1. 1 In this case, each element of a population is assigned to one and only one of several classes or categories. Chapter 11 – Test of Independence - Hypothesis Test for Proportions of a Multinomial Population 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. 2 Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 1. State the null and alternative hypotheses. H 0 : The population follows a multinomial distribution with specified probabilities distribution with specified probabilities for each of the k categories for each of the k categories H 0 : The population follows a multinomial distribution with specified probabilities distribution with specified probabilities for each of the k categories for each of the k categories H a : The population does not follow a multinomial distribution with specified multinomial distribution with specified probabilities for each of the k categories probabilities for each of the k categories H a : The population does not follow a multinomial distribution with specified multinomial distribution with specified probabilities for each of the k categories probabilities for each of the k categories 2. Select a random sample and record the observed frequency, f i, for each of the k categories. frequency, f i, for each of the k categories. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the frequency, e i, in each category by multiplying the category probability by the sample size category probability by the sample size

3. 3 4. Compute the value of the test statistic. 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. 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. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where: Hypothesis Test for Proportions of a Multinomial Population

4. 4 where  is the significance level and there are k - 1 degrees of freedom p -value approach: Critical value approach: Reject H 0 if p -value <  5. Rejection rule: Reject H 0 if Hypothesis Test for Proportions of a Multinomial Population

5. 5 Multinomial Distribution Goodness of Fit Test n 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 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 30 20 35 15

6. 6 n Hypotheses Multinomial Distribution Goodness of Fit Test where: p C = population proportion that purchase a colonial p C = population proportion that purchase a colonial p L = population proportion that purchase a log cabin p L = population proportion that purchase a log cabin p S = population proportion that purchase a split-level p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame p A = population proportion that purchase an A-frame H 0 : p C = p L = p S = p A =.25 H a : The population proportions are not p C =.25, p L =.25, p S =.25, and p A =.25 p C =.25, p L =.25, p S =.25, and p A =.25

7. 7 n Rejection Rule 22 22 7.815 Do Not Reject H 0 Reject H 0 Multinomial Distribution Goodness of Fit Test With  =.05 and k - 1 = 4 - 1 = 3 k - 1 = 4 - 1 = 3 degrees of freedom degrees of freedom if p -value 7.815. Reject H 0 if p -value 7.815.

8. 8 n Expected Frequencies n Test Statistic Multinomial Distribution Goodness of Fit Test e 1 =.25(100) = 25 e 2 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 e 3 =.25(100) = 25 e 4 =.25(100) = 25 = 1 + 1 + 4 + 4 = 10 n Conclusion Using the Critical Value Approach  2 = 10 > 7.815 We reject, at the.05 level of significance, the assumption that there is no home style preference.

9. 9 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?  2 Test for k Proportions Example

10. 10 Price Colonial Log Split-Level A-Frame Price Colonial Log Split-Level A-Frame 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. > $99,000 12 14 16 3 < $99,000 18 6 19 12 Contingency Table (Independence) Test n Example: Finger Lakes Homes (B)

11. 11 n Hypotheses Contingency Table (Independence) Test H 0 : Price of the home is independent of the style of the home that is purchased H a : 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, f ij, for each cell of the contingency table. frequency, f ij, for each cell of the contingency table. 3. Compute the expected frequency, e ij, for each cell.

12. 12 n Expected Frequencies Contingency Table (Independence) Test Price Colonial Log Split-Level A-Frame Total Price Colonial Log Split-Level A-Frame Total < $99K > $99K Total Total 30 20 35 15 100 12 14 16 3 45 18 6 19 12 55

13. 13 Test of Independence: Contingency Tables 5. Determine the rejection rule. Reject H 0 if p -value <  or 4. Compute the test statistic. where  is the significance level and, with n rows and m columns, there are ( n - 1)( m - 1) degrees of freedom.

14. 14 n Rejection Rule Contingency Table (Independence) Test With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if p -value 7.815 =.1364 + 2.2727 +... + 2.0833 = 9.149 n Test Statistic Conclusion Using the Critical Value Approach  2 = 9.145 > 7.8 Conclusion Using the Critical Value Approach  2 = 9.145 > 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. 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. 16 © 2011 Pearson Education, Inc Caution About the  2 Test If the  2 value does not exceed the established critical value of  2, do not 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.