2 test of independence. hypothesis tests categorical data

2 Test of Independence2 Test of Independence

Hypothesis Tests Categorical DataHypothesis Tests Categorical Data

CategoricalData

Z TestZ Test

(Indep.)2 Test(Indep.)

Proportion Independence1 pop.

2 Test

2 or morepop.

2 pop.

CategoricalData

Z TestZ Test

(Indep.)2 Test(Indep.)

Proportion Independence1 pop.

2 Test

2 or morepop.

2 pop.

2 Test of Independence2 Test of Independence

• Shows if a relationship exists between 2 categorical variables– One sample is drawn– Does not show nature of relationship– Does not show causality

• Similar to testing p1 = p2 = … = pc

• Used widely in marketing • Uses contingency (XTAB) table

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

2 Test of Independence Contingency Table

2 Test of Independence Contingency Table

• Shows # observations from 1 sample jointly in 2 categorical variables

Levels of variable 2Levels of variable 2

Levels of variable 1Levels of variable 1

2 Test of Independence Hypotheses & Statistic


• Hypotheses– H0: Variables are independent

– H1: Variables are related (dependent)





• Test statistic

cells all

22

e

eo

f

ff

cells all

22

e

eo

f

ff

Observed frequencyObserved frequency

Expected frequencyExpected frequency





• Test statistic

• Degrees of freedom: (r - 1)(c - 1)

cells all

22

e

eo

f

ff

cells all

22

e

eo

f

ff

Observed frequencyObserved frequency

Expected frequencyExpected frequency

RowsRows Columns Columns

2 Test of Independence Expected Frequencies

2 Test of Independence Expected Frequencies

• Statistical independence means joint probability equals product of marginal probabilities– P(A and B) = P(A)·P(B)

• Compute marginal probabilities• Multiply for joint probability• Expected frequency is sample size times joint

probability

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

Expected Frequencies Calculation

Expected Frequencies Calculation

82·11282·112160160

78·4878·48160160

82·4882·48160160

78·11278·112160160

Expected frequency = (row total*column total)/grand total

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 random sample of 286 consumers if they purchase Diet Pepsi or Diet Coke. At the .05 level, is there evidence of a relationship?

2 Test of Independence Example

2 Test of Independence Example

2 Test of Independence Solution


H0: No Relationship

H1: Relationship

= .05

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

Critical Value(s):

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

20 3.841

Reject

20 3.841

Reject

= .05= .05

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

Expected Frequencies SolutionExpected Frequencies Solutionffee 5 in all cells 5 in all cells

132·170132·170286286

154·170154·170286286

132·116132·116286286

132·154132·154286286

2 Test of Independence Test Statistic Solution

2 Test of Independence Test Statistic Solution

Cell fo fe fo - fe (fo - fe)² (fo - fe)²/ fe

1,1 84 53.5 +30.5 930.25 17.3879

1,2 32 62.5 -30.5 930.25 14.8840

2,1 48 78.5 -30.5 930.25 11.8503

2,2 122 91.5 +30.5 930.25 10.1667

Total 286 286 54.2889

Cell fo fe fo - fe (fo - fe)² (fo - fe)²/ fe

1,1 84 53.5 +30.5 930.25 17.3879

1,2 32 62.5 -30.5 930.25 14.8840

2,1 48 78.5 -30.5 930.25 11.8503

2,2 122 91.5 +30.5 930.25 10.1667

Total 286 286 54.2889



H0: No Relationship

H1: Relationship

= .05

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

Critical Value(s):


Decision:Decision:


20 3.841

Reject

20 3.841

Reject

= .05= .05

54.2889

cells all ef

2e

fo

f2χ



H0: No Relationship

H1: Relationship

= .05

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

Critical Value(s):


Decision:Decision:


Reject at Reject at = .05 = .05

20 3.841

Reject

20 3.841

Reject

= .05= .05

22

54 2889

f f

fo e

e

a fall cells

.

22

54 2889

f f

fo e

e

a fall cells

.



H0: No Relationship

H1: Relationship

= .05

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

Critical Value(s):


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 2889

f f

fo e

e

a fall cells

.

22

54 2889

f f

fo e

e

a fall cells

.


Total 132 154 286


Total 132 154 286

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

2 Test of Independence Thinking Challenge

2 Test of Independence Thinking Challenge

AloneAlone GroupGroup Class Class


Total 88 122 210


Total 44 32 76

You Re-Analyze the DataYou Re-Analyze the Data

Low Low IncomeIncome

High High IncomeIncome

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

This Class...This Class...

• What was the most important thing you learned in class today?

• What do you still have questions about?

• How can today’s class be improved?

Please take a moment to answer the Please take a moment to answer the following questions in writing:following questions in writing:

2 test of independence. hypothesis tests categorical data

Documents

test of independence

relationship h

frequency slide

test of independence

related dependent slide

independent h

significant relationship

nature of relationship