13 - 1 © 2000 prentice-hall, inc. statistics the chi-square test & the analysis of contingency...

13 - 13 - 11

© 2000 Prentice-Hall, Inc.© 2000 Prentice-Hall, Inc.

StatisticsStatistics

The Chi-Square Test & The Chi-Square Test & The Analysis of Contingency TablesThe Analysis of Contingency Tables

Chapter 13Chapter 13

13 - 13 - 22


Learning ObjectivesLearning Objectives

1.1. Explain Explain 22 Test for Proportions Test for Proportions

2.2. Explain Explain 22 Test of Independence Test of Independence

3.3. Solve Hypothesis Testing ProblemsSolve Hypothesis Testing Problems Two or More Population ProportionsTwo or More Population Proportions IndependenceIndependence

13 - 13 - 33


Data TypesData Types

Data

Quantitative Qualitative

Discrete Continuous

Data

Quantitative Qualitative

Discrete Continuous

13 - 13 - 44


Qualitative DataQualitative Data

1.1. Qualitative Random Variables Yield Qualitative Random Variables Yield Responses That ClassifyResponses That Classify Example: Gender (Male, Female)Example: Gender (Male, Female)

2.2. Measurement Reflects # in CategoryMeasurement Reflects # in Category

3.3. Nominal or Ordinal ScaleNominal or Ordinal Scale

4.4. ExamplesExamples Do You Own Savings Bonds? Do You Own Savings Bonds? Do You Live On-Campus or Off-Campus?Do You Live On-Campus or Off-Campus?

13 - 13 - 55


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.

13 - 13 - 66


Chi-Square (Chi-Square (22) Test ) Test for for kk Proportions Proportions

13 - 13 - 77



QualitativeData



2 Test

2 or morepop.

2 pop.

QualitativeData



2 Test

2 or morepop.

2 pop.

13 - 13 - 88


Chi-Square (Chi-Square (22) Test ) Test for for kk Proportions Proportions

1.1. Tests Equality (=) of Proportions OnlyTests Equality (=) of Proportions Only Example: Example: pp11 = .2, = .2, pp22=.3, =.3, pp33 = .5 = .5

2.2. One Variable With Several LevelsOne Variable With Several Levels

3.3. AssumptionsAssumptions Multinomial ExperimentMultinomial Experiment Large Sample SizeLarge Sample Size

All Expected Counts All Expected Counts 5 5

4.4. Uses One-Way Contingency TableUses One-Way Contingency Table

13 - 13 - 99


Multinomial Multinomial ExperimentExperiment

1.1. nn Identical Trial Identical Trial

2.2. kk Outcomes to Each Trial Outcomes to Each Trial

3.3. Constant Outcome Probability, Constant Outcome Probability, ppkk

4.4. Independent TrialsIndependent Trials

5.5. Random Variable is Count, Random Variable is Count, nnkk

6.6. Example: Ask 100 People (Example: Ask 100 People (nn) Which of ) Which of 3 Candidates (3 Candidates (kk) They Will Vote For) They Will Vote For

13 - 13 - 1010


One-Way One-Way Contingency TableContingency Table

1.1. Shows # Observations in Shows # Observations in kk Independent Independent Groups (Outcomes or Variable Levels)Groups (Outcomes or Variable Levels)

13 - 13 - 1111


Candidate

Tom Bill Mary Total

35 20 45 100

Candidate

Tom Bill Mary Total

35 20 45 100

One-Way One-Way Contingency TableContingency Table

1.1. Shows # Observations in Shows # Observations in kk Independent Independent Groups (Outcomes or Variable Levels)Groups (Outcomes or Variable Levels)

Outcomes (Outcomes (kk = 3) = 3)

Number of responsesNumber of responses

13 - 13 - 1212


22 Test for Test for kk Proportions Proportions

Hypotheses & StatisticHypotheses & Statistic

13 - 13 - 1313


1.1. HypothesesHypotheses HH00: : pp11 = = pp1,01,0, , pp22 = = pp2,02,0, ..., , ..., ppkk = = ppkk,0,0

HHaa: Not all : Not all ppii are equal are equal


Hypotheses & StatisticHypotheses & StatisticHypothesized Hypothesized probabilityprobability

