Answer
$\mu$ is between 0.9618 and 1.4066.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{0.7+2.2+...+1.39}{50}=1.1842.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(0.7-1.1842)^2+...+(1.39-1.1842)^2}{49}}=0.5873.$
$\alpha=1-0.99=0.01.$ $\sigma$ is unknown, hence we use the t-distribution with $df=sample \ size-1=50-1=49$ in the table. $t_{\alpha/2}=t_{0.005}=2.678.$ Margin of error:$z_{\alpha/2}\cdot\frac{s}{\sqrt {n}}=2.678\cdot\frac{0.5873}{\sqrt{50}}=0.2224.$ Hence the confidence interval:$\mu$ is between 1.1842-0.2224=0.9618 and 1.1842+0.2224=1.4066.
