1. 1 1 Slide © 2006 Thomson/South-Western Slides Prepared by JOHN S. LOUCKS St. Edward’s University Slides Prepared by JOHN S. LOUCKS St. Edward’s University

2. 2 2 Slide © 2006 Thomson/South-Western Chapter 11 Comparisons Involving Proportions and a Test of Independence n Inferences About the Difference Between Two Population Proportions Two Population Proportions n Test of Independence: Contingency Tables n Hypothesis Test for Proportions of a Multinomial Population of a Multinomial Population

3. 3 3 Slide © 2006 Thomson/South-Western Inferences About the Difference Between Two Population Proportions n Interval Estimation of p 1 - p 2 n Hypothesis Tests About p 1 - p 2

4. 4 4 Slide © 2006 Thomson/South-Western n Expected Value Sampling Distribution of where: n 1 = size of sample taken from population 1 n 2 = size of sample taken from population 2 n 2 = size of sample taken from population 2 n Standard Deviation (Standard Error)

5. 5 5 Slide © 2006 Thomson/South-Western If the sample sizes are large, the sampling distribution If the sample sizes are large, the sampling distribution of can be approximated by a normal probability of can be approximated by a normal probability distribution. distribution. If the sample sizes are large, the sampling distribution If the sample sizes are large, the sampling distribution of can be approximated by a normal probability of can be approximated by a normal probability distribution. distribution. The sample sizes are sufficiently large if all of these The sample sizes are sufficiently large if all of these conditions are met: conditions are met: The sample sizes are sufficiently large if all of these The sample sizes are sufficiently large if all of these conditions are met: conditions are met: n1p1 > 5n1p1 > 5n1p1 > 5n1p1 > 5 n 1 (1 - p 1 ) > 5 n2p2 > 5n2p2 > 5n2p2 > 5n2p2 > 5 n 2 (1 - p 2 ) > 5 Sampling Distribution of

6. 6 6 Slide © 2006 Thomson/South-Western Sampling Distribution of p 1 – p 2

7. 7 7 Slide © 2006 Thomson/South-Western Interval Estimation of p 1 - p 2 n Interval Estimate

8. 8 8 Slide © 2006 Thomson/South-Western Market Research Associates is Market Research Associates is conducting research to evaluate the effectiveness of a client’s new adver- tising campaign. Before the new campaign began, a telephone survey of 150 households in the test market area showed 60 households “aware” of the client’s product. Interval Estimation of p 1 - p 2 n Example: Market Research Associates The new campaign has been initiated with TV and The new campaign has been initiated with TV and newspaper advertisements running for three weeks.

9. 9 9 Slide © 2006 Thomson/South-Western A survey conducted immediately A survey conducted immediately after the new campaign showed 120 of 250 households “aware” of the client’s product. Interval Estimation of p 1 - p 2 n Example: Market Research Associates Does the data support the position Does the data support the position that the advertising campaign has provided an increased awareness of the client’s product?

10. 10 Slide © 2006 Thomson/South-Western Point Estimator of the Difference Between Two Population Proportions = sample proportion of households “aware” of the = sample proportion of households “aware” of the product after the new campaign product after the new campaign = sample proportion of households “aware” of the = sample proportion of households “aware” of the product before the new campaign product before the new campaign p 1 = proportion of the population of households p 1 = proportion of the population of households “aware” of the product after the new campaign “aware” of the product after the new campaign p 2 = proportion of the population of households p 2 = proportion of the population of households “aware” of the product before the new campaign “aware” of the product before the new campaign

11. 11 Slide © 2006 Thomson/South-Western.08 + 1.96(.0510).08 +.10 Interval Estimation of p 1 - p 2 Hence, the 95% confidence interval for the difference Hence, the 95% confidence interval for the difference in before and after awareness of the product is -.02 to +.18. For  =.05, z.025 = 1.96:

12. 12 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 n Hypotheses H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0 Left-tailedRight-tailedTwo-tailed We focus on tests involving no difference between the two population proportions (i.e. p 1 = p 2 )

13. 13 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 Pooled Estimate of Standard Error of Pooled Estimate of Standard Error ofwhere:

14. 14 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 Test Statistic Test Statistic

15. 15 Slide © 2006 Thomson/South-Western Can we conclude, using a.05 level Can we conclude, using a.05 level of significance, that the proportion of households aware of the client’s product increased after the new advertising campaign? Hypothesis Tests about p 1 - p 2 n Example: Market Research Associates

16. 16 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 1. Develop the hypotheses. p -Value and Critical Value Approaches p -Value and Critical Value Approaches H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0 p 1 = proportion of the population of households p 1 = proportion of the population of households “aware” of the product after the new campaign “aware” of the product after the new campaign p 2 = proportion of the population of households p 2 = proportion of the population of households “aware” of the product before the new campaign “aware” of the product before the new campaign

