solution-5-2010

solution-5-2010 - or 1.17 < < 1.23...

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Chap. 6: 6.4 (a), 6.5, 6.13, 6.18(a, b), 6.20, 6.30, 6.32, 6.39, 6.43, 6.48 6.4 (a) 3.06/= .33 mm 6.5 (a) We would predict the SD of the new measurements to be about 3 mm because this is our estimate (based on Exercise 6.4) of the population SD. (b) We would expect the SE of the new measurements to be 3/ .13 mm. 6.13 (a) This statement is false. The confidence interval allows us to make an inference concerning the mean of the entire population. We know that 59.77 < < 61.09. (b) This statement is true. (See part (a).) 6.18 (a) 5,111 ± (2.306)(818/) (4482,5740) or 4,482 < μ < 5,740 units (b) We are 95% confident that the average invertase activity of all fungal tissue incubated at 95% relative humidity for 24 hours is between 4,482 units and 5,740 units. 6.20 = 1.20; s = .14; n = 50. The degrees of freedom are 50 - 1 = 49. From Table 4 with df = 50 (the df value closest to 49) we find that t .05 = 1.676. The 90% confidence interval for μ is ± t .05 1.20 ± 1.676 () (1.17,1.23)
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Unformatted text preview: or 1.17 &lt; &lt; 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 15-19 1 20-24 2 25-29 10 30-34 3 35-39 9 40-44 3 45-49 5 50-54 2 55-59 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 &lt; p &lt; . 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 &lt; p &lt; .087....
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solution-5-2010 - or 1.17 &amp;amp;lt; &amp;amp;lt; 1.23...

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