Answer
$\mu$ is between -25.46 and 106.04. We can see that the outlier really affects the interval, so normally it should be removed when constructing such an interval.
Work Step by Step
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.
