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Chapter 16

# Chapter 16 Textbook Study Guide

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York University

Operations Management and Information System

OMIS 2010

Alan Marshall

Fall

Description

Chapter 16: Chi-Squared Tests
16.1 Introduction
This chapter introduced two popular statistical procedures that use the chi-squared distribution to
conduct tests on qualitative data. At the completion of this chapter, you are expected to know the
following:
l. How to conduct a goodness-of-fit test on data that have been generated by a multinomial experi-
ment.
2. How to conduct a test on data arranged in a contingency table to determine if two classifications
of qualitative data are statistically independent.
16.2 Chi-Squared Goodness-of-Fit Test
This section introduced an extension of the binomial experiment called the multinomial experiment,
in which the outcome of each trial of the experiment can be classified as belonging to one of k
categories, called cells. The goodness-of-fit test described in this section allows us to test hypotheses
concerning the values of the probabilities p (ii= 1, 2, . . . , k), which represent the proportions of
observations that belong to each of the k cells. The null and alternative hypotheses might, for example,
be
H : p = .2, p = .3, p = .5
0 1 2 3
H 1 At least one p is not equal to its specified value.
After a multinomial experiment has been conducted, the test involves a comparison of the observed
frequencies (o i of outcomes that fall into each of the k cells with the frequencies (e i that are expected
for each cell. For a multinomial experiment consisting of n trials, the expected frequencies are calculated
from the hypothesized probabilities p i
ei= np i for i = 1, 2, . . . , k
The test statistic used to compare the observed and expected frequencies is
k 2
2 ( fi ei)
,
i1 ei
which has an approximate chi-squared distribution with k 1 degrees of freedom. Since only relatively
large values of the differences (f e ), and thus of 2 , would cause us to suspect that the values speci-
i i
fied in the null hypothesis are incorrect, the rejection region is always located in the right tail of the chi-
squared distribution.
197 The rule of five requires that t he expected frequency (e ) fir each cell be at least 5. Where
necessary, cells should be combined so that this condition is satisfied.
Example 16.1
The records of an investment banking firm show that, historically, 60% of its clients were primarily
interested in the stock market, 36% in the bonds market, and 4% in the futures market. A recent sample
of 200 clients showed that 132 were primarily interested in stocks, 52 in bonds, and 16 in futures. Is
there sufficient evidence to conclude, at the 1% level of significance, that there has been a shift in the
primary interest of clients?
Solution
The problem objective is the description of a single population (the population of primary interests
of clients), and the data are qualitative, consisting of three categories. The parameters of interest are the
following:
p 1 proportion of clients primarily interested in stocks
p 2 proportion of clients primarily interested in bonds
p 3 proportion of clients primarily interested in futures
Since we want to determine whether these proportions have changed from their historic levels, the null
and alternative hypotheses are
H 0 p 1 .60, p 2 .36, p 3 .04
H :1At least one p is not equal to its specified value.
The test statistic is
3 ( f e )2
2 i i , with d.f. = (k 1) = (3 1) = 2
i e
i
Since the rejection region for this goodness-of-fit test is always in the right tail of the chi-squared
distribution, the rejection region is
2 > 2 2 9.21034
, k 1 .01, 2
In order to evaluate the test statistic, we need to find the expected frequencies. Assuming that the null
hypothesis is true, the expected numbers of clients interested in stocks, bonds, and futures (respectively)
are
e 1 = np1= (200)(.60) = 120
198 e 2= np 2 (200)(.36) = 72
e 3= np 3 (200)(.04) = 8
Finally, the value of the test statistic is
2 2 2
2 = (132 120) (52 72) (16 8) 14.76
120 72 8
Since 14.76 > 9.21034, the value of the test statistic falls into the rejection region. As a result, we reject
H 0 and conclude that there is sufficient evidence to state that there has been a shift in the primary
interest of clients.
When solving this type of problem on your own, you may find it convenient to use a table such as
the one below to compute the value of the test statistic.
Observed Expected
2 ( fi ei)2
Market Frequency f i Frequency e i ( fie )i
ei
Stock 132 120 144 1.20
Bond 52 72 400 5.56
Futures 16 8 64 8.00
Total 200 200 14.76 = 2
Example 16.2
Refer to Example 16.1. What would you conclude if the sample had consisted of only 100 clients, of
whom 66 were primarily interested in stocks, 26 in bonds, and 8 in futures? Use = .10.
Solution
At first sight, it would appear that the hypotheses, the test statistic and the rejection region, remain
the same as in Example 16.1, with only the value of the test statistic and the conclusion requiring a
change. Notice, however, that the new expected frequencies are
e 1= (100)(.60) = 60
e 2= (100)(.36) = 36
e = (100)(.04) = 4
3
199 Since e3< 5, the rule of five requires that we combine the third cell with one of the other two. Lets
choose to combine bonds with futures, leaving us with k = 2 cells. The 2 is redefined to be
p 2 proportion of clients primarily interested in bonds or futures
Thus, p = .36 + .04 = .40. The test now becomes
2
H 0 p 1 .60, p 2 .40
H 1 At least one p is not equal to its specified value.
2 (f e ) 2
Test statistic: 2 i i , with d.f. = (k 1) = (2 1) = 1
e
i i
Rejection region: 2 > ,k 1 .10,1 2.70554
Value of the test statistic:
(66 60)2 (34 40) 2
2 = 1.5
60 40
Conclusion: Do not reject H 0. There is insufficient evidence to conclude that there has been a shift
in the primary interest of clients.
Notice that we are now unable to detect a shift in the clients interests, even though were using a higher
level of significance and the observed frequencies are proportionately the same as in Example 16.1. The
reason is that it is more difficult to reject a null hypothesis with a smaller sample, all other things being
equal.
200

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