Question

A car company claimed that it improved the gas mileage of a particular model of car between 2011 and 2012. A consumer advocacy group obtains a random sample of 16 cars from 2011 and finds the average number of miles driven on exactly 25 gallons of gas in laboratory conditions is 574. 4 miles, with a standard deviation of 43. 1 miles. The consumer advocacy group also obtains a random sample of 24 cars from 2012 and finds the average number of miles driven on exactly 25 gallons of gas in laboratory
conditions is 604. 6 miles, with a standard deviation of 27. 8 miles. The 99% confidence interval for the difference in mean miles driven on 25 gallons of gas (2012 model – 2011 model) is (–3. 95, 64. 35).


Does this confidence interval support the company’s assertion that their gas mileage has increased?


a)Yes, we are 99% confident that the true mean difference in miles driven per 25 gallons is 30. 2 miles.


b)Yes, the fact that most of the values in the confidence interval are positive means that the true mean number of miles driven on 25 gallons is higher in 2012.


c)Yes, the 2012 sample mean of 604. 4 miles per 25 gallons is 30. 2 more miles than the 2011 sample mean of 574. 4, which is an increase of 1. 208 miles per gallon.


d)No, the confidence interval contains 0, so we are 99% confident that the true mean increase in number of miles driven on 25 gallons is 0.


e)No, the confidence interval contains 0, so 0 is a plausible value for the true mean increase in number of miles driven on 25 gallons. An increase of 0 would indicate the gas mileage has not increased

Answer 1

Using the interpretation of the confidence interval, it is found that the correct option is:

e) No, the confidence interval contains 0, so 0 is a plausible value for the true mean increase in number of miles driven on 25 gallons. An increase of 0 would indicate the gas mileage has not increased.

What is the interpretation of a x% confidence interval?

It means that we are x% confident that the population parameter(mean/proportion/standard deviation) is between a and b.

In this question, the 99% confidence interval for the increase is of (-3.95, 64.35). It contains 0 and negative values, hence those are plausible values, and this is not enough evidence that the mean has increased, hence option e is correct.

More can be learned about confidence intervals at brainly.com/question/25890103