Textbook Notes (280,000)

CA (170,000)

York (10,000)

OMIS (50)

OMIS 2010 (30)

Alan Marshall (20)

Chapter 16

# OMIS 2010 Chapter Notes - Chapter 16: Chi-Squared Distribution, Null Hypothesis, Test Statistic

by OC9945

School

York UniversityDepartment

Operations Management and Information SystemCourse Code

OMIS 2010Professor

Alan MarshallChapter

16This

**preview**shows pages 1-3. to view the full**25 pages of the document.**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 pi (i = 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

H0: p1 = .2, p2 = .3, p3 = .5

H1: At least one pi is not equal to its specified value.

After a multinomial experiment has been conducted, the test involves a comparison of the observed

frequencies (oi) of outcomes that fall into each of the k cells with the frequencies (ei) that are expected

for each cell. For a multinomial experiment consisting of n trials, the expected frequencies are calculated

from the hypothesized probabilities pi:

ei = npi, for i = 1, 2, . . . , k

The test statistic used to compare the observed and expected frequencies is

!

"

#

"k

ii

ii

e

ef

1

2

2,

)(

$

which has an approximate chi-squared distribution with k – 1 degrees of freedom. Since only relatively

large values of the differences (fi – ei), and thus of

$

2, would cause us to suspect that the values speci-

fied in the null hypothesis are incorrect, the rejection region is always located in the right tail of the chi-

squared distribution.

197

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

The rule of five requires that the expected frequency (ei) for 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:

p1 = proportion of clients primarily interested in stocks

p2 = proportion of clients primarily interested in bonds

p3 = 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

H0: p1 = .60, p2 = .36, p3 = .04

H1: At least one pi is not equal to its specified value.

The test statistic is

,

)(

3

1

2

2!

"

#

"

ii

ii

e

ef

$

with d.f. = (k – 1) = (3 – 1) = 2

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%,k#1

2"$

.01, 2

2"9.21034

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

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

e

2 = np2 = (200)(.36) = 72

e

3 = np3 = (200)(.04) = 8

Finally, the value of the test statistic is

$ =

2(132 #120)2

120 &(52

#

72)2

72 &(1 6

#

8)2

8"14.76

Since 14.76 > 9.21034, the value of the test statistic falls into the rejection region. As a result, we reject

H0 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

Market Frequency fi Frequency ei

2

)( ef #ii

i

e

ii ef 2

)( #

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

3 = (100)(.04) = 4

199

###### You're Reading a Preview

Unlock to view full version