Question

A meat packaging plant uses a machine that packages chicken livers in seven pound portions. A sample of 96 packages of chicken livers has a standard deviation of 0.37. Construct the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine.

Answer 1

To construct a 95% confidence interval for the standard deviation of the weights of the packages prepared by the machine, use the formula: CI = [sqrt((n-1)*s^2)/sqrt(X^2(n-1)), sqrt((n-1)*s^2)/sqrt(X^2(n-1))].

Step-by-step explanation:

To construct a 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine, we can use the chi-square distribution. The formula for the confidence interval is given by:

CI = [sqrt((n-1)*s^2)/sqrt(X^2(n-1)), sqrt((n-1)*s^2)/sqrt(X^2(n-1))]

where CI is the confidence interval, s is the sample standard deviation, n is the sample size, and X^2 is the chi-square value corresponding to the desired confidence level. In this case, since we want a 95% confidence interval, X^2 at 0.025 significance level and 95 degrees of freedom is approximately 70.062. Substituting the given values, the confidence interval is [sqrt((96-1)*0.37^2)/sqrt(70.062), sqrt((96-1)*0.37^2)/sqrt(70.062)]. You can calculate the values to find the final confidence interval.