1. St. Edward’s University
Slides Prepared by
JOHN S. LOUCKS
St. Edward’s University
© South-Western/Thomson Learning

2. Chapter 12 Tests of Goodness of Fit and Independence
Goodness of Fit Test: A Multinomial Population
Tests of Independence: Contingency Tables
Goodness of Fit Test: Poisson and Normal Distributions

3. Goodness of Fit Test: A Multinomial Population
1. Set up the null and alternative hypotheses.
2. Select a random sample and record the observed
frequency, fi , for each of the k categories.
3. Assuming H0 is true, compute the expected frequency, ei , in each category by multiplying the category probability by the sample size.
continued

4. Goodness of Fit Test: A Multinomial Population
4. Compute the value of the test statistic.
5. Reject H0 if
(where  is the significance level and there are k - 1 degrees of freedom).

5. Example: Finger Lakes Homes (A)
Multinomial Distribution Goodness of Fit Test
Finger Lakes Homes manufactures four models of
prefabricated homes, a two-story colonial, a ranch, a
split-level, and an A-frame. To help in production
planning, management would like to determine if
previous customer purchases indicate that there is a
preference in the style selected.

6. Example: Finger Lakes Homes (A)
Multinomial Distribution Goodness of Fit Test
The number of homes sold of each model for 100
sales over the past two years is shown below.
Model Colonial Ranch Split-Level A-Frame
# Sold

7. Example: Finger Lakes Homes (A)
Multinomial Distribution Goodness of Fit Test
Notation
pC = popul. proportion that purchase a colonial
pR = popul. proportion that purchase a ranch
pS = popul. proportion that purchase a split-level
pA = popul. proportion that purchase an A-frame
Hypotheses
H0: pC = pR = pS = pA = .25
Ha: The population proportions are not
pC = .25, pR = .25, pS = .25, and pA = .25

8. Example: Finger Lakes Homes (A)
Multinomial Distribution Goodness of Fit Test
Expected Frequencies
e1 = .25(100) = e2 = .25(100) = 25
e3 = .25(100) = e4 = .25(100) = 25
Test Statistic =
= 10

9. Example: Finger Lakes Homes (A)
Multinomial Distribution Goodness of Fit Test
Rejection Rule
With  = .05 and
k - 1 = = 3
degrees of freedom
Do Not Reject H0
Reject H0 2
7.81

10. Example: Finger Lakes Homes (A)
Multinomial Distribution Goodness of Fit Test
Conclusion
c2 = 10 > 7.81, so we reject the assumption there is
no home style preference, at the .05 level of significance.

11. Test of Independence: Contingency Tables
1. Set up the null and alternative hypotheses.
2. Select a random sample and record the observed
frequency, fij , for each cell of the contingency table.
3. Compute the expected frequency, eij , for each cell.

12. Test of Independence: Contingency Tables
4. Compute the test statistic.
5. Reject H0 if (where  is the significance level and with n rows and m columns there are
(n - 1)(m - 1) degrees of freedom).

13. Example: Finger Lakes Homes (B)
Contingency Table (Independence) Test
Each home sold can be classified according to price and to style. Finger Lakes Homes’ manager would like to determine if the price of the home and the style of the home are independent variables.
The number of homes sold for each model and price for the past two years is shown below. For convenience, the price of the home is listed as either $65,000 or less or more than $65,000.
Price Colonial Ranch Split-Level A-Frame
< $65,
> $65,

14. Example: Finger Lakes Homes (B)
Contingency Table (Independence) Test
Hypotheses
H0: Price of the home is independent of the style of the home that is purchased
Ha: Price of the home is not independent of the
style of the home that is purchased
Expected Frequencies
Price Colonial Ranch Split-Level A-Frame Total
< $99K
> $99K
Total

15. Example: Finger Lakes Homes (B)
Contingency Table (Independence) Test
Test Statistic
= =
Rejection Rule
With  = .05 and (2 - 1)(4 - 1) = 3 d.f.,
Reject H0 if 2 > 7.81
Conclusion
We reject H0, the assumption that the price of the home is independent of the style of the home that is purchased.

