Chapter 11
Comparisons Involving Proportions
and A Test of Independence
1
Chapter Outline
Goodness of Fit test
Test of Independence
2
Goodness of Fit Test
 Hypothesis test for proportions of a multinomial
population
 It compares the observed and expected frequencies
in each category to test either that all categories
contain the same proportion of values or that each
category contains a user-specified proportion of
values.
3
Goodness of Fit Test
 Procedure
1. Set up the null and alternative hypotheses;
2. Select a random sample and record the
observed frequency, fi, for each of the k
categories;
3. Assuming H0 is true, compute the expected
frequency, ei, in each category by multiplying
the category probability by the sample size.
4
Goodness of Fit Test
 Procedure
4. Compute the value of the test statistic:
2
(
f

e
)
2   i i
ei
i 1
k
where:
fi = observed frequency for category i
ei = expected frequency for category i
k = number of categories
Note: The test statistic has a chi-square distribution with k – 1
degrees of freedom provided that the expected frequencies are 5
or more for all categories.
5
Goodness of Fit Test
 Procedure
5. Rejection rule:
p-value approach: Reject H0 if p-value < 
2
2
Reject
H
if



Critical value approach:
0

where  is the significance level and
there are k - 1 degrees of freedom
6
Goodness of Fit Test

Example: Market Share
In the Scott Market Research example in the textbook,
after company C introduced a new product to the market, a
survey was conducted on 200 customers to study if there is
any change in the market shares. Out of the 200 customers,
48 prefer company A’s product, 98 prefer company B’s,
and 54 prefer company’s C’s. Before the introduction of
the new product by company C, the market shares of the
three companies were:
p A  0.3, pB  0.5, pC  0.2
7
Goodness of Fit Test
 Example: Market Share

Hypotheses
H0: pA = .3; pB = .5; pC = .2
Ha: The population proportions are not
pA = .3; pB = .5; pC = .2
where:
pA = the population market share of company A;
pB = the population market share of company B;
pC = the population market share of company C.
8
Goodness of Fit Test
 Example: Market Share

Rejection rule
Reject H0 if p-value < .05 or 2 > 5.99.
With  = .05 and
k-1=3-1=2
degrees of freedom
Do Not Reject H0
Reject H0
5.99
2
9
Goodness of Fit Test

Example: Market Share
 Expected Frequencies
eA=.3(200)=60; eB=.5(200)=100; eC=.2(200)=40

Observed Frequencies
fA = 48; fB = 98; fC = 54

Test Statistic
2
2
2






48

60
98

100
54

40
2 


60
=
2.4+0.04+4.9
= 7.34
100
40
10
Goodness of Fit Test
 Example: Market Share
 p-Value Approach
Area in Upper Tail
2 Value (df = 2)
.10
4.6
.05
5.99
.025
7.38
.01
9.21
.005
10.60
Because 2 = 7.34 is between 5.99 and 7.38, the area in the
upper tail of the distribution is between .05 and .025.
The p-value <  = .05. We can reject the null hypothesis.
11
Goodness of Fit Test
 Example: Market Share
 Critical Value Approach
2 = 7.34 > 5.99
We reject, at the .05 level of significance,
the assumption that the market shares of companies
A, B, and C remain the same after Company C
introduced a new product.
12
Test of Independence: Contingency Tables
 Hypothesis test for independence between two
variables.
 Similar to a Goodness of Fit test, it computes 2
test statistic based on the observed frequencies and
expected frequencies (assuming the null
hypothesis is true, i.e. the two variables are
independent from each other).
13
Test of Independence
 Procedure
1. Set up the null and alternative hypotheses;
2. Select a random sample and record the
observed frequency, fij, for each cell of the
contingency table;
3. Assuming H0 is true, compute the expected
frequency, eij, for each cell as follows:
(Row i Total)(Column j Total)
eij 
Sample Size
14
Test of Independence
 Procedure
4. Compute the test statistic
2   
i
j
( f ij  eij ) 2
eij
5. Determine the rejection rule
2
2
Reject H0 if p -value <  or   . 
where  is the significance level and,
with n rows and m columns, there are
(n - 1)(m - 1) degrees of freedom.
15
Test of Independence
 Example: Field of Study
The following table shows the results of recent study
regarding gender of individuals and their selected field of
study.
Field of
Study
Medicine
Business
Engineering
Total
Male
80
60
160
300
Female
40
20
40
100
Total
120
80
200
400
16
Test of Independence
 Example: Field of Study

Hypothesis
H0: Field of study is independent of gender
Ha: Field of study is not independent of gender
17
Test of Independence
 Example: Field of Study

Expected Frequencies
Field of
Study
Medicine
Business
Engineering
Total
Male
300*120/400 300*80/400
=90
=60
300*200/400
=150
300
Female
100*120/400 100*80/400
=30
=20
100*200/400
=50
100
200
400
Total
120
80
18
Test of Independence
 Example: Field of Study

Rejection Rule
Reject H0 if p-value < =.05
or
2 > 5.99 [d.f. = (2-1)(3-1)=2].

Test Statistic
2
2




80

90
60

60
2 

90
60
2

40  50

50
= 1.11 + 0 + . . . + 2 = 7.11
19
Test of Independence
 Example: Field of Study
 p-Value Approach
Area in Upper Tail
2 Value (df = 2)
.10
4.6
.05
5.99
.025
7.38
.01
9.21
.005
10.60
Because 2 = 7.11 is between 5.99 and 7.38, the area in the
upper tail of the distribution is between .05 and .025.
The p-value <  = .05. We can reject the null hypothesis.
20
Test of Independence
 Example: Field of Study
 Critical Value Approach
2 = 7.11 > 5.99
We reject, at the .05 level of significance,
the null hypothesis that the selected field of study
is independent of gender.
21