13 - 13 - 1414




2.2. Test StatisticTest Statistic

22

n E n

E ni i

i

afafall cells

22

n E n

E ni i

i

afafall cells



Observed countObserved count

Expected countExpected count

Hypothesized Hypothesized probabilityprobability

13 - 13 - 1515





3.3. Degrees of Freedom: Degrees of Freedom: kk - 1 - 1

22

n E n

E ni i

i

afafall cells

22

n E n

E ni i

i

afafall cells




Expected countExpected count

Number of Number of outcomesoutcomes

Hypothesized Hypothesized probabilityprobability

13 - 13 - 1616


22 Test Basic Idea Test Basic Idea

1.1. Compares Observed Count to Compares Observed Count to Expected Count If Null Hypothesis Expected Count If Null Hypothesis Is TrueIs True

2.2. Closer Observed Count to Expected Closer Observed Count to Expected Count, the More Likely the HCount, the More Likely the H00 Is True Is True Measured by Squared Difference Relative Measured by Squared Difference Relative

to Expected Countto Expected Count Reject Large ValuesReject Large Values

13 - 13 - 1717


Finding Critical Finding Critical Value ExampleValue Example

What is the critical What is the critical 22 value if value if kk = 3, & = 3, & =.05? =.05?

13 - 13 - 1818


Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991



20 20


22 Table Table (Portion)(Portion)

13 - 13 - 1919





20 20



If If nnii = = EE((nnii)), , 22 = 0. = 0.

Do not reject HDo not reject H00

13 - 13 - 2020





20

Reject

20

Reject


= .05= .05


If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2121





20

Reject

20

Reject


= .05= .05


If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2222





20

Reject

20

Reject


= .05= .05


If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2323





20

Reject

20

Reject


= .05= .05


If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2424





20

Reject

20

Reject


= .05= .05


dfdf = = kk - 1 = 2 - 1 = 2

If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2525





20

Reject

20

Reject


= .05= .05


dfdf = = kk - 1 = 2 - 1 = 2

If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2626





20

Reject

20

Reject


= .05= .05


dfdf = = kk - 1 = 2 - 1 = 2

If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2727





20

Reject

20

Reject


= .05= .05


dfdf = = kk - 1 = 2 - 1 = 2

If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2828





20 5.991

Reject

20 5.991

Reject


= .05= .05


dfdf = = kk - 1 = 2 - 1 = 2

If If nnii = = EE((nnii)), , 22 = 0. = 0.


13 - 13 - 2929


As personnel director, you As personnel director, you want to test the perception of want to test the perception of fairness of three methods of fairness of three methods of performance evaluation. Of performance evaluation. Of 180180 employees, employees, 6363 rated rated Method 1Method 1 as fair. as fair. 4545 rated rated Method 2 Method 2 as fair. as fair. 7272 rated rated Method 3 Method 3 as fair. At the as fair. At the .05.05 level, is there a level, is there a differencedifference in in perceptions? perceptions?

22 Test for Test for kk Proportions ExampleProportions Example

13 - 13 - 3030


22 Test for Test for kk Proportions SolutionProportions Solution

13 - 13 - 3131



HH00::

HHaa::

==

nn11 = = nn22 = = nn33 = =

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

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

20

Reject

20

Reject

13 - 13 - 3232



HH00: : pp11 = = pp22 = = pp33 = 1/3 = 1/3

HHaa: : At least 1 is differentAt least 1 is different

==

nn11 = = nn22 = = nn33 = =



Decision:Decision:


20

Reject

20

Reject

13 - 13 - 3333



HH00: : pp11 = = pp22 = = pp33 = 1/3 = 1/3


= = .05.05

nn11 = = 63 63 nn22 = = 45 45 nn33 = = 72 72



Decision:Decision:


20

Reject

20

Reject

13 - 13 - 3434



HH00: : pp11 = = pp22 = = pp33 = 1/3 = 1/3


= = .05.05

nn11 = = 63 63 nn22 = = 45 45 nn33 = = 72 72



Decision:Decision:


20 5.991

Reject

20 5.991

Reject

= .05= .05

13 - 13 - 3535


E n np

E n E n E n

n E n

E n

n n n

i i

i i

i

afaf af af a f

afaf

,

.

