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2. Statistics for Business and Economics
Chapter 9
Categorical Data Analysis
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3. © 2011 Pearson Education, Inc
Contents
9.1 Categorical Data and the Multinomial Experiment
9.2 Testing Category Probabilities: One-Way Table
9.3 Testing Category Probabilities: Two-Way Contingency Table
9.4 A Word of Caution about Chi-Square Tests
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Learning Objectives
Discuss qualitative (i.e., categorical) data with more than two outcomes
Present a chi-square hypothesis test for comparing the category proportions associated with a single qualitative variable–called a one-way analysis
Present a chi-square hypothesis test for relating two qualitative variables–called a two-way analysis
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5. Categorical Data and Multinomial Experiment
9.1
Categorical Data and Multinomial Experiment
:1, 1, 3
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Qualitative Data
Qualitative random variables yield responses that can be classified
Example: gender (male, female)
Qualitative data that fall in more than two categories often result from a multinomial experiment
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7. Properties of the Multinomial Experiment
1. The experiment consists of n identical trials.
2. There are k possible outcomes to each trial. These outcomes are called classes, categories, or cells.
3. The probabilities of the k outcomes, denoted by p1, p2,…, pk, remain the same from trial to trial,where p1 + p2 + … + pk = 1.
4. The trials are independent.
5. The random variables of interest are the cell counts, n1, n2, …, nk, of the number of observations that fall in each of the k classes.
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8. Testing Category Probabilities: One-Way Table
9.2
Testing Category Probabilities: One-Way Table
:1, 1, 3
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9. Multinomial Experiment
In this section, we consider a multinomial experiment with k outcomes that correspond to categories of a single qualitative variable. The results of such an experiment are summarized in a one-way table. The term one-way is used because only one variable is classified. Typically, we want to make inferences about the true proportions that occur in the k categories based on the sample information in the one-way table.
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10. Chi-Square (2) Test for k Proportions
Tests equality (=) of proportions only
Example: p1 = .2, p2=.3, p3 = .5
One variable with several levels
Uses one-way contingency table
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11. One-Way Contingency Table
Shows number of observations in k independent groups (outcomes or variable levels)
Outcomes (k = 3)
Candidate
Tom
Bill
Mary
Total 35 20 45
100
Number of responses
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12. A Test of a Hypothesis about Multinomial Probabilities: One-Way Table
H0: p1 = p1,0, p2 = p2,0, …, pk = pk,0
where p1,0, p2,0, …, pk,0 represent the hypothesized values of the multinomial probabilities.
Ha: At least one of the multinomial probabilities does not equal its hypothesized value.
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13. A Test of a Hypothesis about Multinomial Probabilities: One-Way Table
where Ei = npi,0 is the expected cell count–that is, the expected number of outcomes of type i assuming that H0 is true. The total sample size is n.
where has (k – 1) df.
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14. Conditions Required for a Valid Test: One-way Table
A multinomial experiment has been conducted. This is generally satisfied by taking a random sample from the population of interest.
The sample size n is large. This is satisfied if for every cell, the expected cell count Ei will be equal to 5 or more.
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2 Test Basic Idea
Compares observed count to expected count assuming null hypothesis is true
Closer observed count is to expected count, the more likely the H0 is true
Measured by squared difference relative to expected count
Reject large values
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16. Finding Critical Value Example
What is the critical 2 value if k = 3, and  =.05? c 2
Upper Tail Area DF
.995 …
.95
.05 1
...
0.004
3.841
0.010
0.103
5.991
2 Table (Portion)
If ni = E(ni), 2 = 0. Do not reject H0
Reject H0
 = .05
df = k - 1 = 2
5.991
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17. 2 Test for k Proportions Example
As personnel director, you want to test the perception of fairness of three methods of performance evaluation. Of 180 employees, 63 rated Method 1 as fair, 45 rated Method 2 as fair, 72 rated Method 3 as fair. At the .05 level of significance, is there a difference in perceptions?
