Explanation A 90% confidence interval for the expected yield is as follows: ( a ′ β ^ − t 0.05 . s a ′ ( X ′ X ) − 1 a , a ′ β ^ + t 0.05 . s a ′ ( X ′ X ) − 1 a ) It is given that T 1 = 50 , P = 20 , C = 1 , and T 2 = 200 . From Part (a) in Exercise 11.74, the coded variables can be obtained as follows: T 1 = 50 ⇒ x 1 = − 1 , P = 20 ⇒ x 2 = 1 , C = 1 ⇒ x 3 = − 1 and T 2 = 200 ⇒ x 4 = 1 The matrix ‘ a ′ ’ can be obtained as follows: a ′ = [ 1 − 1 1 − 1 1 ] From Part (b) in Exercise 11.74, the parameter β ^ can be found as follows: β ^ = [ 21.125 − 3.1375 − 1.2125 − 0.1625 − 1.275 ] The value of a ′ β ^ is as follows: a ′ β ^ = [ 1 − 1 1 − 1 1 ] [ 21.125 − 3.1375 − 1.2125 − 0.1625 − 1.275 ] = 21.9375 Thus, the value of a ′ β ^ is 21.9375. From Exercise 11.74, the values of ( X ′ X ) − 1 , s , n , and k can be found as follows: ( X ′ X ) − 1 = [ 0.0625 0 0 0 0 0 0.0625 0 0 0 0 0 0.0625 0 0 0 0 0 0

Mathematical Statistics with Applications

7th Edition

ISBN: 9780495110811

Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer

Publisher: Cengage Learning

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Chapter 11.12, Problem 75E

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Find a 90% confidence interval for the expected yield.

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Mathematical Statistics with Applications

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Find a 90% confidence interval for the expected yield.

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Find a 90% confidence interval for the expected yield.

Chapter 11 Solutions