## Elementary Statistics (12th Edition)

Unsuccesful applicants: The mean can be counted by summing all the data and dividing it by the number of data: $\frac{34+37+...+60}{23}=46.96.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(34-46.96)^2+...+(60-46.96)^2}{22}}=7.22.$ $\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=23-1=22$. $X_{L}^2= X_{0.995}^2=8.643$ $X_{R}^2= X_{0.005}^2=42.796$ Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(22)\cdot 7.22^2}{42.796}}=5.2$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(22)\cdot 7.22^2}{8.643}}=11.5.$ Succesful applicants: The mean can be counted by summing all the data and dividing it by the number of data: $\frac{33+36+...+54}{29}=44.52.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(33-44.52)^2+...+(54-44.52)^2}{22}}=5.03.$ $\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=29-1=28$. $X_{L}^2= X_{0.995}^2=12.461$ $X_{R}^2= X_{0.005}^2=50.933$ Hence the confidence interval:$\sigma$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(28)\cdot 5.03^2}{50.933}}=3.7$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(28)\cdot 5.03^2}{12.461}}=7.5.$ The standard deviations of the groups don't differ too much because the intervals overlap.