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

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CHAPTER 10 Section 10-2 10-1. a) 1) The parameter of interest is the difference in fill volume µ µ 12 . Note that 0 = 0. 2) H 0 : 0 2 1 = µ or 2 1 = 3) H 1 : 0 2 1 or 2 1 4) α = 0.05 5) The test statistic is z xx nn 0 0 1 2 1 2 2 2 = −− + () σσ 6) Reject H 0 if z 0 < z α /2 = 1.96 or z 0 > z α /2 = 1.96 7) x 1 = 16.014 x 2 = 16.006 σ 0.032 σ 0.024 1 = 2 = n 1 = 10 n 2 = 10 63 . 0 10 ) 024 . 0 ( 10 ) 032 . 0 ( ) 006 . 16 014 . 16 ( 2 2 0 = + = z 8) Because –1.96 < 0.63 < 1.96, do not reject the null hypothesis. There is not sufficient evidence to conclude that the two machine fill volumes differ at α = 0.05. b) P- value = 5287 . 0 ) 7357 . 0 1 ( 2 )) 63 . 0 ( 1 ( 2 = = Φ 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 β α = + Φ + Φ 10 ) 024 . 0 ( 10 ) 032 . 0 ( 04 . 0 96 . 1 10 ) 024 . 0 ( 10 ) 032 . 0 ( 04 . 0 96 . 1 2 2 2 2 = ( ) ( ) ( ) 12 . 5 2 . 1 16 . 3 96 . 1 16 . 3 96 . 1 Φ Φ = Φ Φ = 0.1151 0 = 0.1151 Power = 1 0.1151 = 0.8849 d) ( ) ( ) xx z 2 1 2 1 2 2 2 12 12 2 1 2 1 2 2 2 + ≤−≤−+ + αα µµ // 10 ) 024 . 0 ( 10 ) 032 . 0 ( 96 . 1 006 . 16 014 . 16 10 ) 024 . 0 ( 10 ) 032 . 0 ( 96 . 1 006 . 16 014 . 16 2 2 2 1 2 2 + + + 0328 . 0 0168 . 0 2 1 With 95% confidence, we believe the true difference in the mean fill volumes is between 0.0098 and 0.0298. Because 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 = 0.04 ( ) 96 . 12 ) 04 . 0 ( 024 . 0 032 . 0 645 . 1 96 . 1 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + δ z z n Use n 1 = n 2 = 13 10-1
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10-2. 1) The parameter of interest is the difference in breaking strengths µ µ 12 and 0 = 10 2) H 0 : µµ 10 = 3) H 1 : 10 −> 4) α = 0.05 5) The test statistic is z xx nn 0 0 1 2 1 2 2 2 = −− + () σσ 6) Reject H 0 if z 0 > z α = 1.645 7) x 1 = 162.5 x 2 = 155.0 δ = 10 σ 1.0 σ 1.0 1 = 2 = n 1 = 10 n 2 = 12 z 0 22 162 5 1550 10 10 10 12 584 = + =− (. . ) (.) . 8) Because –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 = + Φ Power = 1 – 0.0012 = 0.9988 6 42 . 5 ) 10 12 ( ) 1 1 ( ) 645 . 1 645 . 1 ( ) ( ) ( ) ( 2 2 2 0 2 2 2 1 2 2 = + + = + + = 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 2) H 0 : or 0 −= = 3) H 1 : or 0 −≠ 4) α = 0.05 5) The test statistic is z 0 0 1 2 1 2 2 2 = + 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 3 3 1 = σ 2 = n = 20 n = 20 1 2 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. b) P -value = 0 ) 1 1 ( 2 )) 32 . 6 ( 1 ( 2 = = Φ 10-2
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c) β σσ αα =− + −− − + Φ ∆∆ Φ z nn z // 2 0 1 2 1 2 2 2 2 0 1 2 1 2 2 2 = ΦΦ 196 25 3 20 3 20 3 20 3 20 22 .
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ch10 - CHAPTER 10 Section 10-2 10-1. a) 1) The parameter of...

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