0

1 2 3

22

12

22

32

2 2 2

180 1 3 60

60

60

60

60

60

60

63 60

60

45 60

60

72 60

606 3

all cells

E n np

E n E n E n

n E n

E n

n n n

i i

i i

i

afaf af af a f

afaf

,

.

0

1 2 3

22

12

22

32

2 2 2

180 1 3 60

60

60

60

60

60

60

63 60

60

45 60

60

72 60

606 3

all cells


13 - 13 - 3636



HH00: : pp11 = = pp22 = = pp33 = 1/3 = 1/3


= = .05.05

nn11 = = 63 63 nn22 = = 45 45 nn33 = = 72 72



Decision:Decision:


20 5.991

Reject

20 5.991

Reject

= .05= .05

22 = 6.3 = 6.3

13 - 13 - 3737



HH00: : pp11 = = pp22 = = pp33 = 1/3 = 1/3


= = .05.05

nn11 = = 63 63 nn22 = = 45 45 nn33 = = 72 72



Decision:Decision:


20 5.991

Reject

20 5.991

Reject

= .05= .05

22 = 6.3 = 6.3

Reject at Reject at = .05 = .05

13 - 13 - 3838



HH00: : pp11 = = pp22 = = pp33 = 1/3 = 1/3


= = .05.05

nn11 = = 63 63 nn22 = = 45 45 nn33 = = 72 72



Decision:Decision:



There is evidence of a There is evidence of a difference in proportions difference in proportions 20 5.991

Reject

20 5.991

Reject

= .05= .05

22 = 6.3 = 6.3

13 - 13 - 3939


22 Test of Independence Test of Independence

13 - 13 - 4040



QualitativeData



2 Test

2 or morepop.

2 pop.

QualitativeData



2 Test

2 or morepop.

2 pop.

13 - 13 - 4141


22 Test of Test of IndependenceIndependence

1.1. Shows If a Relationship Exists Between Shows If a Relationship Exists Between 2 Qualitative Variables2 Qualitative Variables One Sample Is DrawnOne Sample Is Drawn Does Does NotNot Show Causality Show Causality

2.2. AssumptionsAssumptions Multinomial ExperimentMultinomial Experiment All Expected Counts All Expected Counts 5 5

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

13 - 13 - 4242


22 Test of Test of Independence 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

13 - 13 - 4343


House LocationHouse Style Urban Rural Total

Split-Level 63 49 112Ranch 15 33 48

Total 78 82 160

House LocationHouse Style Urban Rural Total

Split-Level 63 49 112Ranch 15 33 48

Total 78 82 160


Contingency Table Contingency Table 1.1. Shows # Observations From 1 Sample Shows # Observations From 1 Sample

Jointly in 2 Qualitative VariablesJointly in 2 Qualitative VariablesLevels of variable 2Levels of variable 2

Levels of variable 1Levels of variable 1

13 - 13 - 4444


22 Test of Test of Independence Independence Hypotheses & Hypotheses &

StatisticStatistic1.1. HypothesesHypotheses

HH00: Variables Are Independent : Variables Are Independent

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

13 - 13 - 4545






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

13 - 13 - 4646







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


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

13 - 13 - 4747



Expected CountsExpected Counts1.1. Statistical Independence Means Joint Statistical Independence Means Joint

Probability Equals Product of Marginal Probability Equals Product of Marginal ProbabilitiesProbabilities

