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# hw7s - MSPE PROGRAM ECON 506 FALL 2011 Solution to HW7(The...

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MSPE PROGRAM ECON 506 - FALL 2011 Solution to HW7 (The problem numbers refer to numbers in 7 th edition) Part B: Use likelihood ratio test. The observed test statistics is –2 ln (24/196) = 4.2 The sampling distribution is χ 2 with 1 degree of freedom. At for example α =0.05 the critical value would be 3.84. Thus we reject the null. Part C: 10.2 Note that Y is binomial with parameters n = 20 and p . a. If the experimenter concludes that less than 80% of insomniacs respond to the drug when actually the drug induces sleep in 80% of insomniacs, a type I error has occurred. b. α = P (reject H 0 | H 0 true) = P ( Y ≤ 12 | p = .8) = .032 (using Appendix III). c. If the experimenter does not reject the hypothesis that 80% of insomniacs respond to the drug when actually the drug induces sleep in fewer than 80% of insomniacs, a type II error has occurred. d. β(.6) = P (fail to reject H 0 | H a true) = P ( Y > 12 | p = .6) = 1 – P ( Y ≤ 12 | p = .6) = .416. e. β(.4) = P (fail to reject H 0 | H a true) = P ( Y > 12 | p = .4) = .021. 10.9 a. The simulation is performed with a known p = .5, so rejecting H 0 is a type I error. b. - e. Answers vary. f. This is because of part a. g. - h. Answers vary. 10.18 H 0 : μ = 13.20, H a : μ < 13.20. Using the large sample test for a mean, z = –2.53, and with α = .01, – z .01 = –2.326. So, H 0 is rejected: there is evidence that the company is paying substandard wages.

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10.19 H 0 : μ = 130, H a : μ < 130. Using the large sample test for a mean, z = 40 / 1 . 2 130 6 . 128 - = – 4.22 and with – z .05 = –1.645, H 0 is rejected: there is evidence that the mean output voltage is less than 130. 10.20 H 0 : μ ≥ 64, H a : μ < 64. Using the large sample test for a mean, z = –1.77, and w/ α = .01, z .01 = –2.326. So, H 0 is not rejected: there is not enough evidence to conclude the manufacturer’s claim is false.
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hw7s - MSPE PROGRAM ECON 506 FALL 2011 Solution to HW7(The...

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