## Elementary Statistics (12th Edition)

a)$\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=40-1=39$. $X_{L}^2= X_{0.975}^2=20.707$ $X_{R}^2= X_{0.025}^2=66.766$ Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(39)\cdot 10.3^2}{66.766}}=7.87$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(39)\cdot 10.3^2}{66.766}}=14.14.$ b) $\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=40-1=39$. $X_{L}^2= X_{0.975}^2=20.707$ $X_{R}^2= X_{0.025}^2=66.766$ Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(39)\cdot 11.6^2}{66.766}}=8.87$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(39)\cdot 11.6^2}{66.766}}=15.92.$ c)There seems to be no difference in the variation because the intervals in a) and b) overlap.