Hw_13_Soln_07 - Andrea Gruber Jery Stedinger Fall,07...

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Andrea Gruber & Jery Stedinger Fall ‘07 HOMEWORK # 13 SOULTIONS # 33 Supplemental 15, Devore 7, p 623 (Devore 6 p 694) Observations: 1.7 3.3 5.1 6.9 12.6 14.4 16.4 24.6 26.0 26.5 32.1 37.4 40.1 40.5 41.5 72.4 80.1 86.4 87.5 100.2 Test 0 : 25.0 H μ = % versus : 25.0 a H μ % . The test statistic is Y = the # of obs. that exceed 25. a) Consider rejecting H0 if Y>=15. What is the value of α for this test? [Hint: Think of a “success” as a lifetime that exceeds 25.0. Then Y is the number of successes in the sample.] What kind of a distribution does Y have when 25.0 μ = % ? With “success” as defined, then Y is a binomial with n = 20. To determine the binomial proportion “p” we realize that since 25 is the hypothesized median, 50% of the distribution should be above 25, thus p = .50. From the Binomial Tables (Table A.1) with n = 20 and p = .50, we see that ( 29 ( 29 021 . 979 . 1 14 1 15 = - = - = = Y P Y P α . b) What rejection region of the form Y>=c specifies a test with a significance level as close to 0.05 as possible? Use this region to carry out the test for the given data. [Note: The test statistic is the number of differences Xi-25 that have positive signs, hence the name sign test.] From the same binomial table as in a , we find that ( 29 ( 29 058 . 942 . 1 13 1 14 = - = - = Y P Y P (close as we can get to .05), so c = 14. For this data, we would reject H o at level .058 if 14 Y . Y = (the number of observations in the sample that exceed 25) = 12, and since 12 is not 14 , we fail to reject H o . # 5 Section 15.1, Devore 7, p607 (Devore 6 p 676) Data : Sample: 1 2 3 4 5 6 7 8 9 10 11 12 Grav: 54.7 58.5 66.8 46.1 52.3 74.3 92.5 40.2 87.3 74.8 63.2 68.5 Spec: 55.0 55.7 62.9 45.5 51.1 75.4 89.6 38.4 86.8 72.5 62.3 66.0 Use the Wilcoxon test to decide whether one technique gives on average a different value than the other technique for this type of material . H o : μ D = 0 H a : μ D ≠ 0 For D i = X i – Y i , where X = gravimetric method, Y = spectrophotometric method S + = sum of ranks of positive differences With n = 12 and 05 . = α , from Table A.13, reject H o when S + ≥ 64 or S + ( 29 14 64 2 13 12 = - d i -.3 2.8 3.9 .6 1.2 -1.1 2.9 1.8 .5 2.3 .9 2.5 rank 1 10* 12* 3* 6* 5 11* 7* 2* 8* 4* 9*
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Andrea Gruber & Jery Stedinger Fall ‘07 Actual significance level α = 5.2% We observe S + = 72, and 72 > 64, therefore, we reject H o at the 5% level In fact for 01 . = α , n = 12, the critical value is c = 71, so even at level .01 H o would be rejected.
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  • Fall '08
  • Stedinger
  • Normal Distribution, Statistical hypothesis testing, Andrea Gruber, Jery Stedinger, Jery Stedinger Fall

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