Chapter15 - CHAPTER 15 Section 15.1 1. H 0 : = 100 vs. H a...

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457 CHAPTER 15 Section 15.1 1. We test 100 : 0 = m H vs. 100 : m a H . The test statistic is s + = sum of the ranks associated with the positive values of ) 100 ( - i x , and we reject H o at significance level .05 if 64 + s . (from Table A.13, n = 12, with 026 . 2 / = a , which is close to the desired value of . 025), or if ( 29 14 64 78 64 2 13 12 = - = - + s . i x ) 100 ( - i x ranks 105.6 5.6 7* 90.9 -9.1 12 91.2 -8.8 11 96.9 -3.1 3 96.5 -3.5 5 91.3 -8.7 10 100.1 0.1 1* 105 5 6* 99.6 -0.4 2 107.7 7.7 9* 103.3 3.3 4* 92.4 -7.6 8 S + = 27, and since 27 is neither 64 nor 14 , we do not reject H o . There is not enough evidence to suggest that the mean is something other than 100. 2. We test 25 : 0 = m H vs. 25 : m a H . With n = 5 and 03 . a , reject H o if 15 + s . From the table below we arrive at s + =1+5+2+3 = 11, which is not 15 , so do not reject H o . It is still plausible that the mean = 25. i x ) 25 ( - i x ranks 25.8 0.8 1* 36.6 11.6 5* 26.3 1.3 2* 21.8 -3.2 4 27.2 2.2 3*
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Chapter 15: Distribution-Free Procedures 458 3. We test 39 . 7 : 0 = m H vs. 39 . 7 : m a H , so a two tailed test is appropriate. With n = 14 and 025 . 2 / = a , Table A.13 indicates that H o should be rejected if either 21 84 + or s . The ) 39 . 7 ( - i x ’s are -.37, -.04, -.05, -.22, -.11, .38, -.30, -.17, .06, -.44, .01, -.29, -.07, and -.25, from which the ranks of the three positive differences are 1, 4, and 13. Since 21 18 = + s , H o is rejected at level .05. 4. The appropriate test is 30 : 0 = m H vs. 30 : < m a H . With n = 15, and 10 . = a , reject H o if ( 29 37 83 2 16 15 = - + s . i x ) 30 ( - i x ranks i x ) 30 ( - i x ranks 30.6 0.6 3* 31.9 1.9 5* 30.1 0.1 1* 53.2 23.2 15* 15.6 -14.4 12 12.5 -17.5 13 26.7 -3.3 7 23.2 -6.8 11 27.1 -2.9 6 8.8 -21.2 14 25.4 -4.6 8 24.9 -5.1 10 35 5 9* 30.2 0.2 2* 30.8 0.8 4* S + = 39, which is not 37 , so H o cannot be rejected. There is not enough evidence to prove that diagnostic time is less than 30 minutes at the 10% significance level. 5. The data is paired, and we wish to test 0 : 0 = D H m vs. 0 : D a H m . With n = 12 and 05 . = a , H o should be rejected if either 64 + s or if 14 + s . 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* 72 = + s , and 64 72 , so H o is rejected at level .05. In fact for 01 . = a , the critical value is c = 71, so even at level .01 H o would be rejected.
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Chapter 15: Distribution-Free Procedures 459 6. We wish to test 5 : 0 = D H m vs. 5 : D a H m , where white black D m m m - = . With n = 9 and 05 . a , H o will be rejected if 37 + s . As given in the table below, 37 = + s , which is 37 , so we can (barely) reject H o at level approximately .05, and we conclude that the greater illumination does decrease task completion time by more than 5 seconds. i
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Chapter15 - CHAPTER 15 Section 15.1 1. H 0 : = 100 vs. H a...

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