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:  $(27.81,\ 29.39)$

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 :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $28.6\pm (1.96)\dfrac{2.2}{\sqrt{30}}$

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 = $(27.81,\ 29.39)$