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: Describe the distribution graphical methods. Is it appropriate to analyze these data using methods based on Normal distributions? Explain why or why not. No, it is not appropriate to analyze these data using methods based on Normal distributions. These data are skewed to the right. 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. 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 left. (b) Find the mean. () 45.12 mpg Find the standard deviation. (Round your answer to four decimal places) 4.2854 mpg Find the standard error. (Round your answer to four decimal places.) Find the margin of error for 95% confidence. (Round your answer to tour decimal places.) mpg (c) Report the 95% confidence interval for mu, the mean mpg for this vehicle based on these data. (Round your answer, to four decimal places.) A simple random sample of 100 postal employees is used to test if the average time postal employees have worked for the postal service has changed from the value of 7.5 years recorded 20 years

Explanation / Answer

a) Yes . it is appropriate; there are no ouliers and no particular skewness

b) mean =45.12

std deviation =4.2854

std error =0.9582

for 95% CI and (n-1=19) degree of freedom ; t=2.0930

margin of error =t*Std error =2.0056

c) 95% CI =sample mean -/+ t*std error =43.1144 ; 47.1256

X 50.4 45.1 46.7 39.4 49.7 45.4 52.4 45.1 49.4 44.3 46.3 46.4 39.4 48.4 48.7 36.3 41.5 45.2 38.8 43.5 mean(X) 45.120 std deviation(S) 4.2854 std error =S/(n)1/2 0.9582

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