Question

On this problem, the answer has been worked out, but you must fill in the blanks in the solution.A random sample of 50 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.21 per car. The population standard deviation is $5.92. Estimate the mean bill for all cars from the drive-thru with 99% confidence.

Answer 1

Given the formula;

$\bar{x}\pm z_{\frac{\frac{a}{2}}{2}}(\frac{\sigma}{\sqrt[]{n}})$$\begin{gathered} Given; \\ \bar{x}=18.21 \\ z-\text{score for 99\%= 2.576} \\ \sigma=5.92 \\ n=50 \end{gathered}$

Then, we have;

$18.21\pm2.576(\frac{5.92}{\sqrt[]{50}})\ldots\ldots\ldots\ldots\ldots\text{equation 1}$$\begin{gathered} \text{ \ 18.21}\pm\text{ \ 2.16} \\ \text{ \ 18.21-\ 2.16=\ 16.05} \\ \text{and} \\ \text{\ 18.21+\ 2.16=\ 20.37} \end{gathered}$

Thus, the confidence interval is;

$\text{ \ 16.05}\leq\mu\leq\text{ \ 20.37}$

First box: 18.21

Second box: 2.576

Third box: 5.92

Fourth box: 50

Fifth box: 16.05

Sixth box: 20.37