Question

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

Answer 1

a) Margin of error = 166.311

b) sample size ≥ 62

Explanation:

Given:

Average cost = $1599

Sample size = 50 persons

Standard deviation = $600

Confidence level is 95%

a) Margin of error =
`z(\sigma)/(√(n))`

Now for confidence level of 95% z-value = 1.96

Thus,

Margin of error =
`1.96(600)/(√(50))`

or

Margin of error = 166.311

b) For Margin of error ≤ 150

`z(\sigma)/(√(n))` ≤ 150

or

`1.96(600)/(√(n))` ≤ 150

or

`1.96(600)/(150)` ≤ √n

or

√n ≥ 7.84

or

n ≥ 61.4656

Therefore,

sample size ≥ 62