mont4e_sm_ch11_sec04 - Section 11-4 11-21 a) 1) The...

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11-14 Section 11-4 11-21 a) 1) The parameter of interest is the regressor variable coefficient, β 1 2) H 01 0 : β= 3) H 11 0 : β≠ 4) α = 0.05 5) The test statistic is ) 2 /( 1 / 0 = = n SS SS MS MS f E R E R 6) Reject H 0 if f 0 > f α ,1,12 where f 0.05,1,12 = 4.75 7) Using results from Exercise 11-1 123 . 22 59143 . 137 71429 . 159 59 . 137 ) 057143 . 59 ( 3298017 . 2 ˆ 1 = = = = = = R yy E xy R SS S SS S SS β 63 . 74 12 / 123 . 22 59 . 137 0 = = f 8) Since 74.63 > 4.75 reject H 0 and conclude that compressive strength is significant in predicting intrinsic permeability of concrete at α = 0.05. We can therefore conclude that the model specifies a useful linear relationship between these two variables. Pv a l u e −≅ 0 000002 . b) 8436 1 12 123 22 2 2 . . n SS MS ˆ E E = = = = σ and 2696 . 0 3486 . 25 8436 . 1 ˆ ) ˆ ( 2 1 = = = xx S se σ c) 9043 . 0 3486 . 25 0714 . 3 14 1 8436 . 1 1 ˆ ) ˆ ( 2 2 2 0 = + = + = xx S x n se 11-22 a) 1) The parameter of interest is the regressor variable coefficient, β 1 . 2) H 0 : 3) H 0 : 4) α = 0.05 5) The test statistic is ) n /( SS / SS MS MS f E R E R 2 1 0 = = 6) Reject H 0 if f 0 > f α ,1,18 where f 0.05,1,18 = 4.414 7) Using the results from Exercise 11-2 SS S SS S SS Rx y Ey y R == = =− = ± (. ) ( . ) . ) . . . β 1 12 75 20 0 0041612 141445 05886 886 05886 0143275 2 95 73 18 143275 0 5886 0 0 . / . . f = = 8) Since 73.95 > 4.414, reject H 0 and conclude the model specifies a useful relationship at α = 0.05. a l u e 0 000001 .
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11-15 b) 4 2 1 10 8391 . 4 6 . 33991 00796 . ˆ ) ˆ ( = = = x S se xx σ β 04091 . 0 6 . 33991 9 . 73 20 1 00796 . 1 ˆ ) ˆ ( 2 2 0 = + = + = xx S x n se 11-23 a) The regression equation is Rating Pts = - 5.558 + 12.65 Yds per Att S = 5.71252 R-Sq = 78.7% R-Sq(adj) = 78.0% Analysis of Variance Source DF SS MS F P Regression 1 3378.53 3378.53 103.53 0.000 Error 28 913.72 32.63 Total 29 4292.25 Refer to ANOVA 01 . 0 0 : 0 : 1 1 1 0 = = α H H Because p-value = 0.000 < α = 0.01, reject H 0 . We can conclude that there is a linear relationship between these two variables. b) 63 . 32 ˆ 2 = 243 . 1 106 . 21 63 . 32 ˆ ) ˆ ( 2 1 = = = xx S se 159 . 9 106 . 21 318 . 7 30 1 63 . 32 1 ˆ ) ˆ ( 2 2 2 0 = + = + = xx S x n se c) 1)The parameter of interest is the regressor variable coefficient, β 1 . 2) H 01 10 : β= 3) H 11 10 : β≠ 4) α = 0.01 5) The test statistic is ) ˆ ( ˆ 1 0 , 1 1 0 se t = 6) Reject H 0 if t 0 < t α /2,n-2 where t 0.005,28 = 2.763 or t 0 > t 0.005,28 = 2.763 7) Using the results from Exercise 10-6 134 . 2 243 . 1 10 652 . 12 0 = = t 8) Because 2.134 < 2.763, fail to reject H 0 . There is not enough evidence to conclude that the slope differs from 10 at α = 0.01.
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This note was uploaded on 10/12/2009 for the course IND E 315 taught by Professor Kailashkapur during the Spring '09 term at University of Washington.

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mont4e_sm_ch11_sec04 - Section 11-4 11-21 a) 1) The...

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