Explanation Given info: Reference- Exercise 5: The data provides with the summary data on the modulus of elasticity ( × 10 6 psi ) for lumbar of three different grades. Here, X ¯ 1. = 1.63 , X ¯ 2. = 1.56 , X ¯ 3. = 1.42 , s 1 = 0.27 , s 2 = 0.24 , s 3 = 0.26 . Calculation: Confidence interval for parametric functions : The 100 ( 1 − α ) % confidence interval for a function of μ i ’s, θ = ∑ i c i μ i , where c i ’s are constants for i = 1 , 2 , ... , I , is given as: ∑ i c i x ¯ i ± t α 2 , I ( J − 1 ) MSE ∑ i c i 2 J . Here, MSE is the mean square of error. From Exercise 5, the number of treatments is I = 3 and number of observations per treatment, J = 10 . Thus, I ( J − 1 ) = 3 × ( 10 − 1 ) = 3 × 9 = 27. The analysis in Exercise 5 reveals that, MSE = 0 .066 . Given that, θ = 1 2 ( μ 1 + μ 2 ) − μ 3 = 1 2 μ 1 + 1 2 μ 2 − μ 3 = 0.5 μ 1 + 0.5 μ 2 − μ 3 . Compare this for the general expression θ = ∑ i c i μ i . Thus, here, c 1 = 0.5 , c 2 = 0.5 and c 3 = − 1 . As a result, ∑ i c i 2 = ( 0.5 ) 2 + ( 0.5 ) 2 + ( − 1 ) 2 = 0.25 + 0.25 + 1 = 1.5. Use the available values X ¯ 1 ⋅ = 1.63 , X ¯ 2 ⋅ = 1.56 , X ¯ 3 ⋅ = 1

Probability and Statistics for Engineering and the Sciences

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ISBN: 9781305251809

Author: Jay L. Devore

Publisher: Cengage Learning

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Chapter 10.2, Problem 17E

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Find a 95% t CI for θ=12(μ1+μ2)−μ3.

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Chapter 10 Solutions

Probability and Statistics for Engineering and the Sciences

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Find a 95% t CI for θ = 1 2 ( μ 1 + μ 2 ) − μ 3 .

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