1. Chapter 11 – Test of Independence - Hypothesis Test for Proportions of a Multinomial Population
In this case, each element of a population is assigned to one and only one of several classes or categories.
Such a population is a multinomial population.
On each trial of a multinomial experiment:
One and only one of the outcomes occurs
Each trial is assumed to be independent
The probabilities of the outcomes remain the same for each trial

2. Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population
1. State the null and alternative hypotheses.
H0: The population follows a multinomial
distribution with specified probabilities
for each of the k categories
Ha: The population does not follow a
multinomial distribution with specified
probabilities for each of the k categories
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

3. Hypothesis Test for Proportions of a Multinomial Population
4. Compute the value of the test statistic.
fi = observed frequency for category i
where:
ei = expected frequency for category i
k = number of categories
Note: The test statistic has a chi-square distribution
with k – 1 df provided that the expected frequencies
are 5 or more for all categories.

4. Hypothesis Test for Proportions of a Multinomial Population
5. Rejection rule:
p-value approach:
Reject H0 if p-value < a
Critical value approach:
Reject H0 if
where  is the significance level and
there are k - 1 degrees of freedom

5. Multinomial Distribution Goodness of Fit Test
Example: Finger Lakes Homes (A)
Finger Lakes Homes manufactures four models of prefabricated homes, a two-story colonial, a log cabin, 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.
The number of homes sold of each model for 100
sales over the past two years is shown below.
Model Colonial Log Split-Level A-Frame
# Sold

6. Multinomial Distribution Goodness of Fit Test
Hypotheses
H0: pC = pL = pS = pA = .25
Ha: The population proportions are not
pC = .25, pL = .25, pS = .25, and pA = .25
where:
pC = population proportion that purchase a colonial
pL = population proportion that purchase a log cabin
pS = population proportion that purchase a split-level
pA = population proportion that purchase an A-frame

7. Multinomial Distribution Goodness of Fit Test
Rejection Rule
Reject H0 if p-value < .05 or c2 >
With  = .05 and
k - 1 = = 3
degrees of freedom
Do Not Reject H0
Reject H0 2
7.815

8. Multinomial Distribution Goodness of Fit Test
Expected Frequencies
Test Statistic
e1 = .25(100) = e2 = .25(100) = 25
e3 = .25(100) = e4 = .25(100) = 25
= = 10
Conclusion Using the Critical Value Approach
We reject, at the .05 level of significance,
the assumption that there is no home style
preference.
c2 = 10 > 7.815

9. 2 Test for k Proportions Example
As personnel director, you want to test the perception of fairness of four methods of performance evaluation. Of 240 employees, 60 rated Method 1 as fair, 50 rated Method 2 as fair, 70 rated Method 3 as fair and 60 rated Method 4 as fair. At the .05 level of significance, is there a difference in perceptions?
To check assumptions, use sample proportions as estimators of population proportion:
n1·p = 78·63/78 = 63
n1·(1-p) = 78·(1-63/78) = 15 10

10. Contingency Table (Independence) Test
Example: Finger Lakes Homes (B)
Each home sold by Finger Lakes Homes can be classified according to price and to style. Finger Lakes’ 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 $99,000 or less or more than $99,000.
Price Colonial Log Split-Level A-Frame
< $99,
> $99,

11. 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
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. Contingency Table (Independence) Test
Expected Frequencies
Price Colonial Log Split-Level A-Frame Total
< $99K
> $99K
Total

13. Test of Independence: Contingency Tables
4. Compute the test statistic.
5. Determine the rejection rule.
Reject H0 if p -value < a or
where  is the significance level and, with n rows and m columns, there are (n - 1)(m - 1) degrees of freedom.

14. Contingency Table (Independence) Test
Rejection Rule
With  = .05 and (2 - 1)(4 - 1) = 3 d.f.,
Reject H0 if p-value < .05 or 2 > 7.815
Test Statistic
= =
Conclusion Using the Critical Value Approach c2 = > 7.8
We reject, at the .05 level of significance, the assumption that the price of the home is independent of the style of home that is purchased

15. Caution about the 2 Test
The 2 is one of the most widely applied statistical tools and also one of the most abused statistical tool.
Be certain the experiment satisfies the assumptions.
Be certain the sample is drawn from the correct population.
Avoid using when the expected counts are very small.

16. Caution About the 2 Test
If the 2 value does not exceed the established critical value of 2, do not definitely accept the hypothesis of independence. You risk a Type II error. Avoid concluding that two classifications are independent, even when 2 is small.
If a contingency table 2 value does exceed the critical value, we must be careful to avoid inferring that a causal relationship exists between the classifications. The existence of a causal relationship cannot be established by a contingency table analysis.
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