1. Chapter 16
Chi Squared Tests

2. 16.1 Introduction
Two statistical techniques are presented, to analyze nominal data.
A goodness-of-fit test for the multinomial experiment.
A contingency table test of independence.
Both tests use the c2 as the sampling distribution of the test statistic.

3. 16.2 Chi-Squared Goodness-of-Fit Test
The hypothesis tested involves the 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 that the outcome fall into cell i remains constant for each trial. Moreover, p1 + p2 + … +pk = 1.
Trials of the experiment are independent

4. 16.2 Chi-squared Goodness-of-Fit Test
We test whether there is sufficient evidence to reject a pre-specified set of values for pi.
The hypothesis:
The test builds on comparing actual frequency and the expected frequency of occurrences in all the cells.

5. The multinomial goodness of fit test - Example
Two competing companies A and B have enjoy dominant position in the market. The companies conducted aggressive advertising campaigns.
Market shares before the campaigns were:
Company A = 45%
Company B = 40%
Other competitors = 15%.

6. The multinomial goodness of fit test - Example
Example 16.1 – continued
To study the effect of the campaign on the market shares, a survey was conducted.
200 customers were asked to indicate their preference regarding the product 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.

7. The multinomial goodness of fit test - Example
Example 16.1 – continued Can we conclude at 5% significance level that the market shares were affected by the advertising campaigns?

8. The multinomial goodness of fit test - Example
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 before the campaign?

9. The multinomial goodness of fit test - Example
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:
What actual frequencies
did the sample return?
90 = 200(.45)
80 = 200(.40)
102 82
30 = 200(.15) 16

10. The multinomial goodness of fit test - Example
The statistic is
The rejection region is

11. The multinomial goodness of fit test - Example
Example 16.1 – continued

12. The multinomial goodness of fit test - Example
Example 16.1 – continued
c2 with 2 degrees of freedom
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.
Alpha
P value
5.99
8.18
Rejection region

13. 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 the cells (npi ³ 5).
If the expected frequency in a cell is less than 5, combine it with other cells.

14. 16.3 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.

15. Contingency table c2 test – Example
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.

16. Contingency table c2 test – Example
The observed values
There are two ways to address the problem
If each undergraduate degree
is considered a population, do
these populations differ?
If each classification is considered
a nominal variable, are these two
variables dependent?

17. Contingency table c2 test – Example
Solution
The hypotheses are:
H0: The two variables are independent
H1: The two variables are dependent
Since ei = npi but pi is unknown, we need to estimate the unknown probability from the data, assuming H0 is true.
k is the number of cells in
the contingency table.
The test statistic
The rejection region

18. Estimating the expected frequencies
Undergraduate
MBA Major
Degree
Accounting
Finance
Marketing
Probability 60 BA 60
60/152
BENG 31
31/152
BBA 39 39
39/152
Other 22
22/152 61 44
152
152 61 44 47
152
Probability
61/152
44/152
47/152
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

19. The expected frequencies for a contingency table
The expected frequency of cell of raw i and column j in the contingency table is calculated by
eij =
(Column j total)(Row i total)
Sample size

20. Calculation of the c2 statistic
Solution – continued
Undergraduate
MBA Major
Degree
Accounting
Finance
Marketing BA
31 (24.08)
13 (17.37)
16 (18.55) 60
BENG
8 (12.44)
(8.97)
7 (9.58) 31
BBA
12 (15.65)
10 (11.29)
17 (12.06) 39
Other
(8.83)
5 (6.39)
7 (6.80) 22 61 44 47
152
The expected frequency
c2=
( )2
24.08
( )2
6.39
( )2
6.80
= 14.70
+….+
+….+

21. Contingency table c2 test – Example
Solution – continued
The critical value in our example is:
Conclusion:
Since c2 = > , there is sufficient evidence to infer at 5% significance level that students’ undergraduate degree and MBA students courses selection are dependent.

22. Using the computer
Define a code to specify each nominal value. Input the data in columns one column for each category.
Select the Chi squared / raw data
Option from Data Analysis Plus
under tools. See Xm16-02
Code:
Undergraduate degree 1 = BA
2 = BENG
3 = BBA
4 = OTHERS
MBA Major
1 = ACCOUNTING
2 = FINANCE
3 = MARKETING

23. Required condition Rule of five
The c2 distribution provides an adequate approximation to the sampling distribution under the condition that eij >= 5 for all the cells.
When eij < 5 rows or columns must be added such that the condition is met.
Example
18 (17.9)
23 (22.3)
12 (12.8)
4 (5.1)
7 (6.3)
4 (3.6)
We combine
column 2 and 3

24. 16.5 Chi-Squared test for Normality
The goodness of fit Chi-squared test can be used to determined if data were drawn from any distribution.
The general procedure:
Hypothesize on the parameter values of the distribution we test (i.e. m = m0, s = s0 for the normal distribution).
For the variable tested X specify disjoint ranges that cover all its possible values.
Build a Chi squared statistic that (aggregately) compares the expected frequency under H0 and the actual frequency of observations that fall in each range.
Run a goodness of fit test based on the multinomial experiment.

25. 15.5 Chi-Squared test for Normality
Testing for normality in Example 12.1
For a sample size of n=50 (see Xm12-01) ,the sample mean was with standard error of Can we infer from the data provided that this sample was drawn from a normal distribution with m = and s = 38.83? Use 5% significance level.

26. c2 test for normality Solution
First let us select z values that define each cell (expected frequency > 5 for each cell.)
z1 = -1; P(z < -1) = p1 = .1587; e1 = np1 = 50(.1587) = 7.94
z2 = 0; P(-1 < z< 0) = p2 = .3413; e2 = np2 = 50(.3413) = 17.07
z3 = 1; P(0 < z < 1) = p3 = .3413; e3 = 17.07
P(z > 1) = p4 = .1587; e4 = 7.94
The cell boundaries are calculated from the corresponding z values under H0.
The expected frequencies
can now be
determined for
each cell.
e2 = 17.07
e3 = 17.07
z1 =(x )/38.83 = -1;
x1 =
.1587
.3413
421.55
e1 = 7.94
e4 = 7.94
460.38
499.21

27. c2 test for normality The test statistic c2= = 1.72 + + + (10 - 7.94)2
( )2
7.94
( )2
17.07
( )2
17.07
( )2
7.94
= 1.72 + + +
f3 = 19
e2 = 17.07
e3 = 17.07
f2 = 13
f1 = 10
f4 = 8
e1 = 7.94
e4 = 7.94

28. c2 test for normality The test statistic c2= The rejection region
( )2
7.94
( )2
17.07
( )2
17.07
( )2
7.94
= 1.72 + + +
The rejection region
Conclusion: There is insufficient evidence to conclude at 5% significance level that the data are not normally distributed.