Transcript Chap. 9: The Chi-Square Test & The Analysis of Contingency

Statistics for Business and
Economics
Chapter 9
Categorical Data Analysis
Learning Objectives
1. Explain 2 Test for Proportions
2. Explain 2 Test of Independence
3. Solve Hypothesis Testing Problems
• More Than Two Population Proportions
• Independence
Data Types
Data
Quantitative
Discrete
Continuous
Qualitative
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?
Hypothesis Tests
Qualitative Data
Qualitative
Data
1 pop.
Proportion
More than
2 pop.
Independence
2 pop.
Z Test
Z Test
c2 Test
c2 Test
Chi-Square (2) Test
for k Proportions
Hypothesis Tests
Qualitative Data
Qualitative
Data
1 pop.
Proportion
More than
2 pop.
Independence
2 pop.
Z Test
Z Test
c2 Test
c2 Test
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
2
( )
Chi-Square
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
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
Conditions Required for a Valid
Test: One-way Table
1. A multinomial experiment has been
conducted
2. The sample size n is large: E(ni) is greater
than or equal to 5 for every cell
2 Test for k Proportions
Hypotheses & Statistic
1.
Hypotheses
Hypothesized
probability
H0: p1 = p1,0, p2 = p2,0, ..., pk = pk,0
Ha: At least one pi is different from above
2.
Test Statistic
 ni  E  ni  
  
E  ni 
all cells
2
3.
Observed count
2
Expected count:
E(ni) = npi,0
Degrees of Freedom: k – 1
Number of
outcomes
2 Test Basic Idea
1. Compares observed count to expected count
assuming null hypothesis is true
2. 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
Finding Critical Value
Example
2
What is the critical
If ni = E(ni), 2 = 0.
Do not reject H0
value if k = 3, and
=.05?
Reject H0
= .05
df = k - 1 = 2
0
2
Table
(Portion)
DF .995
1
...
2 0.010
5.991
c2
Upper Tail Area
…
.95
…
… 0.004
…
… 0.103
…
.05
3.841
5.991
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?
2

•
•
•
•
•
Test for k Proportions
Solution
H0: p1 = p2 = p3 = 1/3
Test Statistic:
Ha: At least 1 is different
 = .05
n1 = 63 n2 = 45 n3 = 72
Critical Value(s):
Decision:
Reject H0
= .05
0
5.991
c2
Conclusion:
2

Test for k Proportions
Solution
E  ni   npi ,0
E  n1   E  n2   E  n3   180 1 3  60
 ni  E  ni  
  
E  ni 
all cells
2
2
63  60


2
60
45  60


2
60
72  60


2
60
 6.3
2

•
•
•
•
•
Test for k Proportions
Solution
H0: p1 = p2 = p3 = 1/3
Test Statistic:
2 = 6.3
Ha: At least 1 is different
 = .05
n1 = 63 n2 = 45 n3 = 72
Critical Value(s):
Decision:
Reject H0
Reject at = .05
= .05
0
5.991
c2
Conclusion:
There is evidence of a
difference in proportions
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.
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
Contingency Table Example
(continued)
Sample results organized in a contingency table:
Hand Preference
sample size = n = 300:
120 Females, 12 were
left handed
180 Males, 24 were
left handed
Gender
Left
Right
Female
12
108
120
Male
24
156
180
36
264
300
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
The Chi-Square Test Statistic
The Chi-square test statistic is:
2
χ STAT


all cells
( fo  fe )2
fe
• where:
fo = observed frequency in a particular cell
fe = expected frequency in a particular cell if H0 is
true
2
χ STAT
for the 2 x 2 case has 1 degree of freedom
(Assumed: each cell in the contingency table has
expected frequency of at least 5)
Decision Rule
2
The χ STAT
test statistic approximately follows a chisquared distribution with one degree of freedom
Decision Rule:
2
2
χ

