1. Statistics for Business and Economics
Chapter 9
Categorical Data Analysis

2. Learning Objectives Explain 2 Test for Proportions
Explain 2 Test of Independence
Solve Hypothesis Testing Problems
More Than Two Population Proportions
Independence
As a result of this class, you will be able to ...

3. Data Types
Data
Quantitative
Qualitative
Continuous
Discrete 3

4. Qualitative Data
Qualitative random variables yield responses that classify
Example: gender (male, female)
Measurement reflects number in category
Nominal or ordinal scale
Examples
What make of car do you drive?
Do you live on-campus or off-campus?

5. Hypothesis Tests Qualitative Data
Z Test c 2
Test
Proportion
Independence
1 pop.
More than
2 pop. 5

6. Chi-Square (2) Test for k Proportions

7. Hypothesis Tests Qualitative Data
Z Test c 2
Test
Proportion
Independence
1 pop.
More than
2 pop. 5

8. Multinomial Experiment
n identical trials
k outcomes to each trial
Constant outcome probability, pk
Independent trials
Random variable is count, nk
Example: ask 100 people (n) which of 3 candidates (k) they will vote for

9. 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

10. 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 20

11. Conditions Required for a Valid Test: One-way Table
A multinomial experiment has been conducted
The sample size n is large: E(ni) is greater than or equal to 5 for every cell

12. 2 Test for k Proportions Hypotheses & Statistic
Hypothesized probability
1. Hypotheses
H0: p1 = p1,0, p2 = p2,0, ..., pk = pk,0
Ha: At least one pi is different from above
2. Test Statistic
Observed count
Expected count: E(ni) = npi,0
3. Degrees of Freedom: k – 1
Number of outcomes 24

13. 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

14. Finding Critical Value Example
What is the critical 2 value if k = 3, and  =.05?
If ni = E(ni), 2 = 0. Do not reject H0 c 2
Upper Tail Area DF
.995 …
.95
.05 1
...
0.004
3.841
0.010
0.103
5.991
2 Table (Portion)
Reject H0
 = .05
df = k - 1 = 2
5.991 26

15. 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 10

16. 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:
.05 63 45 72
 = .05 c 2
Reject H0
5.991 11

17. 2 Test for k Proportions Solution12

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
Test Statistic:
Decision:
Conclusion:
2 = 6.3
.05 63 45 72 c 2
Reject H0
Reject at  = .05
5.991
 = .05
There is evidence of a difference in proportions 11

19. Contingency Tables Contingency Tables
Useful in situations involving multiple population proportions
Used to classify sample observations according to two or more characteristics
Also called a cross-classification table.

20. Contingency Table Example
Left-Handed vs. Gender
Dominant Hand: Left vs. Right
Gender: Male vs. Female
2 categories for each variable, so
called a 2 x 2 table
Suppose we examine a sample of
300 children

21. Contingency Table Example
(continued)
Sample results organized in a contingency table:
Gender
Hand Preference
Left
Right
Female 12
108
120
Male 24
156
180 36
264
300
sample size = n = 300:
120 Females, 12 were left handed
180 Males, 24 were left handed

22. 2 Test for the Difference Between Two Proportions
H0: π1 = π2 (Proportion of females who are left
handed is equal to the proportion of
males who are left handed)
H1: π1 ≠ π2 (The two proportions are not the same
hand preference is not independent of gender)
If H0 is true, then the proportion of left-handed females should be the same as the proportion of left-handed males
The two proportions above should be the same as the proportion of left-handed people overall

23. The Chi-Square Test Statistic
The Chi-square test statistic is:
where:
fo = observed frequency in a particular cell
fe = expected frequency in a particular cell if H0 is true
(Assumed: each cell in the contingency table has expected frequency of at least 5)

24. Decision Rule
The test statistic approximately follows a chi-squared distribution with one degree of freedom
Decision Rule:
If , reject H0, otherwise, do not reject H0  2
Do not
reject H0
Reject H0
2α

25. Computing the Average Proportion
The average proportion is:
Here:
120 Females, 12 were left handed
180 Males, 24 were left handed
i.e., of all the children the proportion of left handers is 0.12, that is, 12%

26. Finding Expected Frequencies
To obtain the expected frequency for left handed females, multiply the average proportion left handed (p) by the total number of females
To obtain the expected frequency for left handed males, multiply the average proportion left handed (p) by the total number of males
If the two proportions are equal, then
P(Left Handed | Female) = P(Left Handed | Male) = .12
i.e., we would expect (.12)(120) = 14.4 females to be left handed
(.12)(180) = 21.6 males to be left handed

