ISM_chapter14 - Chapter 14: Analysis of Categorical Data...

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287 Chapter 14: Analysis of Categorical Data 14.1 a. H 0 : p 1 = .41, p 2 = .10, p 3 = .04, p 4 = .45 vs. H a : not H 0 . The observed and expected counts are: A B AB O observed 89 18 12 81 expected 200(.41) = 82 200(.10) = 20 200(.04) = 8 200(.45) = 90 The chi–square statistic is 90 ) 90 81 ( 8 ) 8 12 ( 20 ) 20 18 ( 82 ) 82 89 ( 2 2 2 2 2 + + + = X = 3.696 with 4 –1 = 3 degrees of freedom. Since 2 05 . χ = 7.81473, we fail to reject H 0 ; there is not enough evidence to conclude the proportions differ. b. Using the Applet, p –value = P ( χ 2 > 3.696) = .29622. 14.2 a. H 0 : p 1 = .60, p 2 = .05, p 3 = .35 vs. H a : not H 0 . The observed and expected counts are: admitted unconditionally admitted conditionally refused observed 329 43 128 expected 500(.60) = 300 500(.05) = 25 500(.35) = 175 The chi–square test statistic is 175 ) 175 128 ( 25 ) 25 43 ( 300 ) 300 329 ( 2 2 2 2 + + = X = 28.386 with 3 – 1 = 2 degrees of freedom. Since 2 05 . χ = 7.37776, we can reject H 0 and conclude that the current admission rates differ from the previous records. b. Using the Applet, p –value = P ( χ 2 > 28.386) = .00010. 14.3 The null hypothesis is H 0 : p 1 = p 2 = p 3 = p 4 = 4 1 vs. H a : not H 0 . The observed and expected counts are: lane 1 2 3 4 observed 294 276 238 192 expected 250 250 250 250 The chi–square statistic is 250 ) 250 192 ( ) 250 238 ( ) 250 276 ( ) 250 294 ( 2 2 2 2 2 + + + = X = 24.48 with 4 –1 = 3 degrees of freedom. Since 2 05 . χ = 7.81473, we reject H 0 and conclude that the lanes are not preferred equally. From Table 6, p –value < .005. Note that R can be used by: > lanes <- c(294,276,238,192) > chisq.test(lanes,p = c(.25,.25,.25,.25)) # p is not necessary here Chi-squared test for given probabilities data: lanes X-squared = 24.48, df = 3, p-value = 1.983e-05
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288 Chapter 14: Analysis of Categorical Data Instructor’s Solutions Manual 14.4 The null hypothesis is H 0 : p 1 = p 2 = … = p 7 = 7 1 vs. H a : not H 0 . The observed and expected counts are: SU M T W R F SA observed 24 36 27 26 32 26 29 expected 28.571 28.571 28.571 28.571 28.571 28.571 28.571 The chi–square statistic is 571 . 28 ) 571 . 28 29 ( ) 571 . 28 36 ( ) 571 . 28 24 ( 2 2 2 2 + + + = X = 24.48 with 7 –1 = 6 degrees of freedom. Since 2 05 . χ = 12.5916, we can reject the null hypothesis and conclude that there is evidence of a difference in percentages of heart attacks for the days of the week 14.5 a. Let p = proportion of heart attacks on Mondays. Then, H 0 : p = 7 1 vs. H a : p > 7 1 . Then, p ˆ = 36/200 = .18 and from Section 8.3, the test statistic is 200 ) 7 / 6 )( 7 / 1 ( 7 / 1 18 . = z = 1.50. Since z .05 = 1.645, we fail to reject H 0 . b. The test was suggested by the data, and this is known as “data snooping” or “data dredging.” We should always apply the scientific method: first form a hypothesis and then collect data to test the hypothesis. c. Monday has often been referred to as the most stressful workday of the week: it is the day that is farthest from the weekend, and this realization gets to some people. 14.6 a. j i j i j i np np n E n E n n E = = ) ( ) ( ) (. b. Define the sample proportions n n p i i / ˆ = and n n p j j / ˆ = . Then, j i p p ˆ ˆ is unbiased for p i p j from part a above. c. j i j j i i j i j i j i p np p np p np n n n V n V n n V 2 ) 1 ( ) 1 ( ) , ( Cov 2 ) ( ) ( ) ( + + = + = .
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ISM_chapter14 - Chapter 14: Analysis of Categorical Data...

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