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 based on a sample of 41 persons who had been involved in car crashes and that the population standard deviation is sigma = $550. 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? The input in the box below will not be graded, but may be reviewed and considered by your instructor.

Explanation / Answer

Now the margin of error is given as

Z* sigma/sqrt(n) , where sigma is population standard deviation , n is sample size and Z is the multiplier

so using the z table for 95% confidence we get the value as 1.96

putting the values in the given formula above

we get the value as

1.96*550/sqrt(41) = 168.35

now if we want the margin of error to decrease for the given SD and confidence interval , using the equation again

where MOE = 150 , we can solve for N as

1.96*550/sqrt(N) = 150

(1.96*550/150)^2 = N

so N = 51.64 = 52 approx , hence the sample size must be increased to 52 from 41 to get the desired level of MOE

Hope this helps

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