Answer
μ is between 1034.4 and 1226. The population is all daily receipts of the movie. The interval seems to be good, because it is between 950 and 1800.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{963+1027+...+1204}{10}=1130.2.$
Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(963-1130.2)^2+...+(1204-1130.2)^2}{13}}=117.45.$
α=1−0.99=0.01. σ is 16.36, hence we use the z-distribution with df=sample size−1=10−1=9 in the table. $z_{\alpha/2}=z_{0.005}=2.58.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=2.58\cdot\frac{117.45}{\sqrt{10}}=95.8.$ Hence the confidence interval:μ is between 1130.2-95.8=1034.4 and 1130.2+95.8=1226. The population is all daily receipts of the movie. The interval seems to be good, because it is between 950 and 1800.