χ
If STAT
α , reject H0,
otherwise, do not reject
H0

0
Do not
reject H0
Reject H0
2α
2
Computing the
Average Proportion
The average p  X1  X2  X
n1  n2
n
proportion is:
120 Females, 12 were
left handed
180 Males, 24 were
left handed
Here:
12  24
36
p

 0.12
120  180 300
i.e., of all the children the proportion of left handers is 0.12, that is,
12%
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
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
The Chi-Square Test Statistic
Hand Preference
Left
Right
Observed = 12
Observed = 108
Expected = 14.4 Expected = 105.6
Observed = 24
Observed = 156
Expected = 21.6 Expected = 158.4
Gender
Female
Male
36
The test statistic is:
χ 2STAT 

all cells
264
120
180
300
(f o  f e ) 2
fe
(12  14.4)2 (108  105.6)2 (24  21.6)2 (156  158.4)2




 0.7576
14.4
105.6
21.6
158.4
Decision Rule
2
T he test statisticis χ STAT
 0.7576; χ 02.05 with1 d.f.  3.841
Decision Rule:
2
If χ STAT > 3.841, reject H0,
otherwise, do not reject H0
0.05
0
Do not
reject H0
Reject H0
20.05 = 3.841
2
Here,
2
2
χ STAT
= 0.7576< χ 0.05 = 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
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)
The Chi-Square Test Statistic
The Chi-square test statistic is:
2
χ STAT


all cells
( fo  fe )2
fe
• 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
χ 2STAT for the 2 x c case has (2 - 1)(c - 1)  c - 1 degrees of freedom
(Assumed: each cell in the contingency table has expected
frequency of at least 1)
Computing the
Overall Proportion
X1  X2    Xc X
The overall
p

n1  n2    nc
n
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:
Decision Rule:
2
 χ α2 , reject H0,
If χ STAT
otherwise, do not reject H0
2
χ
Where α is from the chi-
squared distribution with
c – 1 degrees of freedom
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
2 Test of Independence
Hypothesis Tests
Qualitative Data
Qualitative
Data
1 pop.
Proportion
More than
2 pop.
Independence
2 pop.
Z Test
Z Test
c2 Test
c2 Test
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
2 Test of Independence
Contingency Table
Shows number of observations from 1 sample
jointly in 2 qualitative variables
Levels of variable 2
House Style
Split-Level
Ranch
Total
House Location
Urban
Rural
63
49
15
33
78
82
Levels of variable 1
Total
112
48
160
Conditions Required for a
Valid 2 Test: Independence
1. Multinomial experiment has been conducted
2. The sample size, n, is large: Eij is greater than
or equal to 5 for every cell
2 Test of Independence
Hypotheses & Statistic
1. Hypotheses
• H0: Variables are independent
• Ha: Variables are related (dependent)
2. Test Statistic
Observed count
 nij  Eij 
  
Eij
all cells
2
2
Expected
count
3. Degrees of Freedom: (r – 1)(c – 1)
Rows Columns
2

Test of Independence
Expected Counts
1. Statistical independence means joint
probability equals product of marginal
probabilities
2. Compute marginal probabilities and multiply
for joint probability
3. Expected count is sample size times joint
probability
Expected Count Example
Marginal probability = 112
160
House Style
Location
Urban
Rural
Obs.
Obs.
Total
Split–Level
63
49
112
Ranch
15
33
48
Total
78
82
160
Expected Count Example
Marginal probability = 112
160
House Style
Location
Urban
Rural
Obs.
Obs.
Total
Split–Level
63
49
112
Ranch
15
33
48
Total
78
82
160
Marginal probability =
78
160
Expected Count Example
Joint probability =
House Style
112 78
160 160
Marginal probability = 112
160
Location
Urban
Rural
Obs.
Obs.
Total
Split–Level
63
49
112
Ranch
15
33
48
Total
78
82
160
Marginal probability =
78
160
112 78
Expected count = 160· 160 160
= 54.6
Expected Count Calculation
Eij =
112·78
160
House Style
R iC j
n
House Location
Urban
Rural
Obs. Exp. Obs. Exp.
112·82
160
Total
Split-Level
63
54.6
49
57.4
112
Ranch
15
23.4
33
24.6
48
Total
78
78
82
82
48·78
160
160
48·82
160
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?
House Style
Split-Level
Ranch
Total
House Location
Urban
Rural
63
49
15
33
78
82
Total
112
48
160
2 Test of Independence
Solution
•
•
•
•
•
H0: No Relationship
Ha: Relationship
 = .05
df = (2 - 1)(2 - 1) = 1
Critical Value(s):
Test Statistic:
Decision:
Reject H0
= .05
0
3.841
c2
Conclusion:
2 Test of Independence
Solution

