f98Key2b

# f98Key2b - Answer Key for Exam II Version B 1 Sample size n...

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Answer Key for Exam II Version B 1. Sample size n = {Z α / 2 2 * p-hat ( 1 - p-hat ) / E 2 } . n = {1.96 2 *.18*(1-.18) / .06 2 } = 157.51 = 158 Ans. C 2. The number of tankers which did not have spills is 552. The number of tankers which had spills is 148 ( = 700 - 552). P-hat = x / n = ( 148 / 700 ) = .211 Ans. D 3. The 90% C.I. for p is given by P-hat ± Z α / 2 √[p-hat*(1-p-hat)/(n-1)] = .211 ± 1.645 √(.211*.789 / 699) = .1856 to .2364 Ans. B 4. Test statistic Z = (P-hat - P o ) / √[ P o (1- P o ) / n] = (.211-.18) / √[(.18*.82) / 700] = 2.135 Ans. B (difference in answer due to rounding) 5. Since the oil companies’ claim is “not .18,” we need to find the critical (table) value for a two-tailed test corresponding to α/2 = .1/2 = .05. We get Z .05 = 1.645. Ans. A 6. Given that Z* = 2.04 we can find the critical (table) value for the two-tailed corresponding to α/2 = .1/2 = .05, Z .05 = 1.645. Decision is to reject Ho since Z* (2.04) > Z .05 (1.645). Since we reject Ho, there is evidence that the claim of the oil companies is correct. Ans. B

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## This note was uploaded on 03/09/2011 for the course DSCI 3710 taught by Professor Dr.jay during the Spring '08 term at North Texas.

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f98Key2b - Answer Key for Exam II Version B 1 Sample size n...

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