To check assumptions, use sample proportions as estimators of population proportion:
n1·p = 78·63/78 = 63
n1·(1-p) = 78·(1-63/78) = 15
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18. 2 Test for k Proportions Solution
H0:
Ha:
 =
n1 = n2 = n3 =
Critical Value(s):
p1 = p2 = p3 = 1/3
At least 1 is different
.05 63 45 72
 = .05 c 2
Reject H0
5.991
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19. 2 Test for k Proportions Solution
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20. 2 Test for k Proportions Solution
H0:
Ha:
 =
n1 = n2 = n3 =
Critical Value(s):
p1 = p2 = p3 = 1/3
At least 1 is different
Test Statistic:
Decision:
Conclusion:
2 = 6.3
.05 63 45 72
 = .05 c 2
Reject H0
5.991
Reject at  = .05
There is evidence of a difference in proportions
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21. Testing Category Probabilities: Two-Way (Contingency) Table
9.3
Testing Category Probabilities: Two-Way (Contingency) Table
:1, 1, 3
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2 Test of Independence
Shows if a relationship exists between two qualitative variables
One sample is drawn
Does not show causality
Uses two-way contingency table
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23. 2 Test of Independence Contingency Table
Shows number of observations from one sample jointly in sample qualitative variables
Levels of variable 2
Levels of variable 1
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24. Finding Expected Cell Counts for a Two-Way Contingency Table
The estimate of the expected number of observations falling into the cell in row i and column j is given by
where Ri = total for row i, Cj = total for column j, and n = sample size.
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General Form of a Contingency Table Analysis: 2 -Test for Independence
H0: The two classifications are independent.
Ha: The two classifications are dependent.
where
Rejection region: where has (r – 1)(c – 1) df.
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26. Conditions Required for a Valid 2-Test: Contingency Table
A multinomial experiment has been conducted . We may then consider this to be a multinomial experiment with r  c possible outcomes.
The sample size n is large. This is satisfied if for every cell, the expected count Ei will be equal to 5 or more.
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27. 2 Test of Independence Expected Counts
Statistical independence means joint probability equals product of marginal probabilities
Compute marginal probabilities and multiply for joint probability
Expected count is sample size times joint probability e =
Column Tot al
Sample Siz
Row Total a f   f a f
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28. Expected Count Example
Marginal probability =
Location
Urban Rural House Style Obs. Obs. Total
Split–Level
Ranch
Total
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29. Expected Count Example
Marginal probability =
Location
Urban Rural House Style Obs. Obs. Total
Split–Level
Ranch
Total
Marginal probability =
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30. Expected Count Example
Joint probability =
Marginal probability =
Location
Urban Rural House Style Obs. Obs. Total
Split–Level
Ranch
Total
Expected count = 160·
= 54.6
Marginal probability =
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31. Expected Count Calculation
112·
54.6
House Location
112·
57.4
Urban
Rural
House Style
Obs.
Exp.
Obs.
Exp.
Total
Split -
Level 63 49
112
Ranch
48·
23.4 15 33
48·
24.6 48
Total 78 78 82 82
160
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32. 2 Test of Independence Example
As a realtor you want to determine if house style and house location are related. At the .05 level of significance, is there evidence of a relationship?
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33. 2 Test of Independence Solution
H0:
Ha:
 =
df =
Critical Value(s):
No Relationship
Relationship
.05
(2 – 1)(2 – 1) = 1 c 2
Reject H0
3.841
 = .05
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34. 2 Test of Independence Solution
Eij  5 in all cells
112·
112·
48·
48·
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35. 2 Test of Independence Solution
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36. 2 Test of Independence Solution
H0:
Ha:
 =
df =
Critical Value(s):
No Relationship
Relationship
Test Statistic:
Decision:
Conclusion:
2 = 8.41
.05
(2 – 1)(2 – 1) = 1 c 2
Reject H0
3.841
 = .05
Reject at  = .05
There is evidence of a relationship
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37. 2 Test of Independence Thinking Challenge
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 of significance, is there evidence of a relationship?
