Homework 7 Solutions

Homework 7 Solutions - HW 7 SOLUTIONS Confidence Interval...

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HW 7 SOLUTIONS Confidence Interval for p and Chi-Square - 2x2 Tables 1. 6.42. In a study of human blood types in nonhuman primates, a sample of 71 orangutans were tested and 14 were found to be blood type B. a. Construct a 95% confidence interval for the relative frequency of blood type B in the orangutan population. Correct: ± = (14+{1.96 2 /2})/(71+1.96 2 ) = 0.2127; SE = √(0.2127)(0.7873)/(71+1.96 2 ) = √0.0022 = 0.0473. 95% confidence interval 0.2127 ± (1.96)(0.0473) (0.120, 0.305) or 0.120 < p < 0.305. b. Interpret the interval you just computed in part (a). Correct: We are 95% confident that the proportion of type B blood in the orangutan population is no less than 0.120 and no more than 0.305. 2. 6.49(modified). The Luso variety of wheat is resistant to the Hessian fly. In order to understand the genetic mechanism controlling this resistance, an agronomist plans to examine the progeny of a certain cross involving Luso and a nonresistant variety. Each progeny plant will be classified as resistant or susceptible and the agronomist will estimate the proportion of progeny that are resistant. He'd like to report a 95% confidence interval for the proportion of resistant progeny. a. If the researcher believes the proportion of resistant progeny is 0.75, how many progeny does he need to classify in order to guarantee that the standard error of his estimate of this proportion will not exceed .05? Correct: Agresti-Coull Sample Size Determination It is necessary to use p o = 0.75 because the researcher has an idea of what proportion are resistant. Since we want SE less than or equal to 0.05, then we want MOE less than or equal to e o = (1.96)(0.05) = 0.098. Then, since (Z /2 ) 2 = 1.96 2 = 3.8416 and p o = 0.75, the formula gives n {(3.8416)(0.75)(0.25)/(0.098) 2 } - 3.8416 = 75 - 3.8416 = 71.1584 So choose n = 72. b. If the researcher has no idea what the proportion of resistant progeny may be, how many progeny does he need to classify in order to guarantee that the standard error of his estimate of this proportion will not exceed .05? Correct: Agresti-Coull Sample Size Determination It is necessary to use p o = .5 because the proportion of progeny that are resistant is unknown in advance. Since we want SE less than or equal to 0.05, then we want MOE less than or equal to e o = (1.96)(0.05) = 0.098. Then, since (Z /2 ) 2 = 1.96 2 = 3.8416 and p o = 0.5, the formula gives n {(3.8416)(0.5)(0.5)/(0.098) 2 } - 3.8416 = 100 - 3.8416 = 96.1584 So choose n = 97.

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3. 10.58. Experimental studies of cancer often use strains of animals that have a naturally high incidence of tumors. In one such experiment, tumor prone mice were kept in a sterile environment with one group of mice maintained entirely germ free and the other group of mice exposed to the intestinal bacterium
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This note was uploaded on 12/13/2011 for the course STAT 205 taught by Professor Hendrix during the Fall '09 term at South Carolina.

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Homework 7 Solutions - HW 7 SOLUTIONS Confidence Interval...

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