Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 7 - Estimates and Sample Sizes - 7-3 Estimating a Population Mean - Page 352: 22

Answer

$\mu$ is between 1009.5 and 1250.9. 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=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(963-1130.2)^2+...+(1027-1130.2)^2}{9}}=117.45.$ $\alpha=1-0.99=0.01.$ $\sigma$ is unknown, hence we use the t-distribution with $df=sample \ size-1=10-1=9$ in the table. $t_{\alpha/2}=t_{0.005}=3.25.$ Margin of error:$t_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=3.25\cdot\frac{117.45}{\sqrt{10}}=120.7$ Hence the confidence interval:$\mu$ is between 1130.2-120.7=1009.5 and 1130.2+120.7=1250.9. The population is all daily receipts of the movie. The interval seems to be good, because it is between 950 and 1800.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.