Ch_5_4th_Ed

Ch_5_4th_Ed - CHAPTER 5 Section 5-2 5-1 a The parameter of interest is the difference in fill volume μ μ 1 2 H μ μ 1 2 = or μ μ 1 2 = H 1 μ

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Unformatted text preview: CHAPTER 5 Section 5-2 5-1. a) The parameter of interest is the difference in fill volume, μ μ 1 2- H : μ μ 1 2- = or μ μ 1 2 = H 1 : μ μ 1 2- ≠ or μ μ 1 2 ≠ The test statistic is z x x n n 1 2 1 2 1 2 2 2 =-- + ( ) ∆ σ σ x 1 = 16.015 x 2 = 16.005 δ = 0 σ 1 = 0.02 σ 2 = 0.025 n 1 = 10 n 2 = 10 z 2 2 16 015 16 005 0 02 10 0 025 10 0 99 =-- + = ( . . ) ( . ) ( . ) . P-value = 2 1 0 99 2 1 08389 0 3222 ( ( . )) ( . ) .- =- = Φ . Since p-value is greater than 0.05, we do not reject the null hypothesis. b) Power = 1- β , where β σ σ σ σ α α =-- + ---- + Φ ∆ ∆ Φ ∆ ∆ z n n z n n / / 2 1 2 1 2 2 2 2 1 2 1 2 2 2 = +-- Φ- +- Φ 10 ) 025 . ( 10 ) 02 . ( 04 . 96 . 1 10 ) 025 . ( 10 ) 02 . ( 04 . 96 . 1 2 2 2 2 = ( 29 ( 29 ( 29 ( 29 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 = 0.977 c) ( 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 ( 29 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. d) Assume the sample sizes are to be equal, use α = 0.05, β = 0.01, and ∆ = 0.04 ( 29 ( 29 ( 29 ( 29 , 08 . 2 ) 04 . ( ) 025 . ( ) 02 . ( 33 . 2 96 . 1 z z n 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + 2245 δ σ σ β α n = 11.79, use n 1 = n 2 = 12 5-2. The parameter of interest is the difference in breaking strengths, μ μ 1 2- and ∆ = 10 1 H : μ μ 1 2 10- = or μ μ 1 2 = H 1 : μ μ 1 2 10- or μ μ 1 2 The test statistic is z x x n n 1 2 1 2 1 2 2 2 =-- + ( ) ∆ σ σ = 1 x 162.7 x 2 = 155.4 δ = 10 σ 1 = 1.0 σ 2 = 1.0 n 1 = 10 n 2 = 12 31 . 6 12 ) . 1 ( 10 ) . 1 ( 10 ) 4 . 155 7 . 162 ( z 2 2- = +-- = P-value = 1 1 )) 31 . 6 ( 1 ( 2245- =- Φ- . Since p-value is bigger than 0.05, we do not reject the null hypothesis. 5-3. a) 1) The parameter of interest is the difference in mean burning rate, μ μ 1 2- 2) H : μ μ 1 2- = or μ μ 1 2 = 3) H 1 : μ μ 1 2- ≠ or μ μ 1 2 ≠ 4) α = 0.05 5) The test statistic is z x x n n 1 2 1 2 1 2 2 2 =-- + ( ) ∆ σ σ 6) Reject H if z < - z α /2 = - 1.96 or z > z α /2 = 1.96 7) x 1 = 18.02 x 2 = 24.37 δ = 0 σ 1 = 3 σ 2 = 3 n 1 = 20 n 2 = 20- = +-- = 20 ) 3 ( 20 ) 3 ( ) 37 . 24 02 . 18 ( z 2 2 6.70 8) Since - 6.70 < - 1.96 reject the null hypothesis and conclude the mean burning rates do not differ significantly at α = 0.05....
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This note was uploaded on 12/11/2009 for the course CSE IEE taught by Professor Chattin during the Summer '09 term at ASU.

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Ch_5_4th_Ed - CHAPTER 5 Section 5-2 5-1 a The parameter of interest is the difference in fill volume μ μ 1 2 H μ μ 1 2 = or μ μ 1 2 = H 1 μ

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