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 33 persons who had been involved in car crashes and that the population standard deviation is σ = $450.

What is the margin of error for a 95% confidence interval? Round your answer to two decimal places.

Answer 1

Given that, the sample size ( n ) = 33

population standard deviation $(\sigma)$ = $450

A 95% confidence level has significance level $(\alpha)$ = 0.05 and critical value is, $Z_{\alpha/2} = 1.96$

We want to find, the margin of error ( E ),

$E = Z_{\alpha/2} * \frac { \sigma}{\sqrt { n }} = 1.96 * \frac { 450}{\sqrt { 33}} = 153.54$

Margin of error is 153.54