A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds. Unfortunately, the company has a problem with the variability of the weight. In a sample of 11 of the bowling balls the sample standard deviation was found to be 0.71 pounds. Construct a 95% confidence interval for the variance of the bowling ball weight. Assume normality.

Answer: [tex]0.2461<\sigma^2<1.5511[/tex]Step-by-step explanation:Given : A company produces a women's bowling ball that is supposed to weigh exactly 14 pounds. Sample size : n=11Degree of freedom =n-1=10Sample standard deviation : s= 0.71 poundsSignificance level for 95% confidence interval :[tex]\alpha=1-0.95=0.05[/tex]We assume that the bowling ball weight is normally distributed.Using chi-square distribution table, the required critical values are :-[tex]\chi^2_{df, \alpha/2}=20.48[/tex][tex]\chi^2_{df, 1-\alpha/2}=3.25[/tex]Then, the 95% confidence interval for the variance of the bowling ball weight will be :[tex]\dfrac{s^2(n-1)}{\chi_{\alpha/2}}<\sigma^2<\dfrac{s^2(n-1)}{\chi_{1-\alpha/2}}[/tex][tex]=\dfrac{(0.71)^2(10)}{20.48}<\sigma^2<\dfrac{(0.71)^2(10)}{3.25}\\\\=0.2461<\sigma^2<1.5511[/tex]∴ The 95% confidence interval for the variance of the bowling ball weight will be : [tex]0.2461<\sigma^2<1.5511[/tex]