17. 17 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 2. Specify the level of significance.  =.05 3. Compute the value of the test statistic. p -Value and Critical Value Approaches p -Value and Critical Value Approaches

18. 18 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 5. Determine whether to reject H 0. We cannot conclude that the proportion of households aware of the client’s product increased after the new campaign. 4. Compute the p –value. For z = 1.56, the p –value =.0594 Because p –value >  =.05, we cannot reject H 0. p –Value Approach p –Value Approach

19. 19 Slide © 2006 Thomson/South-Western Hypothesis Tests about p 1 - p 2 Critical Value Approach Critical Value Approach 5. Determine whether to reject H 0. Because 1.56 < 1.645, we cannot reject H 0. For  =.05, z.05 = 1.645 4. Determine the critical value and rejection rule. Reject H 0 if z > 1.645 We cannot conclude that the proportion of households aware of the client’s product increased after the new campaign.

20. 20 Slide © 2006 Thomson/South-Western Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 1. Set up the null and alternative hypotheses. 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.

21. 21 Slide © 2006 Thomson/South-Western Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 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. f i = observed frequency for category i e i = expected frequency for category i k = number of categories where:

22. 22 Slide © 2006 Thomson/South-Western Hypothesis (Goodness of Fit) Test for Proportions of a Multinomial Population 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

23. 23 Slide © 2006 Thomson/South-Western Multinomial Distribution Goodness of Fit Test n Example: Finger Lakes Homes (A) Finger Lakes Homes manufactures Finger Lakes Homes manufactures four models of prefabricated homes, four models of prefabricated homes, a two-story colonial, a log cabin, a a two-story colonial, a log cabin, a split-level, and an A-frame. To help split-level, and an A-frame. To help in production planning, management in production planning, management would like to determine if previous would like to determine if previous customer purchases indicate that there customer purchases indicate that there is a preference in the style selected. is a preference in the style selected.

24. 24 Slide © 2006 Thomson/South-Western Split- A- Split- A- Model Colonial Log Level Frame # Sold 30 20 35 15 The number of homes sold of each The number of homes sold of each model for 100 sales over the past two years is shown below. Multinomial Distribution Goodness of Fit Test n Example: Finger Lakes Homes (A)

25. 25 Slide © 2006 Thomson/South-Western 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

26. 26 Slide © 2006 Thomson/South-Western 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.

27. 27 Slide © 2006 Thomson/South-Western 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

28. 28 Slide © 2006 Thomson/South-Western Multinomial Distribution Goodness of Fit Test n Conclusion Using the p -Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = 10 is between 9.348 and 11.345, the Because  2 = 10 is between 9.348 and 11.345, the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.025 and.01..025 and.01. Area in Upper Tail.10.05.025.01.005  2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838

29. 29 Slide © 2006 Thomson/South-Western n Conclusion Using the Critical Value Approach Multinomial Distribution Goodness of Fit Test We reject, at the.05 level of significance, We reject, at the.05 level of significance, the assumption that there is no home style preference.  2 = 10 > 7.815

30. 30 Slide © 2006 Thomson/South-Western Test of Independence: Contingency Tables 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.

31. 31 Slide © 2006 Thomson/South-Western 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.

32. 32 Slide © 2006 Thomson/South-Western Each home sold by Finger Lakes 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. Contingency Table (Independence) Test n Example: Finger Lakes Homes (B)

33. 33 Slide © 2006 Thomson/South-Western Price Colonial Log Split-Level A-Frame Price Colonial Log Split-Level A-Frame The number of homes sold for 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)

34. 34 Slide © 2006 Thomson/South-Western n Hypotheses Contingency Table (Independence) Test H 0 : Price of the home is independent of the style of the home that is purchased 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 style of the home that is purchased

35. 35 Slide © 2006 Thomson/South-Western 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

36. 36 Slide © 2006 Thomson/South-Western 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

37. 37 Slide © 2006 Thomson/South-Western n Conclusion Using the p -Value Approach The p -value < . We can reject the null hypothesis. The p -value < . We can reject the null hypothesis. Because  2 = 9.145 is between 7.815 and 9.348, the Because  2 = 9.145 is between 7.815 and 9.348, the area in the upper tail of the distribution is between area in the upper tail of the distribution is between.05 and.025..05 and.025. Area in Upper Tail.10.05.025.01.005  2 Value (df = 3) 6.251 7.815 9.348 11.345 12.838 Contingency Table (Independence) Test

38. 38 Slide © 2006 Thomson/South-Western n Conclusion Using the Critical Value Approach Contingency Table (Independence) Test We reject, at the.05 level of significance, 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.  2 = 9.145 > 7.815

39. 39 Slide © 2006 Thomson/South-Western End of Chapter 11