Answer
$\mu$ is between 4.67 and 28.05. The population is all daily receipts of the movie. A problem can be that it is a convenience sample and hence we cannot deduct anything to the population.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{58+22+...+4}{14}=16.36.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(58-16.36)^2+...+(4-16.36)^2}{13}}=14.52.$ $\alpha=1-0.99=0.01.$ $\sigma$ is 16.36, hence we use the z-distribution with $df=sample \ size-1=14-1=13$ in the table. $t_{\alpha/2}=t_{0.005}=3.012.$ Margin of error:$t_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=3.012\cdot\frac{14.52}{\sqrt{14}}=11.69.$ Hence the confidence interval:$\mu$ is between 16.36-11.69=4.67 and 16.36+11.69=28.05 The population is all daily receipts of the movie. A problem can be that it is a convenience sample and hence we cannot deduct anything to the population.
