A commercial farm uses a machine that packages carrots in eighteen ounce portions. A sample of 7 packages of carrots has a standard deviation of 0.19. Construct the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine. Round your answers to two decimal places.

Answer: [tex]0.12< \sigma<0.42[/tex]Step-by-step explanation:Confidence interval for standard deviation is given by :-[tex]\sqrt{\dfrac{s^2(n-1)}{\chi^2_{\alpha/2}}}< \sigma<\sqrt{\dfrac{s^2(n-1)}{\chi^2_{1-\alpha/2}}}[/tex]Given : Confidence level : [tex]1-\alpha=0.95[/tex]⇒[tex]\alpha=0.05[/tex]Sample size : n= 7Degree of freedom = 6    (df= n-1) sample standard deviation : s= 0.19Critical values by using chi-square distribution table :[tex]\chi^2_{\alpha/2, df}}=\chi^2_{0.025, 6}}=14.4494\\\\\chi^2_{1-\alpha/2, df}}=\chi^2_{0.975, 6}}=1.2373[/tex]Confidence interval for standard deviation of the weights of the packages prepared by the machine is given by :-[tex]\sqrt{\dfrac{ 0.19^2(6)}{14.4494}}< \sigma<\sqrt{\dfrac{ 0.19^2(6)}{1.2373}}[/tex][tex]\Rightarrow0.12243< \sigma<0.418400[/tex][tex]\approx0.12< \sigma<0.42[/tex]Hence, the 95% confidence interval to estimate the standard deviation of the weights of the packages prepared by the machine.  : [tex]0.12< \sigma<0.42[/tex]