A frozen food company uses a machine that packages blueberries in seven pound portions. A sample of 7373 packages of blueberries has a variance of 0.300.30. Construct the 99%99% confidence interval to estimate the variance of the weights of the packages prepared by the machine. Round your answers to two decimal places.

Answer: [tex]0.20<\sigma^2<0.48[/tex]Step-by-step explanation:Given : A A sample of 73 packages of blueberries has a variance of 0.30. i.e. n= 73 and [tex]s^2=0.30[/tex]Confidence level: [tex]\alpha=0.99[/tex] ⇒Significance level : [tex]\alpha=1-0.99=0.01[/tex]Critical values using chi-square distribution:-[tex]\chi^2_{\alpha/2, n-1}=\chi^2_{0.005, 72}=106.65[/tex] [tex]\chi^2_{1-\alpha/2, n-1}=\chi^2_{0.995, 72}=44.84[/tex]99% confidence interval for population variance will be :-[tex]\dfrac{s^2(n-1)}{\chi^2_{\alpha/2, n-1}}<\sigma^2<\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2, n-1}}\\\\\\\dfrac{(0.30)(72)}{106.65}<\sigma^2<\dfrac{(0.30)(72)}{44.84}\\\\ \approx0.20<\sigma^2<0.48[/tex]Hence, the  99% confidence interval to estimate the variance of the weights of the packages prepared by the machine: [tex]0.20<\sigma^2<0.48[/tex]