Computers in some vehicles calculate various quantities related to performance.

Question

Computers 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 equipped 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. There are no outliers and the data is skewed to the right. No, it is not appropriate to analyze these data using methods based on Normal distributions. These data are skewed to the left. Yes, it is appropriate to analyze these data using methods based on Normal distributions. There are no outliers and no particular skewness. No, it is not appropriate to analyze these data using methods based on Normal distributions. There are extreme outliers. No, it is not appropriate to analyze these data using methods based on Normal distributions. These data are skewed to the right. (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 appropriate,........there are no outliers and no particluar skewness

b)

mean =43.17

std deviation=4.3074

std error =0.9632

for 95% CI and 19 df ; t=2.0930

margin of error =t*std error =2.0159

95% confidence interval =sample mean -/+ t*Std error =41.1541 ; 45.1859

X 44.9 48 43.6 43.2 37.4 50.5 46.2 36.7 42.2 43.1 48.3 47.4 37.5 41.6 44.6 34.3 43.3 46.7 39.5 44.4 mean(X) 43.170 std deviation(S) 4.307 std error =S/(n)1/2 0.963

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