Question

A mileage test is conducted for a new car model. Thirty randomly selected cars are driven for a month and the mileage is measured for cach. The mean mileage for the sample is 28.6 miles per gallon (mpg) and the sample standard deviation is 2.2 mpg Estimate a 95% confidence interval for the mean mpg in the entire population of that car model.

Answer 1

Answer:
`(27.81,\ 29.39)`

Step-by-step explanation:

Given : Sample size : n= 30 , it means it is a large sample (n≥ 30), so we use z-test .

Significance level :
`\alpha: 1-0.95=0.05`

Critical value:
`z_(\alpha/2)=1.96`

Sample mean :
`\overline{x}=28.6`

Standard deviation :
`\sigma=2.2`

The formula to find the confidence interval is given by :-

`\overline{x}\pm z_(\alpha/2)(\sigma)/(√(n))`

i.e.
`28.6\pm (1.96)(2.2)/(√(30))`

i.e.
`28.6\pm 0.787259889321`

`\approx28.6\pm 0.79=(28.6-0.79,28.6+0.79)=(27.81,\ 29.39)`

Hence, the 95% confidence interval for the mean mpg in the entire population of that car model =
`(27.81,\ 29.39)`