## Elementary Statistics (12th Edition)

a) The mean can be counted by summing all the data and dividing it by the number of data: $\frac{6.5+6.6+...+7.7}{10}=7.15.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(6.5-7.15)^2+...+(7.7-7.15)^2}{9}}=0.4767.$ $\alpha=1-0.95=0.05.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=10-1=9$. $X_{L}^2= X_{0.975}^2=2.7$ $X_{R}^2= X_{0.025}^2=19.023$ Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(9)\cdot 0.4767^2}{19.023}}=0.33$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(9)\cdot 0.4767^2}{2.7}}=0.87.$ b) The mean can be counted by summing all the data and dividing it by the number of data: $\frac{4.2+5.4+...+10}{10}=7.15.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(4.2-7.15)^2+...+(10-7.15)^2}{9}}=1.8216.$ $\alpha=1-0.95=0.05.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=10-1=9$. $X_{L}^2= X_{0.975}^2=2.7$ $X_{R}^2= X_{0.025}^2=19.023$ Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(9)\cdot 1.8216^2}{19.023}}=1.25$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(9)\cdot 1.8216^2}{2.7}}=3.33.$ c) The two confidence intervals in a) and b) don't overlap, hence the answer is yes. The single-line system has a smaller standard deviation, although they both have the same mean, hence it is better.