A frozen food company uses a machine that packages okra in six ounce portions. A sample of 54 packages of okra has a variance of 0.44. Construct the 98% confidence interval to estimate the variance of the weights of the packages prepared by the machine. Round your answers to two decimal places. Lower endpoint and Upper endpoint

Answer: Interval would be (0.29,0.58).Step-by-step explanation:Since we have given n = Sample size = 54 Variance = 0.44we need to find 98% confidence interval to estimate the variance.So, we will use "Chi square distribution"For this , we will find df = n-1=54-1=53Interval would be [tex]\dfrac{(n-1)s^2}{\chi^2_{\frac{\alpha }{2}}}<\sigma^2<\dfrac{(n-1)^2s^2}{\chi^2_{1-\frac{\alpha }{2}}}[/tex][tex]\alpha =1-0.98=0.02\\\\\dfrac{\alpha }{2}=\dfrac{0.02}{2}=0.01\\\\So, \chi^2_{0.01,53}=79.84[/tex]similarly,[tex]1-\dfrac{\alpha}{2}=1-0.01=0.99\\\\ \chi^2_{1-\frac{\alpha }{2},df}=\chi^2_{0.99,53}=40.308[/tex]So, it becomes,[tex]\dfrac{53\times 0.44}{79.84}<\sigma^2<\dfrac{53\times 0.44}{40.308}\\\\=0.29<\sigma^2<0.58[/tex]Hence, interval would be (0.29,0.58).