1. 1 1 Slide Chapter 11 Comparisons Involving Proportions n Inference about the Difference Between the Proportions of Two Populations Proportions of Two Populations n A Hypothesis Test for Proportions of a Multinomial Population Multinomial Population n Test of Independence: Contingency Tables H o : p 1 - p 2 = 0 H a : p 1 - p 2 = 0

2. 2 2 Slide Inferences About the Difference Between the Proportions of Two Populations n Sampling Distribution of n Interval Estimation of p 1 - p 2 n Hypothesis Tests about p 1 - p 2

3. 3 3 Slide Sampling Distribution of n Expected Value n Standard Deviation n Distribution Form If the sample sizes are large ( n 1 p 1, n 1 (1 - p 1 ), n 2 p 2, and n 2 (1 - p 2 ) are all greater than or equal to 5), the sampling distribution of can be approximated by a normal probability distribution.

4. 4 4 Slide Interval Estimation of p 1 - p 2 : Large-Sample Case n Interval Estimate n Point Estimator of

5. 5 5 Slide Example: MRA MRA (Market Research Associates) is conducting research to evaluate the effectiveness of a client’s new advertising 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. The new campaign has been initiated with TV and newspaper advertisements running for three weeks. A survey conducted immediately after the new campaign showed 120 of 250 households “aware” of the client’s product. Does the data support the position that the advertising campaign has provided an increased awareness of the client’s product?

6. 6 6 Slide n Point Estimator of the Difference Between the Proportions of Two Populations 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 = 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 Example: MRA

7. 7 7 Slide n Interval Estimate of p 1 - p 2 : Large-Sample Case For  =.05, z.025 = 1.96: For  =.05, z.025 = 1.96:.08 + 1.96(.0510).08 +.10.08 +.10 or -.02 to +.18 Conclusion Conclusion At a 95% confidence level, the interval estimate of the difference between the proportion of households aware of the client’s product before and after the new advertising campaign is -.02 to +.18. At a 95% confidence level, the interval estimate of the difference between the proportion of households aware of the client’s product before and after the new advertising campaign is -.02 to +.18. Example: MRA

8. 8 8 Slide Hypothesis Tests about p 1 - p 2 n Hypotheses H 0 : p 1 - p 2 < 0 H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0 H a : p 1 - p 2 > 0 n Test statistic n Point Estimator of where p 1 = p 2 where:

9. 9 9 Slide n Hypothesis Tests about p 1 - p 2 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? 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 HypothesesHypotheses H 0 : p 1 - p 2 < 0 H 0 : p 1 - p 2 < 0 H a : p 1 - p 2 > 0 H a : p 1 - p 2 > 0 Example: MRA

10. 10 Slide n Hypothesis Tests about p 1 - p 2 Rejection Rule Reject H 0 if z > 1.645Rejection Rule Reject H 0 if z > 1.645 Test StatisticTest Statistic Conclusion Do not reject H 0.Conclusion Do not reject H 0. Example: MRA

11. 11 Slide A 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. 3. Assuming H 0 is true, compute the expected frequency, e i, in each category by multiplying the category probability by the sample size. 4. Compute the value of the test statistic. 5. Reject H 0 if (where  is the significance level and there are k - 1 degrees of freedom).

12. 12 Slide Example: Finger Lakes Homes (A) n 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. 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 Model Colonial Ranch Split-Level A-Frame # Sold 30 20 35 15 # Sold 30 20 35 15

13. 13 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Let: p C = population proportion that purchase a colonial p R = population proportion that purchase a ranch p S = population proportion that purchase a split-level p A = population proportion that purchase an A-frame HypothesesHypotheses H 0 : p C = p R = p S = p A =.25 H a : The population proportions are not p C =.25, p R =.25, p S =.25, and p A =.25 p R =.25, p S =.25, and p A =.25

14. 14 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test Expected FrequenciesExpected Frequencies e 1 =.25(100) = 25 e 2 =.25(100) = 25 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 Test StatisticTest Statistic = 1 + 1 + 4 + 4 = 1 + 1 + 4 + 4 = 10 = 10

15. 15 Slide n Multinomial Distribution Goodness of Fit Test Rejection RuleRejection Rule With  =.05 and With  =.05 and k - 1 = 4 - 1 = 3 degrees of k - 1 = 4 - 1 = 3 degrees of freedom freedom ConclusionConclusion We reject the assumption there is no home style We reject the assumption there is no home style preference, at the.05 level of significance. preference, at the.05 level of significance. Example: Finger Lakes Homes (A) 22 22 7.815 Do Not Reject H 0 Reject H 0

16. 16 Slide Example: Finger Lakes Homes (A) n Multinomial Distribution Goodness of Fit Test p -Value p -Value Using the Chi-Square Distribution table (Table 3 in Appendix B), we can approximate the p -value for this problem. Our test statistic of 10.0 is greater than    for  =.025. Thus, we know the p -value for this problem is no more than.025. Using the Chi-Square Distribution table (Table 3 in Appendix B), we can approximate the p -value for this problem. Our test statistic of 10.0 is greater than    for  =.025. Thus, we know the p -value for this problem is no more than.025. Minitab and Excel would provide a precise p - value. Minitab and Excel would provide a precise p - value.

17. 17 Slide 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. 3. Compute the expected frequency, e ij, for each cell. 4. Compute the test statistic. 5. Reject H 0 if (where  is the significance level and with n rows and m columns there are ( n - 1)( m - 1) degrees of freedom).

18. 18 Slide Example: Finger Lakes Homes (B) n 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. 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 Price Colonial Ranch Split-Level A-Frame < $65,000 18 6 19 12 < $65,000 18 6 19 12 > $65,000 12 14 16 3 > $65,000 12 14 16 3

19. 19 Slide n Contingency Table (Independence) Test HypothesesHypotheses H 0 : Price of the home is independent of the style of the home that is purchased 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 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 Expected FrequenciesExpected Frequencies Price Colonial Ranch Split-Level A-Frame Total Price Colonial Ranch Split-Level A-Frame Total < $65K 18 6 19 12 55 > $65K 12 14 16 3 45 Total 30 20 35 15 100 Total 30 20 35 15 100 Example: Finger Lakes Homes (B)

20. 20 Slide n Contingency Table (Independence) Test Test StatisticTest Statistic =.14 + 2.27 +... + 2.08 = 8.00 =.14 + 2.27 +... + 2.08 = 8.00 Rejection RuleRejection Rule With  =.05 and (2 - 1)(4 - 1) = 3 d.f., With  =.05 and (2 - 1)(4 - 1) = 3 d.f., Reject H 0 if  2 > 7.81 Reject H 0 if  2 > 7.81 ConclusionConclusion We reject H 0, the assumption that the price of the home is independent of the style of the home that is purchased. We reject H 0, the assumption that the price of the home is independent of the style of the home that is purchased. Example: Finger Lakes Homes (B)

21. 21 Slide The End of Chapter 11