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

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Unformatted text preview: 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) 2) H : 2 1 =-μ or 2 1 = 3) H 1 : 2 1 ≠-or 2 1 ≠ 4) α = 0.05 5) The test statistic is z x x n n 1 2 1 2 1 2 2 2 =--+ ( ) Δ σ σ 6) Reject H if z < -z α /2 = -1.96 or z > 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 . 10 ) 025 . ( 10 ) 02 . ( ) 005 . 16 015 . 16 ( 2 2 = +-= 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 =--+ & ¡ ¢ ¢ ¢ ¢ ¢ £ ¤ ¥ ¥ ¥ ¥ ¥----+ & ¡ ¢ ¢ ¢ ¢ ¢ £ ¤ ¥ ¥ ¥ ¥ ¥ Φ Δ Δ Φ Δ Δ 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 = ( ) ( ) ( ) ( ) 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 10-2 ( )( ) ( ) ( ) , 35 . 8 ) 04 . ( ) 025 . ( ) 02 . ( 645 . 1 96 . 1 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + ≅ δ 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 = 10 2) H : 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 1 2 1 2 1 2 2 2 =--+ ( ) Δ σ σ 6) Reject H if z > 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 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 . 03 . 3 12 1 10 1 ) 10 12 ( 645 . 1 =-Φ = & & & & ¡ ¢ £ £ £ £ ¤ ¥ +--Φ , Power = 1 – 0.0012 = 0.9988 & 1 6 42 . 5 ) 10 12 ( ) 1 1 ( ) 645 . 1 645 . 1 ( ) ( ) ( ) ( 2 2 2 2 2 2 1 2 2 ≅ =-+ + = Δ-Δ + + = 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 : μ μ 1 2-= or μ μ 1 2 = 10-3 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 =--+ ( ) Δ σ σ...
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This note was uploaded on 04/09/2008 for the course MTHSC 301 taught by Professor Any during the Spring '08 term at Clemson.

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

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