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# ch10 - CHAPTER 10 Section 10-2 10-1 a 1 The parameter of...

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10-1 CHAPTER 10 Section 10-2 10-1. a) 1) The parameter of interest is the difference in fill volume, μ μ 1 2 - ( note that Δ 0 =0) 2) H 0 : 0 2 1 = - μ μ or 2 1 μ μ = 3) H 1 : 0 2 1 - μ μ or 2 1 μ μ 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 0 1 2 1 2 2 2 = - - + ( ) Δ 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 = 16.015 x 2 = 16.005 σ 1 = 0.02 σ 2 = 0.025 n 1 = 10 n 2 = 10 99 . 0 10 ) 025 . 0 ( 10 ) 02 . 0 ( ) 005 . 16 015 . 16 ( 2 2 0 = + - = z 8) since -1.96 < 0.99 < 1.96, do not reject the null hypothesis and conclude there is no evidence that the two machine fill volumes differ at α = 0.05. b) P-value = 2 1 0 99 2 1 08389 0 3222 ( ( . )) ( . ) . - = - = Φ c) Power = 1 - β , where β σ σ σ σ α α = - - + ° ± ² ² ² ² ² ³ ´ µ µ µ µ µ - - - - + ° ± ² ² ² ² ² ³ ´ µ µ µ µ µ Φ Δ Δ Φ Δ Δ z n n z n n / / 2 0 1 2 1 2 2 2 2 0 1 2 1 2 2 2 = µ µ µ µ µ ´ ³ ² ² ² ² ² ± ° + - - Φ - µ µ µ µ µ ´ ³ ² ² ² ² ² ± ° + - Φ 10 ) 025 . 0 ( 10 ) 02 . 0 ( 04 . 0 96 . 1 10 ) 025 . 0 ( 10 ) 02 . 0 ( 04 . 0 96 . 1 2 2 2 2 = ( ) ( ) ( ) ( ) 91 . 5 99 . 1 95 . 3 96 . 1 95 . 3 96 . 1 - Φ - - Φ = - - Φ - - Φ = 0.0233 - 0 = 0.0233 Power = 1 - 0.0233 = 0.9967 d) ( ) ( ) 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 / / ( ) ( ) 16 015 16 005 196 0 02 10 0 025 10 16 015 16 005 196 0 02 10 0 025 10 2 2 1 2 2 2 . . . ( . ) ( . ) . . . ( . ) ( . ) - - + - - + + μ μ - - 0 0098 0 0298 1 2 . . μ μ With 95% confidence, we believe the true difference in the mean fill volumes is between - 0.0098 and 0.0298. Since 0 is contained in this interval, we can conclude there is no significant difference between the means. e) Assume the sample sizes are to be equal, use α = 0.05, β = 0.05, and Δ = 0.04

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10-2 ( ) ( ) ( ) ( ) , 35 . 8 ) 04 . 0 ( ) 025 . 0 ( ) 02 . 0 ( 645 . 1 96 . 1 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + δ σ σ β α z z n n = 9, use n 1 = n 2 = 9 10-2. 1) The parameter of interest is the difference in breaking strengths, μ μ 1 2 - and Δ 0 = 10 2) H 0 : μ μ 1 2 10 - = 3) H 1 : μ μ 1 2 10 - > 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 0 1 2 1 2 2 2 = - - + ( ) Δ σ σ 6) Reject H 0 if z 0 > z α = 1.645 7) x 1 = 162.5 x 2 = 155.0 δ = 10 σ 1 = 1.0 σ 2 = 1.0 n 1 = 10 n 2 = 12 z 0 2 2 162 5 1550 10 10 10 10 12 584 = - - + = - ( . . ) ( . ) ( . ) . 8) Since -5.84 < 1.645 do not reject the null hypothesis and conclude there is insufficient evidence to support the use of plastic 1 at α = 0.05. 10-3 β = ( ) 0012 . 0 03 . 3 12 1 10 1 ) 10 12 ( 645 . 1 = - Φ = µ µ µ µ ´ ³ ² ² ² ² ± ° + - - Φ , Power = 1 – 0.0012 = 0.9988 ° 1 6 42 . 5 ) 10 12 ( ) 1 1 ( ) 645 . 1 645 . 1 ( ) ( ) ( ) ( 2 2 2 0 2 2 2 1 2 2 = - + + = Δ - Δ + + = σ σ β α z z n Yes, the sample size is adequate 10-4. a) 1) The parameter of interest is the difference in mean burning rate, μ μ 1 2 - 2) H 0 : μ μ 1 2 0 - = or μ μ 1 2 =
10-3 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 = - - + ( ) Δ 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 = 18 x 2 = 24 σ 1 = 3 σ 2 = 3 n 1 = 20 n 2 = 20 32 . 6 20 ) 3 ( 20 ) 3 ( ) 24 18 ( 2 2 0 - = + - = z 8) Since - 6.32 < - 1.96 reject the null hypothesis and conclude the mean burning rates differ significantly at α = 0.05. b) P-value = 0 ) 1 1 ( 2 )) 32 . 6 ( 1 ( 2 = - = Φ - c) β σ σ σ σ α α = - - + ° ± ² ² ² ² ² ³ ´ µ µ µ µ µ - - - - + ° ± ² ² ² ² ² ³ ´ µ µ µ µ µ Φ Δ Δ Φ Δ Δ z n n z n n / / 2 0 1 2 1 2 2 2 2 0 1 2 1 2 2 2 = Φ Φ 196 2 5 3 20 3 20 196 2 5 3 20 3 20 2 2 2 2 . . ( ) ( ) . . ( ) ( ) - + ° ± ² ² ² ² ² ³ ´ µ µ µ µ µ - - - + ° ± ² ² ² ² ² ³ ´ µ µ µ µ µ = ( ) ( ) ( ) ( ) Φ Φ Φ Φ 196 2 64 196 2 64 0 68 4 6 . . . . . . - - - - = - - - = 0.24825 - 0 = 0.24825 d) ( ) ( ) 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 / / ( ) ( ) 18 24 196 3 20 3 20 18 24 196 3 20 3 20 2 2 1 2 2 2 - - + - - + + .

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