Question

2. Billy needs to buy a car. He is interested in a certain model but he is worried about its gas consumption. Billy gets data from the car constructor who claims having run independent tests about the fuel economy of this model, expressed below in miles per gallon: 21.1, 23.4, 35.9, 25.5, 25.9, 37.0, 28.2, 16.1, 20.2. The car constructor claims that fuel economy measurements follow a normal distribution with a standard deviation of 7 miles per gallon. (a) Compute the sample estimate of the mean fuel economy. (b) Find the 95% confidence interval for the mean fuel economy, based on the car constructor’s claim. (c) With the same data, what would be the confidence level (1 − α) associated with a confidence interval of length (i.e. twice the margin of error) of 6.4 miles per gallon? (d) Compute the 95% confidence interval again, but now using the sample variance instead of the constructor’s value. How does this interval compare to the one based on the constructor’s claim?

Answer 1

a)

sample mean = 25.9222

b) n = 9

confidence interval

( - z* / , + z* / )

z = 1.96

(25.9222 -1.96 * 7/sqrt(9) , 25.9222 +1.96 * 7/sqrt(9) )

=( 21.3489, 30.4955 )

c)

interval length = 6.4

margin of error = 3.2

(25.9222 - 3.2 , 25.9222 + 3.2)

=( 22.7222 , 29.1222 )

d)

var(x) =

48.30944

95 percent confidence interval:
 20.57960 31.26485