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

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

13 - 13 - 4848


Expected Count Expected Count ExampleExample

13 - 13 - 4949


LocationUrban Rural

House Style Obs. Obs. Total

Split-Level 63 49 112

Ranch 15 33 48

Total 78 82 160

LocationUrban Rural



Ranch 15 33 48

Total 78 82 160


13 - 13 - 5050


LocationUrban Rural



Ranch 15 33 48

Total 78 82 160

LocationUrban Rural



Ranch 15 33 48

Total 78 82 160


112 112 160160

Marginal probability = Marginal probability =

13 - 13 - 5151


LocationUrban Rural



Ranch 15 33 48

Total 78 82 160

LocationUrban Rural



Ranch 15 33 48

Total 78 82 160


112 112 160160

78 78 160160



13 - 13 - 5252


LocationUrban Rural



Ranch 15 33 48

Total 78 82 160

LocationUrban Rural



Ranch 15 33 48

Total 78 82 160


112 112 160160

78 78 160160



Joint probability = Joint probability = 112 112 160160

78 78 160160

13 - 13 - 5353


LocationUrban Rural



Ranch 15 33 48

Total 78 82 160

LocationUrban Rural



Ranch 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

13 - 13 - 5454


Expected Count Expected Count CalculationCalculation

13 - 13 - 5555



Expected count = Row total Column total

Sample sizea fa f


Sample sizea fa f

13 - 13 - 5656


House LocationUrban Rural

House Style Obs. Exp. Obs. Exp. Total

Split-Level 63 54.6 49 57.4 112

Ranch 15 23.4 33 24.6 48

Total 78 78 82 82 160

House LocationUrban Rural

House Style Obs. Exp. Obs. Exp. Total

Split-Level 63 54.6 49 57.4 112

Ranch 15 23.4 33 24.6 48

Total 78 78 82 82 160


112·82 112·82 160160

48·78 48·78 160160

48·82 48·82 160160

112·78 112·78 160160


Sample sizea fa f


Sample sizea fa f

13 - 13 - 5757


Diet PepsiDiet Coke No Yes TotalNo 84 32 116Yes 48 122 170

Total 132 154 286


Total 132 154 286

You’re a marketing research analyst. You ask a You’re a marketing research analyst. You ask a random sample of random sample of 286286 consumers if they consumers if they purchase Diet Pepsi or Diet Coke. At the purchase Diet Pepsi or Diet Coke. At the .05.05 level, is there evidence of a level, is there evidence of a relationshiprelationship??


ExampleExample

13 - 13 - 5858



SolutionSolution

13 - 13 - 5959



SolutionSolutionHH00: :

HHaa: :

= =

df = df =



Decision:Decision:


20

Reject

20

Reject

13 - 13 - 6060



SolutionSolutionHH00: : No Relationship No Relationship

HHaa: : Relationship Relationship

= =

df = df =



Decision:Decision:


20

Reject

20

Reject

13 - 13 - 6161





= = .05.05

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



Decision:Decision:


20

Reject

20

Reject

13 - 13 - 6262





= = .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

13 - 13 - 6363


Diet PepsiNo Yes

Diet Coke 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

Diet PepsiNo Yes

Diet Coke 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

170·132 170·132 286286

170·154 170·154 286286

116·132 116·132 286286

154·132 154·132 286286


SolutionSolution

13 - 13 - 6464


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


SolutionSolution

13 - 13 - 6565





= .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

13 - 13 - 6666





= .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

13 - 13 - 6767





= .05= .05

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



Decision:Decision:



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

13 - 13 - 6868



Total 132 154 286


Total 132 154 286

OK. There is a statistically significant OK. There is a statistically significant relationship between purchasing Diet Coke & relationship between purchasing Diet Coke & Diet Pepsi. So what do you think the Diet Pepsi. So what do you think the relationship is? Aren’t they competitors?relationship is? Aren’t they competitors?


Thinking ChallengeThinking Challenge

13 - 13 - 6969


You Re-Analyze the You Re-Analyze the DataData

13 - 13 - 7070



Total 44 32 76


Total 44 32 76


High High IncomeIncome

13 - 13 - 7171



Total 88 122 210


Total 88 122 210


Total 44 32 76


Total 44 32 76


Low Low IncomeIncome

High High IncomeIncome

13 - 13 - 7272


True Relationships*True Relationships*

Apparent Apparent relationrelation

Underlying Underlying causal relationcausal relation

Control or Control or intervening variable intervening variable

(true cause)(true cause)

Diet Diet CokeCoke

Diet Diet PepsiPepsi

13 - 13 - 7474


ConclusionConclusion

1.1. Explained Explained 22 Test for Proportions Test for Proportions

2.2. Explained Explained 22 Test of Independence Test of Independence

3.3. Solved Hypothesis Testing ProblemsSolved Hypothesis Testing Problems Two or More Population ProportionsTwo or More Population Proportions IndependenceIndependence

End of Chapter

Any blank slides that follow are blank intentionally.

13 - 1 © 2000 prentice-hall, inc. statistics the chi-square test & the analysis of contingency...

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