Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

in some vehicles calculate various quantities related to performance. One of the

ID: 3179098 • Letter: I

Question

in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg), For one vehicle in this way, the mpg were recorded each time the gas tank was filled, and the computer was then reset. Here are the mpg values for a random sample of 20 of these records: (a) Describe the distribution using graphical methods. Is it appropriate to analyze these data using methods based on Normal distributions? Explain why or why not. Yes, it is appropriate to analyze these data using methods based on Normal distributions. These data are skewed to the right. Yes, it is appropriate to analyze theses data using methods based on normal distributions. There are no outliers and the data is skewness. No, it is not appropriate to analyze these data using methods based on Normal distributions. These are extreme outliers. No, it is not appropriate to analyze these data using methods based on Normal distributions. These data are skewed to the left. (b) Find the mean. () _____ mpg Find the standard deviation. (Round your answer to four decimal places.) _____mpg Find the standard error. (Round your answer to four decimal places.) _____ mpg Find the margin of error for 95% confidence. (Round your answer to four decimal places.) ______ mpg (c) Report the 95% confidence interval for mu, the mean mpg for this vehicle based on these data. (Round your answers to four decimal places.) (_____ mpg, ______ mpg)

Explanation / Answer

a) Yes , it is... option C

b) mean =42.15

c)std deviation =4.3082

d)std error =0.9633

e)for 19 degree of freedom and 0.05 level; t=2.093

hence margin of error =t*std error =2.0163

f)confidence interval =sample mean +/- margin of error =40.1337 ; 44.1663

X 47.300 45.300 43.800 43.200 42.300 42.100 43.600 33.300 49.600 42.600 47.000 46.400 42.100 36.500 45.600 36.500 41.200 38.500 35.700 40.400 mean(X) 42.150 std deviation(S) 4.308 std error =S/(n)1/2 0.963