1. Chapter Fifteen McGraw-Hill/Irwin
© 2005 The McGraw-Hill Companies, Inc., All Rights Reserved.

2. 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

3. Characteristics of the Chi-Square Distribution
Chi-Square Applications
The major characteristics of the chi-square distribution are:
It is positively skewed
It is non-negative
There is a family of chi-square distributions
Characteristics of the Chi-Square Distribution

4. df = 3
df = 5
df = 10 c2
c2 distribution

5. Goodness-of-Fit Test: Equal Expected Frequencies
Let f0 and fe be the observed and expected frequencies respectively.
H1: There is a difference between the observed and the expected frequencies.
H0: There is no difference between the observed and 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
Goodness-of-Fit Test: Equal Expected Frequencies

6. The following information shows the number of employees absent by day of the week at a large a manufacturing plant. At the .01 level of significance, is there a difference in the absence rate by day of the week?
Day of Week
Number Absent
Monday
120
Tuesday 45
Wednesday 60
Thursday 90
Friday
130
Total
445
Example 1 continued

7. Step 1: State the null and alternate hypotheses
H0: There is no difference between the observed and expected frequencies.
H1: There is a difference between the observed and the expected frequencies
Step 2: Select the level of significance.
This is given in the problem as .01.
Step 3: Select the test statistic.
It is the chi-square distribution.
Example 1 continued

8. Step 4: Formulate the decision rule.
Assume equal expected frequency as given in the problem
fe = ( )/5=89
The degrees of freedom: (5-1)=4
The critical value of c2 is Reject the null and accept the alternate if
Computed c2 >
or p< .01
EXAMPLE 1 continued

9. Step Five: Compute the value of chi-square and make a decision.
Day Frequency Expected (fo – fe)2/fe
Monday
Tuesday
Wednesday
Thursday
Friday
Total
The p(c2 > 60.9) = or essentially 0.
Example 1 continued

10. Because the computed value of chi-square, 60
Because the computed value of chi-square, 60.90, is greater than the critical value, 13.28, the p of < .01, H0 is rejected.
We conclude that there is a difference in the number of workers absent by day of the week.
Example 1 continued

11. Goodness-of-fit Test: Unequal Expected Frequencies
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?
Example 2

12. Step 4: H0 is rejected if c2 >9.837, df=3, or if p < a of .02
Step 3: The test statistic is the chi-square.
Step 2: The significance level given is .02.
Step 1: H0: The distribution has not changed
H1: The distribution has changed.
Example 2 continued

13. Calculate the expected frequencies Calculate chi-square values.
Married: (.639)500 = 319.5
Widowed: (.077)500 = 38.5
Divorced: (.069)500 =
Single: (.215)500 = 107.5
Calculate the expected frequencies
Calculate chi-square values.
Example 2 continued

14. Step 5: c2 = , p(c2 > ) = .497.
The null hypothesis is not rejected. The distribution regarding marital status in Philadelphia is not different from the rest of the United States.
Example 2 continued

15. Contingency Table Analysis
Chi-square can be used to test for a relationship between two nominal scaled variables, where one variable is independent of the other.
A contingency table is used to investigate whether two traits or characteristics are related.
Contingency Table Analysis

16. Contingency table analysis
Each observation is classified according to two criteria.
We use the usual hypothesis testing procedure.
The degrees of freedom are equal to:
(number of rows-1)(number of columns-1).
The expected frequency is computed as:
Expected Frequency = (row total)(column total)
grand total
Contingency table analysis

17. 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 are related?
Example 3

18. Step 1: H0: Gender and location are not related.
H1: 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.
Step 4: The degrees of freedom equal (r-1)(c-1) or 2. The critical c2 at 2 d.f. is If computed c2 >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.

19. Observed frequencies (fo )
Gender
Work
Home
Other
Total
Male 60 20 10 90
Female 30 80 50
150
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.
Example 3 continued

20. Expected frequencies (fe )
Gender
Work
Home
Other
Total
Male
(80)(90)150
= 48
(50)(90)
150
=30
(20)(90)
=12 90
Female
(80)(60)
=32
(50)(60)
=20
(20)(60) =8 60 80 50 20
Example 3 continued

21. c2: (fo – fe)2/ fe Gender Work Home Other Total c2 Male (60-48)2 48
(20-30)2 30
(10-12)2 12
6.667
Female
(20-32)2 32
(30-20)2 20
(12-10)2 10
10.000
Total
16.667
Example 3 continued

22. The p(c2 > ) =
Since the c2 of > 9.21, p of < .01, 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