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Unformatted text preview: or 1.17 < μ < 1.23 mm. 6.30 Guessed SD = 1.5 in The SE should be no more than .25 in, so n must satisfy 1 .5 n .25 ≤ which yields n 36. ≥ 6.32 (a) There were 36 cells, but only seven guinea pigs, so there is a hierarchical structure in the data, which suggests that the observations are not independent. (b) Number of branch segments Frequency 1519 1 2024 2 2529 10 3034 3 3539 9 4044 3 4549 5 5054 2 5559 1 Total 36 15 35 55 2 4 6 8 10 25 45 Number of branch segments Frequency 6.39 = (28 + 2)/(580 + 4) = .051; SE = = .009. The 95% confidence interval is .051 ± (1.96)(.009) or (.033,.069) or .033 < p < . 069. •6.43 The required n must satisfy the inequality Desired SE ≤ or .04. ≤ It follows that ≤ or n+4 or 150 n+4, so n 146. ≤ ≤ ≥ 6.48 = (103+2)/(1438+4) = .073; SE = .073(1.073) 1438 + 4 = .0069. 95% confidence interval: .073 ± (1.96)(.0069) or (.059,.087) or .059 < p < .087....
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This note was uploaded on 12/09/2010 for the course BIOL 442 taught by Professor Brewster,r during the Spring '08 term at UMBC.
 Spring '08
 Brewster,R
 Developmental Biology

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