Explanation Here, n = 5 and α = 0.1 . The sample variance is calculated as follows: x ¯ = 785 + 805 + 790 + 793 + 802 5 = 795 s 2 = ( 785 − 795 ) 2 + ( 805 − 795 ) 2 + ( 790 − 795 ) 2 + ( 793 − 795 ) 2 + ( 802 − 795 ) 2 5 − 1 = 69.5 From Table 6: Percentage Points of the χ 2 Distributions, the value corresponding to χ α 2 , n − 1 = 0.05 , 4 2 = 9.4877 and the value corresponding to χ 1 − α 2 , n − 1 = 0.95 , 4 2 = 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 8, Problem 115SE

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Chapter 8 Solutions

Mathematical Statistics with Applications

Ch. 8.2 - Using the identity ()=[E()]+[E()]=[E()]+B(), Show...Ch. 8.2 - a. If is an unbiased estimator for , what is B()?...Ch. 8.2 - Suppose that is an estimator for a parameter and...Ch. 8.2 - Refer to Exercise 8.1. a. If is an unbiased...Ch. 8.2 - Refer to Exercises 8.1 and consider the unbiased...Ch. 8.2 - Suppose that E(1)=E(2)=, V(1)=12, and V(2)=22....Ch. 8.2 - Consider the situation described in Exercise 8.6....Ch. 8.2 - Suppose that Y1, Y2, Y3 denote a random sample...Ch. 8.2 - Suppose that Y1, Y2,, Yn constitute a random...Ch. 8.2 - The number of breakdowns per week for a type of...

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