Explanation Given info: Reference- Exercise 38: The data relates to the DNA contents (mg/g liver) in the livers of rats. There are 5 treatments in the form of diets of five different types of carbohydrate, each treatment having observations on 6 rats. Here, x ¯ 1. = 2.58 , x ¯ 2. = 2.63 , x ¯ 3. = 2.13 , x ¯ 4. = 2.41 and x ¯ 5. = 2.49 . Moreover, ∑ ∑ x i j 2 = 183.4 . The carbohydrate types are starch, sucrose, fructose, glucose and maltose. 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 38, the number of treatments is I = 5 and number of observations per treatment, J = 6 . Thus, I ( J − 1 ) = 5 × ( 6 − 1 ) = 5 × 5 = 25. The analysis in Exercise 38 reveals that, MSE = 0 .1076 . Given that, θ = μ 1 − 1 4 ( μ 2 + μ 3 + μ 4 + μ 5 ) = μ 1 − 1 4 μ 2 − 1 4 μ 3 − 1 4 μ 4 − 1 4 μ 5 = μ 1 − 0.25 μ 2 − 0.25 μ 3 − 0.25 μ 4 − 0.25 μ 5 Compare this for the general expression θ = ∑ i c i μ i . Thus, here, c 1 = 1 , c 2 = − 0.25 , c 3 = − 0.25 , c 4 = − 0.25 and c 5 = − 0.25 . As a result, ∑ i c i 2 = ( 1 ) 2 + ( − 0.25 ) 2 + ( − 0.25 ) 2 + ( − 0.25 ) 2 + ( − 0.25 ) 2 = 1 + 0.0625 + 0.0625 + 0.0625 + 0.0625 = 1.25. Use the available values x ¯ 1 ⋅ = 2.58 , x ¯ 2 ⋅ = 2.63 , x ¯ 3 ⋅ = 2

Probability and Statistics for Engineering and the Sciences

9th Edition

ISBN: 9781305251809

Author: Jay L. Devore

Publisher: Cengage Learning

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Chapter 10, Problem 39SE

To determine

Find a 95% t CI for θ=μ1−14(μ2+μ3+μ4+μ5) that measures the difference between the average DNA content for the starch diet and the combined average for the four other diets.

Remark whether the resulting interval includes zero.

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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 = μ 1 − 1 4 ( μ 2 + μ 3 + μ 4 + μ 5 ) that measures the difference between the average DNA content for the starch diet and the combined average for the four other diets. Remark whether the resulting interval includes zero.

Solution Summary: The author explains the data relating to the DNA contents (mg/g liver) in the livers of rats.

Videos

Find a 95% t CI for θ=μ1−14(μ2+μ3+μ4+μ5) that measures the difference between the average DNA content for the starch diet and the combined average for the four other diets.

Remark whether the resulting interval includes zero.

Chapter 10 Solutions

Find a 95% t CI for θ = μ 1 − 1 4 ( μ 2 + μ 3 + μ 4 + μ 5 ) that measures the difference between the average DNA content for the starch diet and the combined average for the four other diets. Remark whether the resulting interval includes zero.

Solution Summary: The author explains the data relating to the DNA contents (mg/g liver) in the livers of rats.

Videos

Find a 95% t CI for θ=μ1−14(μ2+μ3+μ4+μ5) that measures the difference between the average DNA content for the starch diet and the combined average for the four other diets. Remark whether the resulting interval includes zero.

Chapter 10 Solutions

Find a 95% t CI for θ=μ1−14(μ2+μ3+μ4+μ5) that measures the difference between the average DNA content for the starch diet and the combined average for the four other diets.

Remark whether the resulting interval includes zero.