Question

a. A random sample of 43 cars in the drive-thru of a popular fast food restaurant revealed an average bill of $18.58 per car. The population standard deviation is $6.22. Estimate the mean bill for all cars from the drive-thru with 97% confidence. Round intermediate and final answers to two decimal places.

Answer 1

Given

`\begin{gathered} n=43 \\ Mean\text{ = \$18.58} \\ \sigma=\text{ \$6.22} \end{gathered}`

Solution

Formula

`\text{Confident interval =M }\pm\frac{Z\sigma}{\sqrt[]{n}}`

where

`\begin{gathered} M=\text{ mean or Average} \\ Z-score=Z_(97)=2.17 \\ n=43 \end{gathered}`

Substitute the parameters into the Confident Interval formula

`\text{Confident interval =18.58}\pm\frac{2.17*6.22}{\sqrt[]{43}}`

Then we calculate the Addition and subtraction

First the Addition

`\begin{gathered} \text{Confident interval =18.58+}\frac{2.17*6.22}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58+}\frac{13.4974}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58+}(13.4974)/(6.5574) \\ \text{Confident interval =18.58+}2.05833 \\ \text{Confident interval =}20.63833342 \\ \\ \text{Confident interval =}20.64\text{ two decimal places} \end{gathered}`

Then now for subtraction

`\begin{gathered} \text{Confident interval =18.58-}\frac{2.17*6.22}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58-}\frac{13.4974}{\sqrt[]{43}} \\ \\ \text{Confident interval =18.58-}(13.4974)/(6.5574) \\ \text{Confident interval =18.58-}2.05833 \\ \text{Confident interval =}16.5216658 \\ \\ \text{Confident interval =16.52 two decimal places} \end{gathered}`

The final answer

`(16.52,\text{ 20.64)}`