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# ch05 - CHAPTER 5 Section 5-2 5-1 a 1 The parameter of...

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CHAPTER 5 Section 5-2 5-1. a) 1) The parameter of interest is the difference in fill volume, µ µ 1 2 2) H 0 : or µ µ 1 2 0 = µ µ 1 2 = 3) H 1 : or µ µ 1 2 0 µ µ 1 2 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 1 2 1 2 2 2 = + ( ) σ σ 0 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 δ = 0 σ 0.02 0.025 1 = σ 2 = n 1 = 10 n 2 = 10 z 0 2 2 16 015 16 005 0 0 02 10 0 025 10 0 99 = + = ( . . ) ( . ) ( . ) . 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 = , where 1 − β β σ σ σ σ α α = + + Φ Φ z n n z n n / / 2 0 1 2 1 2 2 2 2 0 1 2 1 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 ( ) ( ) Φ ( ) ( ) = 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 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 = 0.01, and = 0.04 1

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( ) ( ) ( ) ( ) , 08 . 2 ) 04 . 0 ( ) 025 . 0 ( ) 02 . 0 ( 33 . 2 96 . 1 z z 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + δ σ σ β α n n = 11.79, use n 1 = n 2 = 12 5-2. 1) The parameter of interest is the difference in breaking strengths, µ µ 1 2 and 0 = 10 2) H 0 : µ µ 1 2 10 = or µ µ 1 2 = 3) H 1 : µ µ or 1 2 10 > µ µ 1 2 > 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 1 2 1 2 2 2 = + ( ) σ σ 0 6) Reject H 0 if z 0 > z α = 1.645 7) = 1 x 162.7 x 2 = 155.4 δ = 10 σ 1.0 σ 1.0 1 = 2 = n 1 = 10 n 2 = 12 31 . 6 12 ) 0 . 1 ( 10 ) 0 . 1 ( 10 ) 4 . 155 7 . 162 ( z 2 2 0 = + = 8) Since –6.31 < 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. 5-3. a) 1) The parameter of interest is the difference in mean burning rate, µ µ 1 2 2) H 0 : or µ µ 1 2 0 = µ µ 1 2 = 3) H 1 : or µ µ 1 2 0 µ µ 1 2 4) α = 0.05 5) The test statistic is z x x n n 0 1 2 1 2 1 2 2 2 = + ( ) σ σ 0 6) Reject H 0 if z 0 < z α /2 = 1.96 or z 0 > z α /2 = 1.96 7) x 1 = 18.02 x 2 = 24.37 δ = 0 σ 3 3 1 = σ 2 = n 1 = 20 n 2 = 20 = + = 20 ) 3 ( 20 ) 3 ( 0 ) 37 . 24 02 . 18 ( z 2 2 0 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. b) P-value = 0 ) 0 ( 2 )) 70 . 6 ( ( 2 = = Φ c) β σ σ σ σ α α + + Φ Φ z n n z n n / / 2 0 1 2 1 2 2 2 2 0 1 2 1 2 2 2 = 2
= Φ Φ 196 2 5 3 20 3 20 196 2 5 3 20 3 20 2 2 2 2 .

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ch05 - CHAPTER 5 Section 5-2 5-1 a 1 The parameter of...

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