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A rivet is to be inserted into a hole. A random sample n=15 of parts is...

A rivet is to be inserted into a hole. A random sample n=15 of parts is selected, and the hole diameter is measured. The sample standard deviation of the hole diameter measurements is s=0.008 millimeters. Construct a 99% lower confidence bound for σ2 using MATLAB step by step . Please screenshot the MATLAB screen

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Answer #1

We know that the S is used as a point estimator of the population variance, sigma squared. Therefore, the confidence interval of sigma squared is depend on the sampling distribution of (n-1)S/sigma squared. And the sampling distribution of (n-1)S/sigma squaredis a chi square distribution with (n-1 ) degree of freedom. Thus the confidence interval for sigma squared is given as follows



       (n-1)s2 (n-1)s2 2

Following is the matlab code to find the 99 % confidence interval:

>> n=15; %sample size
s=0.008; % sample standard deviation
alpha = 0.01;
chi2q_up = chi2inv((1-alpha/2),n-1); % calculating 1-alpha/2 th quantile for chi sqaure distribution
chi2q_lw = chi2inv(alpha/2,n-1); % calculating alpha/2 th quantile for chi sqaure distribution
conf_int_lower = ((n-1)*(s^2))/chi2q_up; % lower limit of Confidence interval
conf_int_upper = ((n-1)*(s^2))/chi2q_lw; % lower limit of Confidence interval
Confidence_Interval = [conf_int_lower, conf_int_upper]; % Confidence interval
disp(Confidence_Interval);
   1.0e-03 *

    0.0286    0.2199

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