f98Key2a - Answer Key for Exam II Version A 1. Sample size...

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Answer Key for Exam II Version A 1. Sample size n = {Z α / 2 2 * p-hat ( 1 - p-hat ) / E 2 } . n = {2.33 2 *.18*(1-.18) / .06 2 } = 222.58 = 223 ) Ans. A 2. The number of tankers that did not have spills is 482. The number of tankers that had spills is 118 ( = 600 - 482). P-hat = (x / n) = ( 118 / 600 ) = .197 Ans. D 3. The 90% C.I. for p is given by P-hat ± Z α / 2 √[p-hat*(1-p-hat)/(n-1)] = .197 ± 1.645 √(.197*.803 / 599) = .1703 to .2237 Ans. C 4. Test statistic Z = (P-hat - P o ) / √[ P o (1- P o ) / n] = (.197-.18) / √[(.18*.82) / 600] = 1.084 Ans. C 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 = .05/2 = .025. We get Z .025 = 1.96. Ans. B 6. Given that Z* = 1.065 we can find the critical (table) value for the two-tailed corresponding to α/2 = .05/2 = .025, Z .025 = 1.96. Decision is fail to reject Ho since Z* (1.065) < Z .025 (=1.96). Since we fail to reject Ho, there is insufficient evidence that the claim of the oil companies is
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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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f98Key2a - Answer Key for Exam II Version A 1. Sample size...

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