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Unformatted text preview: 132 Instructors Solutions Manual for Statistics for Business & Economics, 5 th Edition Chapter 9: Hypothesis Testing 9.1 a. European perspective: : H Genetically modified food stuffs are not safe 1 : H They are safe b. U.S. farmer perspective: : H Genetically modified food stuffs are safe 1 : H They are not safe 92 : B G H T T No difference in the total number of votes between Bush and Gore 1 : B G H T T Bush with more votes 93 : H No change interest rates in interest rates is warranted 1 : H Reduce interest rates to stimulate the economy 94 : A B H : There is no difference in the percentage of underfilled cereal packages 1 : A B H < : Lower percentage after the change 95 : 16 H ; 1 : 16 H < ; reject H if Z .10 < 1.28. 15.84 16 .4 16 Z = = 1.6, therefore, Reject H at the 10% level. 96 : 50 H ; 1 : 50 H < ; reject H if Z .10 < 1.28 48.2 50 3 9 Z = = 1.8, therefore, Reject H at the 10% level. 97 a. : 3 H = ; 1 : 3 H ; reject H if Z .05 > 1.645 3.07 3 .4 64 Z = = 1.4, therefore, Do Not Reject H at the 5% level. b. pvalue = 1 F Z (1.4) = 1  .9192 = .0808 c. the pvalue would be higher the graph should show that the pvalue now corresponds to the area in both of the tails of the distribution whereas before it was the area in one of the tails. d. A onesided alternative is more appropriate since we are not interested in detecting possible low levels of impurity, only high levels of impurity. 133 Instructors Solutions Manual for Statistics for Business & Economics, 5 th Edition 98 1 : 3; : 3; H H < 2.4 3 1.8 100 Z = = 3.33, pvalue = 1 F Z (3.33) = 1  .9996 = .0004, therefore, reject H at levels in excess of .04% 99 1 : 4; : 4; H H = reject H if 2.575 > Z .005 > 2.575 4.27 4 1.32 1562 Z = = 8.08, pvalue = 1 F Z (8.08) = 1 1.0000 = .0000, therefore, reject H at any common level of alpha. 910 1 : 0; : 0; H H = .078 0 .201 76 Z = = 3.38, pvalue = 2[1 F Z (3.38)] = 2[1  .9996] = .0008, therefore, reject H at levels in excess of .08% 911 1 : 3; : 3; H H reject H if Z .01 > 2.33 3.31 3 .7 172 Z = = 5.81, pvalue = 1 F Z (5.81) = 1 1.0000 = .0000, therefore, reject H at any common level of alpha. 912 1 : 0; : 0; H H = < 2.91 0 11.33 170 Z = = 3.35, pvalue = 1 F Z (3.35) = 1 .9996 = .0004, therefore, reject H at any common level of alpha. 913 1 : 125.32; : 125.32; H H = reject H if t 15, .05/2  > 2.131 131.78 125.32 25.4 16 t = = 1.017, therefore, do not reject H at the .05 level. 9.14a. No, the 95% confidence level provides for 2.5% of the area in either tail. This does not correspond to a onetailed hypothesis test with an alpha of 5% which has 5% of the area in one of the tails....
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This note was uploaded on 03/25/2009 for the course FIN FIN534 taught by Professor Stevenjordan during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 StevenJordan

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