(b) Compute the 95% confidence interval estimate of the population variance.A confidence interval for the population variance o? will have the following form where n is the sample sizeand s? is the sample variance. The x2 values correspond to a chi-square distribution with n - 1 degrees ofafreedom where " and 1 - " are the upper tail areas.2(n – 1)s2X1- a/2(n - 1)s2X a/2Recall that the value of alpha is found by setting the confidence level equal to (1 - a) and solving for a. A95% confidence interval is to be found. Expressing 95% as a probability gives 0.95, so we have 1 - a = 0.95.aTherefore, a = 0.05, so " =and 122The sample size is n = 20, so the degrees of freedom is n - 1 =SubmitSkip (you cannot come back)

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(b) Compute the 95% confidence interval estimate of the population variance. A confidence interval for the population variance o? will have the following form where n is the sample size and s2 is the sample variance. The x2 values correspond to a chi-square distribution with n - 1 degrees of freedom where and 1 - a are the upper tail areas. 2 (n - 1)s? (n – 1)s2 so2 s X1- a/2 X a/2 Recall that the value of alpha is found by setting the confidence level equal to (1 – a) and solving for a. A 95% confidence interval is to be found. Expressing 95% as a probability gives 0.95, so we have 1 - a = 0.95. a Therefore, a = 0.05, so . 2 a and 1 - 2 The sample size is n = 20, so the degrees of freedom is n - 1 =

(b) Compute the 95% confidence interval estimate of the population variance. A confidence interval for the population variance o? will have the following form where n is the sample size and s2 is the sample variance. The x2 values correspond to a chi-square distribution with n - 1 degrees of freedom where and 1 - a are the upper tail areas. 2 (n - 1)s? (n – 1)s2 so2 s X1- a/2 X a/2 Recall that the value of alpha is found by setting the confidence level equal to (1 – a) and solving for a. A 95% confidence interval is to be found. Expressing 95% as a probability gives 0.95, so we have 1 - a = 0.95. a Therefore, a = 0.05, so . 2 a and 1 - 2 The sample size is n = 20, so the degrees of freedom is n - 1 =

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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