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_{((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