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 50 persons who had been involved in car crashes and that the population standard deviation is σ = $600. What is the margin of error for a 95% confidence interval? Round your answer to two decimal places.

Answer 1

Answer:

Error = 166.308459

Explanation:

Given that,

Standard deviation (SD) = $600

Sample (n) = 50 persons

Error(E) = ?

For a 95% confidence interval,

Z-value from the z-table is 1.959963985

$E=\frac{z\times SD}{\sqrt{n}}$

$E=\frac{1.959963985\times 600}{\sqrt{50}}$

$E=\frac{1,175.97839}{7.07106781}$

         = 166.308459

Therefore, the margin of error for a 95% confidence interval is 166.308459.