Chapter10

# Chapter10 - CHAPTER 10 Section 10.1 1. a. Ho will be...

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295 CHAPTER 10 Section 10.1 1. a. H o will be rejected if 06 . 3 15 , 4 , 05 . = F f (since I – 1 = 4, and I ( J – 1 ) = (5)(3) = 15 ). The computed value of F is 44 . 2 2 . 1094 3 . 2673 = = = MSE MSTr f . Since 2.44 is not 06 . 3 , H o is not rejected. The data does not indicate a difference in the mean tensile strengths of the different types of copper wires. b. 06 . 3 15 , 4 , 05 . = F and 36 . 2 15 , 4 , 10 . = F , and our computed value of 2.44 is between those values, it can be said that .05 < p-value < .10. 2. Type of Box x s 1 713.00 46.55 2 756.93 40.34 3 698.07 37.20 4 682.02 39.87 Grand mean = 712.51 ( 29 ( 29 ( 29 [ 2 2 2 51 . 712 07 . 698 51 . 712 93 . 756 51 . 712 00 . 713 1 4 6 - + - + - - = MSTr ( 29 ] 0604 . 223 , 6 51 . 712 02 . 682 2 = - + ( 29 ( 29 ( 29 [ ( 29 ] 9188 . 691 , 1 87 . 39 20 . 37 34 . 40 55 . 46 4 1 2 2 2 2 = + + + = MSE 678 . 3 9188 . 691 , 1 0604 . 223 , 6 = = = MSE MSTr f 10 . 3 20 , 3 , 05 . = F 3.678 > 3.10, so reject H o . There is a difference in compression strengths among the four box types.

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The Analysis of Variance 296 3. With = i m true average lumen output for brand i bulbs, we wish to test 3 2 1 0 : m m m = = H versus : a H at least two s i ' m are unequal. 60 . 295 2 2 . 591 ˆ 2 = = = B MSTr s , 30 . 227 21 3 . 4773 ˆ 2 = = = W MSE s , so 30 . 1 30 . 227 60 . 295 = = = MSE MSTr f For finding the p-value, we need degrees of freedom I – 1 = 2 and I ( J – 1) = 21. In the 2 nd row and 21 st column of Table A.9, we see that 57 . 2 30 . 1 21 , 2 , 10 . = < F , so the p-value > .10. Since .10 is not < .05 , we cannot reject H o . There are no differences in the average lumen outputs among the three brands of bulbs. 4. ( 29 08 . 166 19 . 5 32 = = = x IJ x , so ( 29 95 . 49 32 08 . 166 91 . 911 2 = - = SST . ( 29 ( 29 ] [ 38 . 20 19 . 5 36 . 6 ... 19 . 5 39 . 4 8 2 2 = - + + - = SSTr , so 57 . 29 38 . 20 95 . 49 = - = SSE . Then 43 . 6 28 57 . 29 3 38 . 20 = = f . Since 95 . 2 43 . 6 28 , 2 , 05 . = F , 4 3 2 1 0 : m m m m = = = H is rejected at level .05. There are differences between at least two average flight times for the four treatments. 5. = i m true mean modulus of elasticity for grade i (i = 1, 2, 3). We test 3 2 1 0 : m m m = = H vs. : a H at least two s i ' m are unequal. Reject H o if 49 . 5 27 , 2 , 01 . = F f . The grand mean = 1.5367, ( 29 ( 29 ( 29 ] [ 1143 . 5367 . 1 42 . 1 5367 . 1 56 . 1 5367 . 1 63 . 1 2 10 2 2 2 = - + - + - = MSTr ( 29 ( 29 ( 29 [ ] 0660 . 26 . 24 . 27 . 3 1 2 2 2 = + + = MSE , 73 . 1 0660 . 1143 . = = = MSE MSTr f . Fail to reject H o . The three grades do not appear to differ. 6. Source Df SS MS F Treatments 3 509.112 169.707 10.85 Error 36 563.134 15.643 Total 39 1,072.256 51 . 4 30 , 3 , 01 . 36 , 3 , 01 . = F
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## This note was uploaded on 03/30/2008 for the course STAT 211 taught by Professor Parzen during the Spring '07 term at Texas A&M.

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Chapter10 - CHAPTER 10 Section 10.1 1. a. Ho will be...

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