The Wall Street Journal reported that automobile crashes cost the United States

Question

The Wall Street Journal reported that automobile crashes cost the United States $162 billion annually (The Wall Street Journal, March 5, 2008). The average cost per person for crashes in the Tampa, Florida, area was reported to be $1599. Suppose this average cost was basec on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is sigma = $600. What is the margin of error for a 95% confidence interval? What would you recommend if the study required a margin of error of $150 or less?

Explanation / Answer

WE NEED TO FIND THE 95% MARGIN OF ERROR

= 0.05, so z/2 =Z0.025 = 1.96 from the table of Normal distribution.

THE FORMULA TO BE USED = 1.96*STANDARD DEVATION/SQRT(N)

N = 50

STANDARD DEVIATION = 600

HENCE MARGIN OF ERROR = 1.96*600/SQRT(50)

= 166.31

B) IF WE NEED TO HAVE MARGIN OF ERROR = 150 OR LESS

WE NEED TO FINS THE N

THEREFORE

150 = 1.96*600/SQRT(N)

THEREFORE

SQRT(N) = 1.96*600/150

SQUARING BOTH SIDES

N = (1.96*600/150)^2

N = 61.46 = 61

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