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chap17_2010 - Chapter 17 Estimation and Hypothesis Tests...

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Chapter 17 Estimation and Hypothesis Tests: Two Populations
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Two Independent Populations 2 2 1 1 2 2 n1 n2
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Difference in Means: Variance Known 2 1 2 1 μ μ μ - = - X X 2 2 2 1 2 1 2 1 n n X X σ σ σ + = -
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Testing Hypothesis on the Difference in Means: Variance Known Alt. Hypothesis P-value Rejection Criterion H1:   0 P(z>z0)+P(z<- z0) z0 > z1-/2 or z0 < z/2 H1:  > 0 P(z>z0) z0 > z1- H1:  < 0 P(z<-z0) z0 < z Null Hypothesis: H0: 1 - 2 = 0 Test statistic: 2 2 2 1 2 1 0 2 1 0 ) ( n n X X Z σ σ + - - =
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Type II Error and Sample Size ) ( ) ( 2 2 2 1 2 1 0 2 2 2 1 2 1 0 1 2 2 n n z n n z σ σ σ σ β α α + - - Φ - + - - Φ = - 2 0 2 2 2 1 2 1 1 ) ( ) ( ) ( 2 - + + - - σ σ β α z z n
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Type II Error (OC Curve) Two-sided, =.05 Montgomery, D.C., Runger G.C., “Applied Statistics and Probability for Engineers”, 5th Ed., 2010, John Wiley
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Claim: Drying times of 2 different paints are same n1=n2=10, 1= 2 =8, =.05 Null Hypothesis: H0: 1 - 2 = 0 Alt. Hypothesis: H1: 1 > 2 Test statistic: Rejection region: z.95= 1.645 P-value = P(z>2.52)= .0059 Reject H0 .95 .05 C Non-rejection Region Rejection Region 9 1.645 Testing Hypothesis on the Difference of Means with Variance Known – Example 112 121 2 1 = = x x 52 . 2 10 8 10 8 ) 112 121 ( ) ( 2 2 2 2 2 1 2 1 0 2 1 0 = + - = + - - = n n X X Z σ σ
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Confidence Interval for Difference in Means: Variance Known Confidence Interval One-sided Upper Confidence Bound One-sided Lower Confidence Bound 2 2 2 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 1 2 2 ) ( ) ( n n z x x n n z x x σ σ μ μ σ σ α α + + - - + + - - 2 2 2 1 2 1 1 2 1 2 1 ) ( n n z x x σ σ μ μ α + + - - - 2 1 2 2 2 1 2 1 2 1 ) ( μ μ σ σ α - + + - n n z x x
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Choice of Sample Size If n1 = n2 = n ) ( ) ( 2 2 2 1 2 0 1 2 σ σ α + - = - z n
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Tensile Strength n1=10 n2=12, 1=1 2 =1.5, =.10 Confidence Interval on the Difference of Means with Variance Known – Example 5 . 74 6 . 87 2 1 = = x x 2 2 2 1 2 1 1 2 1 2 1 2 2 2 1 2 1 2 1 2 2 ) ( ) ( n n z x x n n z x x σ σ μ μ σ σ α α + + - - + + - - 12 5 . 1 10 1 ) 645 . 1 ( ) 5 . 74 6 . 87 ( 12 5 . 1 10 1 ) 645 . 1 ( ) 5 . 74 6 . 87 ( 2 2 2 1 2 2 + + - - + - + - μ μ 98 . 13 22 . 12 2 1 - μ μ
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Two Independent Populations Variance Unknown 1 2 n1 n2 Case 1: 12=22=2 2 ) , min( ) , max( 2 1 2 1 s s s s
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Difference in Means: Variance Unknown (but equal) Pooled Estimator of 2 Student – t 2 1 2 1 μ μ μ - = - X X 2 ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 - + - + - = n n s n s n s p
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Testing Hypothesis on the Difference in Means: Variance Unknown (but equal) Null Hypothesis: H0: 1 - 2 = 0 Test statistic: Alt. Hypothesis P-value Rejection Criterion H1:   0 2*P(t>|t0|) t0 > t/2,n1+n2-2 or t0 < -t/2, n1+n2-2 H1:  > 0 P(t>t0) t0 > t, n1+n2-2 H1:  < 0 P(t<-t0) t0 <- t, n1+n2-2 2 1 0 2 1 0 1 1 ) ( n n s X X T p + - - =
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B12 in Energy drink n1=14, n2=16, s1=4.7, s2=3.9, =.01 Null Hypothesis: H0: 1 - 2 = 0 Alt. Hypothesis: H1: 1 ≠ 2 Test statistic: Rejection region: ± t.005,28= ±2.763 P-value = 2P(t>|3.822|)= .000676 Reject H0 Errors in book!
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