27. Observed vs. Expected Frequencies
Gender
Hand Preference
Left
Right
Female
Observed = 12
Expected = 14.4
Observed = 108
Expected = 105.6
120
Male
Observed = 24
Expected = 21.6
Observed = 156
Expected = 158.4
180 36
264
300

28. The Chi-Square Test Statistic
Gender
Hand Preference
Left
Right
Female
Observed = 12
Expected = 14.4
Observed = 108
Expected = 105.6
120
Male
Observed = 24
Expected = 21.6
Observed = 156
Expected = 158.4
180 36
264
300
The test statistic is:

29. Decision Rule 2 Decision Rule:
If > 3.841, reject H0, otherwise, do not reject H0
Here,
= < = 3.841,
so we do not reject H0 and conclude that there is not sufficient evidence that the two proportions are different at  = 0.05
0.05 2
Do not
reject H0
Reject H0
20.05 = 3.841

30. 2 Test for Differences Among More Than Two Proportions
Extend the 2 test to the case with more than two independent populations:
H0: π1 = π2 = … = πc
H1: Not all of the πj are equal (j = 1, 2, …, c)

31. The Chi-Square Test Statistic
The Chi-square test statistic is:
Where:
fo = observed frequency in a particular cell of the 2 x c table
fe = expected frequency in a particular cell if H0 is true
(Assumed: each cell in the contingency table has expected
frequency of at least 1)

32. Computing the Overall Proportion
The overall proportion is:
Expected cell frequencies for the c categories are calculated as in the 2 x 2 case, and the decision rule is the same:
Where is from the chi-squared distribution with c – 1 degrees of freedom
Decision Rule:
If , reject H0, otherwise, do not reject H0

33. The Marascuilo Procedure
Used when the null hypothesis of equal proportions is rejected
Enables you to make comparisons between all pairs
Start with the observed differences, pj – pj’, for all pairs (for j ≠ j’) . . .
. . .then compare the absolute difference to a calculated critical range

34. 2 Test of Independence

35. Hypothesis Tests Qualitative Data
Z Test c 2
Test
Proportion
Independence
1 pop.
More than
2 pop. 5

36. 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

37. 2 Test of Independence Contingency Table
Shows number of observations from 1 sample jointly in 2 qualitative variables
Levels of variable 2
Levels of variable 1 40

38. Conditions Required for a Valid 2 Test: Independence
Multinomial experiment has been conducted
The sample size, n, is large: Eij is greater than or equal to 5 for every cell

39. 2 Test of Independence Hypotheses & Statistic
H0: Variables are independent
Ha: Variables are related (dependent)
Test Statistic
Observed count
Expected count
Degrees of Freedom: (r – 1)(c – 1)
Rows Columns 41

40. 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

41. Expected Count Example
Marginal probability =
Location
Urban Rural House Style Obs. Obs. Total
Split–Level
Ranch
Total 43

42. Expected Count Example
Marginal probability =
Location
Urban Rural House Style Obs. Obs. Total
Split–Level
Ranch
Total
Marginal probability = 43

43. 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 = 43

44. 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 43

45. 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? 44

46. 2 Test of Independence Solution
H0:
Ha:
 =
df =
Critical Value(s):
No Relationship
Relationship
Test Statistic:
Decision:
Conclusion:
.05
(2 - 1)(2 - 1) = 1 c 2
Reject H0
3.841
 = .05 47

47. 2 Test of Independence Solution
Eij  5 in all cells
112·
112·
48·
48· 45

48. 2 Test of Independence Solution12

49. 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
Reject at  = .05
3.841
 = .05
There is evidence of a relationship 47

50. 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 44

51. 2 Test of Independence Solution*
H0:
Ha:
 =
df =
Critical Value(s):
No Relationship
Relationship
Test Statistic:
Decision:
Conclusion:
.05
(2 - 1)(2 - 1) = 1 c 2
Reject H0
3.841
 = .05 47

52. 2 Test of Independence Solution*
Eij  5 in all cells
116·
154·
170·
170· 45

53. 2 Test of Independence Solution*12

54. 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
Reject at  = .05
3.841
 = .05
There is evidence of a relationship 47

55. 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 48

56. 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 49

57. 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 50

58. Numbers don’t think - People do!
Moral of the Story*
Numbers don’t think - People do!
© T/Maker Co. 51

59. Conclusion Explained 2 Test for Proportions
Explained 2 Test of Independence
Solved Hypothesis Testing Problems
More Than Two Population Proportions
Independence
As a result of this class, you will be able to ...