MBS-ProblemSolutions-Ch12

MBS-ProblemSolutions - Chapter 12 Tests of Goodness of Fit and Independence Learning Objectives 1 Know how to conduct a goodness of fit test 2 Know

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Unformatted text preview: Chapter 12 Tests of Goodness of Fit and Independence Learning Objectives 1. Know how to conduct a goodness of fit test. 2. Know how to use sample data to test for independence of two variables. 3. Understand the role of the chi-square distribution in conducting tests of goodness of fit and independence. 4. Be able to conduct a goodness of fit test for cases where the population is hypothesized to have a multinomial, a Poisson, or a normal distribution. 5. For a test of independence, be able to set up a contingency table, determine the observed and expected frequencies, and determine if the two variables are independent. 6. Be able to use p-values based on the chi-square distribution. Solutions: 12 - 1 Chapter 12 1. a. Expected frequencies: e 1 = 200 (.40) = 80, e 2 = 200 (.40) = 80 e 3 = 200 (.20) = 40 Actual frequencies: f 1 = 60, f 2 = 120, f 3 = 20 2 2 2 2 (60 80) (120 80) (20 40) 80 80 40 400 1600 400 80 80 40 5 20 10 35 χ--- = + + = + + = + + = k- 1 = 2 degrees of freedom Using the 2 χ table with df = 2, 2 χ = 35 shows the p-value is less than .005. Using Excel, the p-value corresponding to 2 χ = 35 is approximately 0. p-value ≤ .01, reject H b. .01 χ = 9.210 Reject H if 2 χ ≥ 9.210 2 χ = 35, reject H 2. Expected frequencies: e 1 = 300 (.25) = 75, e 2 = 300 (.25) = 75 e 3 = 300 (.25) = 75, e 4 = 300 (.25) = 75 Actual frequencies: f 1 = 85, f 2 = 95, f 3 = 50, f 4 = 70 2 2 2 2 2 (85 75) (95 75) (50 75) (70 75) 75 75 75 75 100 400 625 25 75 75 75 75 1150 75 15.33 χ---- = + + + = + + + = = k- 1 = 3 degrees of freedom Using the 2 χ table with df = 3, 2 χ = 15.33 shows the p-value is less than .005. Using Excel, the p-value corresponding to 2 χ = 15.33 is .0016. p-value ≤ .05, reject H 12 - 2 Tests of Goodness of Fit and Independence The population proportions are not the same. 3. H = p ABC = .29, p CBS = .28, p NBC = .25, p IND = .18 H a = The proportions are not p ABC = .29, p CBS = .28, p NBC = .25, p IND = .18 Expected frequencies: 300 (.29) = 87, 300 (.28) = 84 300 (.25) = 75, 300 (.18) = 54 e 1 = 87, e 2 = 84, e 3 = 75, e 4 = 54 Actual frequencies: f 1 = 95, f 2 = 70, f 3 = 89, f 4 = 46 2 2 2 2 2 (95 87) (70 84) (89 75) (46 54) 87 84 75 54 6.87 χ---- = + + + = k- 1 = 3 degrees of freedom Using the 2 χ table with df = 3, 2 χ = 6.87 shows the p-value is between .05 and .10. Using Excel, the p-value corresponding to 2 χ = 6.87 is .0762. p-value > .05, do not reject H . There has not been a significant change in the viewing audience proportions. 4. Observed Expected Hypothesized Frequency Frequency Category Proportion ( f i ) ( e i ) ( f i- e i ) 2 / e i Brown 0.30 177 151.8 4.18 Yellow 0.20 135 101.2 11.29 Red 0.20 79 101.2 4.87 Orange 0.10 41 50.6 1.82 Green 0.10 36 50.6 4.21 Blue 0.10 38 50.6 3.14 Totals: 506 29.51 k- 1 = 5 degrees of freedom Using the 2 χ table with df = 5, 2 χ = 29.51 shows the p-value is less than .005....
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MBS-ProblemSolutions - Chapter 12 Tests of Goodness of Fit and Independence Learning Objectives 1 Know how to conduct a goodness of fit test 2 Know

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