Chi Squared Tests
Introduction
• Two statistical techniques are presented. Both
are used to analyze nominal data.
– A goodness-of-fit test for a multinomial experiment.
– A contingency table test of independence.
• The test statistics in both cases follow the c2
distribution.
Chi-Squared Goodness-of-Fit Test
• The hypothesis tested involves the “success” probabilities
p1, p2, …, pk.of a multinomial distribution.
• The multinomial experiment is an extension of the binomial
experiment.
– There are n independent trials.
– The outcome of each trial can be classified into one of k
categories, called cells.
– The probability pi for an outcome to fall into cell i remains
constant for each trial. By assumption,
p1 + p2 + … +pk = 1.
– Trials in the experiment are independent.
• Our objective is to find out whether there is sufficient
evidence to reject a pre-specified set of values for pi .
• The hypotheses:
H 0 : p1  a1, p2  a2 , ..., pk  ak
H1 : At least one pi  ai
• The test builds on comparing actual frequency
and the expected frequency of occurrences in all
 cells.
An Example
• Example 16.1
– Two competing companies A and B have been
dominant players in the market. Both companies
conducted recent advertising campaigns on their
products.
– Market shares before the campaigns were:
• Company A = 45%
• Company B = 40%
• Other competitors = 15%.
• Example 16.1 – continued
– To study the effect of the campaigns on the market shares, a
survey was conducted.
– 200 customers were asked to indicate their preference
regarding the products advertised.
– Survey results:
• 102 customers preferred the company A’s product,
• 82 customers preferred the company B’s product,
• 16 customers preferred the competitors product.
• Example 16.1 – continued
Can we conclude at 5% significance level that
the market shares were affected by the
advertising campaigns?
• Solution
–
–
–
–
The population investigated is the brand preferences.
The data are nominal (A, B, or other)
This is a multinomial experiment (three categories).
The question of interest: Are p1, p2, and p3 different
after the campaign from their values prior to the
campaigns?
• The hypotheses are:
H0: p1 = .45, p2 = .40, p3 = .15
H1: At least one pi changed.
The expected frequency for each
category (cell) if the null hypothesis
is true is shown below:
90 = 200(.45)
80 = 200(.40)
What actual frequencies
did the sample return?
102
82
1
2
1
3
2
30 = 200(.15)
3
16
• The statistic is:
2
(
f

e
)
i
c2   i
ei
i1
k
where ei  npi
Intuitively, this measures the extent of differences
between the observed and the expected frequencies.
• The
rejection region is:
c  c ,k 1
2
2
• Example 16.1 – continued
k
c2 

i1
(102  90)2 (82  80)2 (16  30)2


 8.18
90
80
30
2
c2 ,k 1  c.05,31
 5.99147
The p  value  P( c 2  8.18)  .01679
[this come from Excel :  CHIDIST(8.18,2)]

• Example 16.1 – continued
2 with 2 degrees of freedom
c
0.025
0.02
0.015
0.01
Alpha
0.005
P value
0
0
2
4
5.996
88.18 10
Rejection region
Conclusion: Since 8.18 > 5.99, there is sufficient
evidence at 5% significance level to reject the null
hypothesis. At least one of the probabilities pi is
different. Thus, at least two market shares have
changed.
12
Required Conditions – The Rule of Five
• The test statistic used to perform the test is only
approximately Chi-squared distributed.
• For the approximation to apply, the expected cell
frequency has to be at least 5 for all cells
(npi  5).
• If the expected frequency in a cell is less than 5,
combine it with other cells.
Chi-squared Test of a Contingency Table
• This test is used to test whether…
– two nominal variables are related?
– there are differences between two or more
populations of a nominal variable?
• To accomplish the test objectives, we need to
classify the data according to two different
criteria.
• The idea is also based on goodness of fit.
• Example 16.2
– In an effort to better predict the demand for courses
offered by a certain MBA program, it was hypothesized
that students’ academic background affect their choice
of MBA major, thus, their courses selection.
– A random sample of last year’s MBA students was
selected. The following contingency table summarizes
relevant data.
Degree
BA
BENG
BBA
Other
Accounting
31
8
12
10
61
Finance
13
16
10
5
44
Marketing
16
7
17
7
47
60
31
60
39
152
The observed values
There are two ways to view this problem
If each classification is considered
a nominal variable, are these two
variables dependent?
If each undergraduate degree
is considered a population, do
these populations differ?
• Solution
–
Since ei = npi but pi is
unknown, we need to
The hypotheses are:
estimate the unknown
H0: The two variables are independent probability from the data,
H1: The two variables are dependent assuming H0 is true.
– The test statistic
k
c 
2

i1
( fi  e i )2
ei
k is the number of cells in
the contingency table.
– The rejection region
c2  c2,(r 1)( c 1)
Estimating the expected frequencies
Undergraduate
Degree
Accounting
BA
BENG
BBA
Other
6161
Probability
61/152
MBA Major
Finance Marketing
44
44
44/152
6060
31
3939
22
47
47/152
Probability
60/152
31/152
39/152
22/152
152
152
Under the null hypothesis the two variables are independent:
P(Accounting and BA) = P(Accounting)*P(BA) = [61/152][60/152].
The number of students expected to fall in the cell “Accounting - BA” is
eAcct-BA = n(pAcct-BA) = 152(61/152)(60/152) = [61*60]/152 = 24.08
The number of students expected to fall in the cell “Finance - BBA” is
eFinance-BBA = npFinance-BBA = 152(44/152)(39/152) = [44*39]/152 = 11.29
• The expected frequency of cell of row i and
column j in the contingency table is calculated
by:
(Column j total)(Row i total)
eij =
Sample size
Calculation of the c2 statistic
2
(
f

e
)
i
c2   i
ei
i1
k
• Solution – continued
Undergraduate
Degree
Accounting
31 (24.08)
24.08
BA
k
BENG 2 8 (12.44)
BBA 31 24.08
12 (15.65)
Other
10 (8.83)
i61
1
31 24.08
c 
31
24.08
31
c2=
24.08
MBA Major
Finance
Marketing
13 (17.37) 2 16 (18.55)
16
(8.97)
7 (9.58)
i
i
10 (11.29) 17 (12.06)
(6.39) 77 6.80
(6.80)
55 6.39
i
44
47
(f  e )
 e

5 6.39
The expected frequency
5 6.39
60
31
39
22
152
7 6.80
7 6.80
7 6.80
5 6.39
(31 - 24.08)2
(5 - 6.39)2
(7 - 6.80)2
=
+….+
+….+
24.08
6.39
6.80
14.70
• Solution – continued
– The critical value in our example is:
2
c2 ,(r1)(c1)  c.05,(4
1)(31)  12.5916
• Conclusion:

Since c2 = 14.70 > 12.5916, there
is sufficient evidence to infer at 5% significance
level that students’ undergraduate degree
and MBA students courses selection
are dependent.
Using the computer
Select the Chi squared / raw data
Option from Data Analysis Plus
under tools. See Xm16-02
Define a code to specify each nominal
value. Input the data in columns one
column for each category.
Code:
Undergraduate degree
1 = BA
2 = BENG
3 = BBA
4 = OTHERS
MBA Major
1 = ACCOUNTING
2 = FINANCE
3 = MARKETING
Degree MBA Major
3
1
1
1
1
1
1
1
2
2
1
3
.
.
.
.
Contingency Table
1
2
3 Total
1
31
13
16
60
2
8
16
7
31
3
12
10
17
39
4
10
5
7
22
Total 61
44
47 152
Test Statistic CHI-Squared = 14.7019
P-Value = 0.0227