Question

Fuel Efficiency of Cars and Trucks Since 1975 the average fuel efficiency of U.S. cars and light trucks (SUVs) has increased from 13.5 to 25.8 mpg, an increase of over 90%! A random sample of 40 cars from a large community got a mean mileage of 28.1 mpg per vehicle. The population standard deviation is 4.7 mpg. Estimate the true mean gas mileage with 95% confidence.

Answer 1

Answer:

$28.1-1.96\frac{4.7}{\sqrt{40}}=26.64$    

$28.1+ 1.96\frac{4.7}{\sqrt{40}}=29.56$    

We are confident at 95% of confidence that the true mean for the mpg is between 26.64 and 29.56. And since the lower limit from the confidence interval is highert than 25.8 then we can conclude that we have a significant increase from 1975

Step-by-step explanation:

Information given

$\bar X= 28.1$ represent the sample mean

$\mu$ population mean

$\sigma =4.7$ represent the population standard deviation

n=40 represent the sample size  

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

The Confidence level is is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$, and we can calculate the critical value using the normal standard distribution and we got $z_{\alpha/2}=1.96$

And replacing we got:

$28.1-1.96\frac{4.7}{\sqrt{40}}=26.64$    

$28.1+ 1.96\frac{4.7}{\sqrt{40}}=29.56$    

We are confident at 95% of confidence that the true mean for the mpg is between 26.64 and 29.56. And since the lower limit from the confidence interval is highert than 25.8 then we can conclude that we have a significant increase from 1975