Elementary Statistics (12th Edition)

The mean can be counted by summing all the data and dividing it by the number of data: $\frac{300+6.5+...+17.5}{10}=40.75.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(300-40.75)^2+...+(17.5-40.75)^2}{9}}=91.28.$ $\alpha=1-0.95=0.05.$ $\sigma$ is unknown, hence we use the t-distribution with $df=sample \ size-1=10-1=9$ in the table. $t_{\alpha/2}=t_{0.025}=2.262.$ Margin of error:$t_{\alpha/2}\cdot\frac{s}{\sqrt {n}}=2.262\cdot\frac{91.28}{\sqrt{10}}\approx65.29.$ Hence the confidence interval:$\mu$ is between 40.75-65.29=-25.46 and 40.75+65.29=106.04. We can see that the outlier really affects the interval, so normally it should be removed when constructing such an interval.