Chapter_5_Recommended_Homework_Solutions

Chapter_5_Recommended_Homework_Solutions - Chapter 5...

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Chapter 5 Recommended Homework Solutions 5-4. x 1 = 30.61 x 2 = 30.34 σ 1 = 0.10 σ 2 = 0.15 n 1 = 12 n 2 = 10 a) 90% two-sided confidence interval: ( 29 ( 29 x x z n n x x z n n 1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 - - + - - + + α α σ σ μ μ σ σ / / ( 29 10 ) 15 . 0 ( 12 ) 10 . 0 ( 645 . 1 34 . 30 61 . 30 10 ) 15 . 0 ( 12 ) 10 . 0 ( 645 . 1 ) 34 . 30 61 . 30 ( 2 2 2 1 2 2 + + - - + - - μ μ 361 . 0 179 . 0 2 1 - μ μ We are 90% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.179 and 0.361 fl. oz. b) 95% two-sided confidence interval: ( 29 ( 29 x x z n n x x z n n 1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 - - + - - + + α α σ σ μ μ σ σ / / ( 29 10 ) 15 . 0 ( 12 ) 10 . 0 ( 96 . 1 34 . 30 61 . 30 10 ) 15 . 0 ( 12 ) 10 . 0 ( 96 . 1 ) 34 . 30 61 . 30 ( 2 2 2 1 2 2 + + - - + - - μ μ 379 . 0 161 . 0 2 1 - μ μ We are 95% confident that the mean fill volume for machine 1 exceeds that of machine 2 by between 0.161 and 0.379 fl. oz. Comparison of parts a and b: As the level of confidence increases, the interval width also increases (with all other values held constant). c) 95% upper-sided confidence interval: ( 29 μ μ σ σ α 1 2 1 2 1 2 1 2 2 2 - - + + x x z n n ( 29 10 ) 15 . 0 ( 12 ) 10 . 0 ( 645 . 1 34 . 30 61 . 30 2 2 2 1 + + - - μ μ 361 . 0 2 1 - μ μ With 95% confidence, we believe the fill volume for machine 1 exceeds the fill volume of machine 2 by no more than 0.361 fl. oz. 5-10. Catalyst 1 Catalyst 2 x 1 = 63.56 x 2 = 67.81 σ 1 = 3 σ 2 = 3 n 1 = 10 n 2 = 10 a) 95% confidence interval on μ μ 1 2 - , the difference in mean active concentration ( 29 ( 29 x x z n n x x z n n 1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 - - + - - + + α α σ σ μ μ σ σ / / ( 29 2 2 2 2 1 2 (3) (3) (3) (3) (63.56 67.81) 1.96μ μ 63.56 67.81 1.96 10 10 10 10 - - + - - + + 1 2 6.88μ μ 1.62 - - -
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b) Yes, since the 95% confidence interval did not contain the value 0, we would conclude the mean active concentration depends on the choice of catalyst. 5-12. 1) The parameter of interest is the difference in mean active concentration, μ μ 1 2 - 2) H 0 : μ μ 1 2 0 - = or μ μ 1 2 = 3) H 1 : μ μ 1 2 0 - or μ μ 1 2 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 0 1 2 1 2 2 2 = - - + ( ) σ σ 6) Reject H 0 if z 0 < - z α /2 = - 1.96 or z 0 > z α /2 = 1.96 7) x 1 = 63.56 x 2 = 67.81 δ = 0 σ 1 = 3 σ 2 = 3 n 1 = 10 n 2 = 10 0 2 2 (63.56 67.81) 0 3.17 (3) (3) 10 10 z - - = = - + 8) Since - 3.17 < - 1.96 reject the null hypothesis and conclude the mean active concentrations do differ significantly at α = 0.05. P-value = 2 (1 (3.17)) 2(1 0.99924) 0.00152 = - = The conclusions reached by the confidence interval of problem 5-10 and the test of hypothesis conducted here are the same. A two-sided confidence interval can be thought of as representing the “acceptance region” of a hypothesis test, given that the level of significance is the same for both procedures. Thus if the value δ falls outside the confidence interval, it is the same result as rejecting the null hypothesis. 5-17. a) 1) The parameter of interest is the difference in mean etch rate, μ μ 1 2 - 2) H 0 : μ μ 1 2 0 - = or μ μ 1 2 = 3) H 1 : μ μ 1 2 0 - or μ μ 1 2 4) α = 0.05 5) The test statistic is t x x s n n p 0 1 2 0 1 2 1 1 = - - + ( ) 6) Reject the null hypothesis if t 0 < - + - t n n α / , 2 2 1 2 where - t 0 025 18 . , = - 2.101 or t 0 > t n n α / , 2 2 1 2 + - where t 0 025 18 . , = 2.101 7) x 1 = 9.97 x 2 = 10.4 0 = 0 s n s n s n n p = - + - + - ( ) ( ) 1 1 2 2 2 2 1 2 1 1 2 s 1 = 0.422 s 2 = 0.231 = + = 10 0 422 10 0 231 18 0 340 2 2 ( . ) ( . ) .
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