16. Goodness of Fit Test: Poisson Distribution
1. Set up the null and alternative hypotheses.
2. Select a random sample and
a. Record the observed frequency, fi , for each of the
k values of the Poisson random variable.
b. Compute the mean number of occurrences, .
3. Compute the expected frequency of occurrences, ei , for each value of the Poisson random variable.
continued

17. Goodness of Fit Test: Poisson Distribution
4. Compute the value of the test statistic.
5. Reject H0 if
(where  is the significance level and there are k - 2 degrees of freedom).

18. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
In studying the need for an additional entrance to a city parking garage, a consultant has recommended an approach that is applicable only in situations where the number of cars entering during a specified time period follows a Poisson distribution.

19. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
A random sample of 100 one-minute time intervals resulted in the customer arrivals listed below. A statistical test must be conducted to see if the assumption of a Poisson distribution is reasonable.
# Arrivals
Frequency

20. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Hypotheses
H0: Number of cars entering the garage during
a one-minute interval is Poisson distributed.
Ha: Number of cars entering the garage during a one-minute interval is not Poisson distributed

21. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Estimate of Poisson Probability Function
otal Arrivals = 0(0) + 1(1) + 2(4) (1) = 600
Total Time Periods = 100
Estimate of  = 600/100 = 6
Hence,

22. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Expected Frequencies
x f (x ) xf (x ) x f (x ) xf (x )
Total

23. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Observed and Expected Frequencies
i fi ei fi - ei
0 or 1 or
10 or more

24. Example: Troy Parking Garage
Poisson Distribution Goodness of Fit Test
Test Statistic
Rejection Rule
With  = .05 and k - p - 1 = = 7 d.f. (where k = number of categories and p = number of population parameters estimated),
Reject H0 if 2 > 14.07
Conclusion
We cannot reject H0. There’s no reason to doubt the assumption of a Poisson distribution.

25. Goodness of Fit Test: Normal Distribution
1. Set up the null and alternative hypotheses.
2. Select a random sample and
a. Compute the mean and standard deviation.
b. Define intervals of values so that the expected
frequency is at least 5 for each interval.
c. For each interval record the observed frequencies
3. Compute the expected frequency, ei , for each interval.
continued

26. Goodness of Fit Test: Normal Distribution
4. Compute the value of the test statistic.
5. Reject H0 if
(where  is the significance level
and there are k - 3 degrees of freedom).

27. Example: Victor Computers
Normal Distribution Goodness of Fit Test
Victor Computers manufactures and sells a
general purpose microcomputer. As part of a study to evaluate sales personnel, management wants to determine if the annual sales volume (number of units sold by a salesperson) follows a normal probability distribution.

28. Example: Victor Computers
Normal Distribution Goodness of Fit Test
A simple random sample of 30 of the salespeople was taken and their numbers of units sold are below.
(mean = 71, standard deviation = 18.54)

29. Example: Victor Computers
Normal Distribution Goodness of Fit Test
Hypotheses
H0: The population of number of units sold has a normal distribution with mean and standard deviation
Ha: The population of number of units sold does not have a normal distribution with mean 71 and standard deviation

30. Example: Victor Computers
Normal Distribution Goodness of Fit Test
Interval Definition
To satisfy the requirement of an expected
frequency of at least 5 in each interval we
will divide the normal distribution into 30/5 = 6
equal probability intervals.

31. Example: Victor Computers
Normal Distribution Goodness of Fit Test
Interval Definition
Areas
= 1.00/6
= .1667
53.02 71
88.98 = (18.54)
63.03
78.97

32. Example: Victor Computers
Normal Distribution Goodness of Fit Test
Observed and Expected Frequencies
i fi ei fi - ei
Less than
53.02 to
63.03 to
71.00 to
78.97 to
More than
Total 30 30

33. Example: Victor Computers
Normal Distribution Goodness of Fit Test
Test Statistic
Rejection Rule
With  = .05 and k - p - 1 = = 3 d.f.,
Reject H0 if 2 > 7.81
Conclusion
We cannot reject H0. There is little evidence to
support rejecting the assumption the population is normally distributed with  = 71 and  =

34. End of Chapter 12