Statistical techniques in business ecohomics chap015

TWO Conduct a test of hypothesis comparing an observed set of frequencies to an expected set of frequencies.. Goodness-of-Fit Test: Equal Expected FrequenciesLet f0 and fe be the obse

Chapter Fifteen

Nonparametric Methods: Chi-Square Applications

GOALS

When you have completed this chapter, you

will be able to:

ONE

List the characteristics of the Chi-square distribution.

TWO

Conduct a test of hypothesis comparing an observed set of

frequencies to an expected set of frequencies.

THREE

Conduct a hypothesis test to determine whether two

classification criteria are related

Goals

Characteristics of the Chi-Square

Goodness-of-Fit Test: Equal Expected Frequencies

Let f0 and fe be the observed and expected

frequencies respectively

H0 : There is no difference between the observed and expected frequencies

H1 : There is a difference

between the observed and

the expected frequencies.

The test statistic is:

The critical value is a chi-square value with (k-1)

degrees of freedom, where k is the number of categories

H0 : There is no difference between the

observed and expected frequencies

H1 : There is a difference between the

observed and the expected frequencies

This is given in the problem as 01

It is the chi-square distribution.

EXAMPLE 1 continued

Assume equal expected frequency as given in the

problem

f e = (120+45+60+90+130)/5=89

The degrees of freedom: (5-1)=4

The critical value of 2 is 13.28 Reject the null and

accept the alternate if

Example 1 continued

We conclude that there is a difference in the number of workers absent

by day of the week.

Because the computed value of chi-square, 60.90, is greater than the critical value, 13.28, the p of 000000000001877

< 01, H0 is rejected

Example 2

The U.S Bureau of the

Census indicated that 63.9%

of the population is married,

7.7% widowed, 6.9%

divorced (and not

re-married), and 21.5% single

(never been married) A

sample of 500 adults from

the Philadelphia area showed

that 310 were married, 40

widowed, 30 divorced, and

120 single At the 02

significance level can we

conclude that the

Philadelphia area is different

from the U.S as a whole?

Frequencies

Example 2 continued

Step 4: H0 is rejected if 2 >9.837, df=3, or if

p of 02

Step 1: H0: The distribution has not changed

H1: The distribution has changed

Step 2: The significance level given is 02

Step 3: The test statistic is the chi-square

f 0 f e (f f0  e) /2 f e

Step 5: 2 = 2.3814, p(2 > 2.3814) = 497.

Example 2 continued

The null hypothesis is not rejected The

distribution regarding marital status in

Philadelphia is not different from the rest of the

United States

characteristics are related.

Chi-square can be

used to test for a

relationship between

two nominal scaled

variables, where one

Each observation is classified according to two

criteria.

We use the usual hypothesis testing procedure.

(number of rows-1)(number of columns-1).

Expected Frequency = (row total)(column total)

grand total

Contingency Table Analysis

Contingency table analysis

Is there a relationship between

the location of an accident and

the gender of the person

involved in the accident? A

sample of 150 accidents

reported to the police were

classified by type and gender

At the 05 level of significance,

can we conclude that gender

and the location of the accident

Contingency Table Analysis

Step 4: The degrees of freedom equal (r-1)(c-1) or 2

The critical 2 at 2 d.f is 9.21 If computed 2 >9.21,

or if p < 01, reject the null and accept the alternate

Step 5: A data table and the following contingency

table are constructed

Step 1: H0: Gender and location are not related

H 1: Gender and location are related

Step 2: The level of significance is set at 01

Step 3: the test statistic is the chi-square distribution

Example 3 continued

The expected frequency for the work-male

intersection is computed as (90)(80)/150=48.

Similarly, you can compute the expected

frequencies for the other cells.

(20)(90)150

=20

(20)(60)150

=8

60

Example 3 continued

The p(2 > 16.667) = 00024.

reject the null and conclude that there is a

relationship between the location of an accident and the gender of the person involved.

Example 3 concluded