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HW 3 Solution

# HW 3 Solution - 10-6 10-18 10-32 10-44 10-56 10-62 10-72...

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10-6, 10-18, 10-32, 10-44, 10-56, 10-62, 10-72, 10-74 10-6 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 = 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 = 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) Because - 6.32 < - 1.96 reject the null hypothesis and conclude the mean burning rates differ significantly at α = 0.05. P -value = 0 ) 1 1 ( 2 )) 32 . 6 ( 1 ( 2 = - = Φ - b) ( 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 18 24 196 3 20 3 20 18 24 196 3 20 3 20 2 2 1 2 2 2 - - + - - + + . ( ) ( ) . ( ) ( ) μ μ - - ≤ - 7 86 414 1 2 . . μ μ We are 95% confident that the mean burning rate for solid fuel propellant 2 exceeds that of propellant 1 by between 4.14 and 7.86 cm/s. c) + - - - Φ - + - - Φ = 2 2 2 1 2 1 0 2 / 2 2 2 1 2 1 0 2 / n n z n n z σ σ σ σ β α α = Φ Φ 196 2 5 3 20 3 20 196 2 5 3 20 3 20 2 2 2 2 . . ( ) ( ) . . ( ) ( ) - + - - - + = ( 29 ( 29 ( 29 ( 29 Φ Φ Φ Φ 196 2 64 196 2 64 0 68 4 6 . . . . . . - - - - = - - - = 0.24825 - 0 = 0.24825 d) Assume the sample sizes are to be equal, use α = 0.05, β = 1-power=0.1, and = 4

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( 29 ( 29 ( 29 ( 29 12 ) 4 ( 3 3 28 . 1 96 . 1 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + 2245 δ σ σ β α z z n Use n 1 = n 2 = 12 10-18 a) 1) The parameter of interest is the difference in mean impact strength, μ μ 1 2 - , with 0 = 0 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 2 2 2 1 2 1 0 2 1 0 ) ( n s n s x x t + - - = 6) Reject the null hypothesis if t 0 < - t α ν , where 23 , 05 . 0 t = 1.714 since 23 72 . 23 1 1 2 2 2 2 1 2 1 2 1 2 2 2 2 1 2 1 2245 = - + - + = ν ν n n s n n s n s n s (truncated) 7) x 1 = 290 x
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