 - 1) = 1



Decision:Decision:


20 3.841

Reject

20 3.841

Reject

= .05= .05


HIV No Yes

STDs HX Obs. Exp. Obs. Exp. Total

No 84 53.5 32 62.5 116

Yes 48 78.5 122 91.5 170

Total 132 132 154 154 286

HIV No Yes

STDs HX Obs. Exp. Obs. Exp. Total

No 84 53.5 32 62.5 116

Yes 48 78.5 122 91.5 170

Total 132 132 154 154 286

EE((nnijij)) 5 in all 5 in all

cellscells

170x132 170x132 286286

170x154 170x154 286286

116x132 116x132 286286

154x116 154x116 286286



2

2

11 11

2

11

12 12

2

12

22 22

2

22

2 2 284 53 5

53 5

32 62 5

62 5

122 915

91554 29

n E n

E n

n E n

E n

n E n

E n

n E n

E n

ij ij

ij

.

.

.

.

.

..

c hc h

a fa f

a fa f

a fa f

all cells

2

2

11 11

2

11

12 12

2

12

22 22

2

22

2 2 284 53 5

53 5

32 62 5

62 5

122 915

91554 29

n E n

E n

n E n

E n

n E n

E n

n E n

E n

ij ij

ij

.

.

.

.

.

..

c hc h

a fa f

a fa f

a fa f

all cells






= .05= .05

dfdf = (2 - 1)(2 - 1) = 1 = (2 - 1)(2 - 1) = 1



Decision:Decision:


20 3.841

Reject

20 3.841

Reject

= .05= .05

22 = 54.29 = 54.29





= .05= .05

dfdf = (2 - 1)(2 - 1) = 1 = (2 - 1)(2 - 1) = 1



Decision:Decision:


Reject at Reject at = .05 = .05

20 3.841

Reject

20 3.841

Reject

= .05= .05

22 = 54.29 = 54.29





= .05= .05

dfdf = (2 - 1)(2 - 1) = 1 = (2 - 1)(2 - 1) = 1



Decision:Decision:


Reject at Reject at = .05 = .05

There is evidence of a There is evidence of a relationshiprelationship20 3.841

Reject

20 3.841

Reject

= .05= .05

22 = 54.29 = 54.29


22 Test of Independence Test of Independence SAS CODESSAS CODES

DataData dis; dis;input STDs HIV count;input STDs HIV count;cards;cards;1 1 84 1 1 84 1 2 321 2 322 1 482 1 482 2 1222 2 122;;runrun;;

ProcProc freqfreq data=dis order=data; data=dis order=data; weight Count;weight Count; tables STDs*HIV/tables STDs*HIV/chisqchisq;;runrun;;


22 Test of Independence Test of Independence SAS OUTPUTSAS OUTPUT

Statistics for Table of STDs by HIV

Statistic DF Value Prob ------------------------------------------------------- Chi-Square 1 54.1502 <.0001 Likelihood Ratio Chi-Square 1 55.7826 <.0001 Continuity Adj. Chi-Square 1 52.3871 <.0001 Mantel-Haenszel Chi-Square 1 53.9609 <.0001 Phi Coefficient 0.4351 Contingency Coefficient 0.3990 Cramer's V 0.4351

EPI809/Spring 2008 1 Chapter 10 Hypothesis testing: Categorical Data Analysis.

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