Eij
5 in all cells
112·78
160
House Style
House Location
Urban
Rural
Obs. Exp. Obs. Exp.
112·82
160
Total
Split-Level
63
54.6
49
57.4
112
Ranch
15
23.4
33
24.6
48
Total
78
78
82
82
48·78
160
160
48·82
160
2 Test of Independence
Solution
 nij  Eij 
  
Eij
all cells
2
2
n11  E11 


n12  E12 


2
E11
E12
63  54.6


2
54.6
n22  E22 


2
2

49  57.4


E22
2
57.4
33  24.6


2

24.6
 8.41
2 Test of Independence
Solution
•
•
•
•
•
H0: No Relationship
Ha: Relationship
 = .05
df = (2 - 1)(2 - 1) = 1
Critical Value(s):
Reject H0
= .05
0
3.841
c2
Test Statistic:
2
= 8.41
Decision:
Reject at
= .05
Conclusion:
There is evidence of a
relationship
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 Coke
No
Yes
Total
Diet Pepsi
No
Yes
84
32
48
122
132
154
Total
116
170
286
2 Test of Independence
Solution*
•
•
•
•
•
H0: No Relationship
Ha: Relationship
 = .05
df = (2 - 1)(2 - 1) = 1
Critical Value(s):
Test Statistic:
Decision:
Reject H0
= .05
0
3.841
c2
Conclusion:
2 Test of Independence
Solution*

Eij
5 in all cells
116·132
286
Diet Coke
Diet Pepsi
154·132
286
No
Yes
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
170·132
286
170·154
286
2 Test of Independence
Solution*
 nij  Eij 
  
Eij
all cells
2
2
n11  E11 


n12  E12 


2
E11
E12
84  53.5


2
53.5
n22  E22 


2
2

32  62.5


E22
2
62.5
122  91.5


2

91.5
 54.29
2 Test of Independence
Solution*
•
•
•
•
•
H0: No Relationship
Ha: Relationship
 = .05
df = (2 - 1)(2 - 1) = 1
Critical Value(s):
Reject H0
= .05
0
3.841
c2
Test Statistic:
2
= 54.29
Decision:
Reject at
= .05
Conclusion:
There is evidence of a
relationship
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 Coke
No
Yes
Total
Diet Pepsi
No
Yes
84
32
48
122
132
154
Total
116
170
286
You Re-Analyze the Data
High Income
Diet Coke
No
Yes
Total
Diet Pepsi
No
Yes
4
30
40
2
44
32
Total
34
42
76
Diet Pepsi
No
Yes
80
2
8
120
88
122
Total
82
128
210
Low Income
Diet Coke
No
Yes
Total
True Relationships*
Diet
Coke
Underlying causal
relation
Control or intervening
variable (true cause)
Apparent
relation
Diet
Pepsi
Moral of the Story*
Numbers don’t
think - People do!
© 1984-1994 T/Maker Co.
Conclusion
1. Explained 2 Test for Proportions
2. Explained 2 Test of Independence
3. Solved Hypothesis Testing Problems
•
•
More Than Two Population Proportions
Independence