Diet Pepsi
Diet Coke No
Yes
Total No 84 32
116
Yes 48
122
170
Total
132
154
286
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38. 2 Test of Independence Solution
H0:
Ha:
 =
df =
Critical Value(s):
No Relationship
Relationship
.05
(2 – 1)(2 – 1) = 1 c 2
Reject H0
3.841
 = .05
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39. 2 Test of Independence Solution*
Eij  5 in all cells
116·
154(116) 286
170·
170·
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40. 2 Test of Independence Solution
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41. 2 Test of Independence Solution
H0:
Ha:
 =
df =
Critical Value(s):
No Relationship
Relationship
Test Statistic:
Decision:
Conclusion:
2 = 54.29
.05
(2 – 1)(2 – 1) = 1 c 2
Reject H0
3.841
 = .05
Reject at  = .05
There is evidence of a relationship
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42. 2 Test of Independence Thinking Challenge 2
There is a statistically significant relationship between purchasing Diet Coke and Diet Pepsi. So what do you think the relationship is? Aren’t they competitors?
Diet Pepsi
Diet Coke No
Yes
Total No 84 32
116
Yes 48
122
170
Total
132
154
286
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43. You Re-Analyze the Data
High Income
Diet Pepsi
Diet Coke No
Yes
Total No 4 30 34
Yes 40 2 42
There is a spurious relationship between purchasing Diet Coke & Diet Pepsi. Income is an intervening or control variable & is the true cause.
The analysis here uses only descriptive statistics.
For low income, consumers are price conscious. Either they can’t afford to buy either or they buy whatever is on sale.
For high income, consumers buy depending on preference regardless of price.
Total 44 32 76
Low Income
Diet Pepsi
Diet Coke No
Yes
Total No 80 2 82
Yes 8
120
128
Total 88
122
210
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44. Control or intervening variable (true cause)
True Relationships
Diet Coke
There is a spurious relationship between purchasing Diet Coke & Diet Pepsi. Income is an intervening or control variable & is the true cause.
The analysis here uses only descriptive statistics.
For low income, consumers are price conscious. Either they can’t afford to buy either or they buy whatever is on sale.
For high income, consumers buy depending on preference regardless of price.
Underlying causal relation
Apparent relation
Control or intervening variable (true cause)
Diet Pepsi
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45. Numbers don’t think - People do!
Moral of the Story
Numbers don’t think - People do!
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© T/Maker Co. 51

46. A Word of Caution about Chi-Square Tests
9.4
A Word of Caution about Chi-Square Tests
:1, 1, 3
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47. Caution about the 2 Test
The 2 is one of the most widely applied statistical tools and also one of the most abused statistical tool.
Be certain the experiment satisfies the assumptions.
Be certain the sample is drawn from the correct population.
Avoid using when the expected counts are very small.
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48. Caution about the 2 Test
If the 2 value does not exceed the established critical value of 2 , do not accept the hypothesis of independence. You risk a Type II error. Avoid concluding that two classifications are independent, even when 2 is small.
If a contingency table 2 value does exceed the critical value, we must be careful to avoid inferring that a causal relationship exists between the classifications. The existence of a causal relationship cannot be established by a contingency table analysis.
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49. © 2011 Pearson Education, Inc
Key Ideas
Multinomial Data
Qualitative data that fall into more than two categories (or classes)
As a result of this class, you will be able to...
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50. © 2011 Pearson Education, Inc
Key Ideas
Properties of a Multinomial Experiment
1. n identical trials
2. k possible outcomes
3. probabilities of the k outcomes (p1, p2, …, pk) remain the same from trial to trial, where p1 + p2 + … + pk = 1
4. trials are independent
5. variables of interest: cell counts (i.e., number of observations falling into each outcome category), denoted n1, n2, …, nk
As a result of this class, you will be able to...
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51. © 2011 Pearson Education, Inc
Key Ideas
One-Way Table
Summary table for a single qualitative variable
Two-Way (Contingency) Table
Summary table for two qualitative variables
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Key Ideas
Chi-Square (2) Statistic
used to test category probabilities in one-way and two-way tables
Chi-Square tests for independence
should not be used to infer a causal relationship between 2 Qualitative Variables
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Key Ideas
Conditions Required for Valid 2-Tests
1. multinomial experiment
2. sample size n is large (expected cell counts are all greater than or equal to 5)
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