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

Answer:

a) Margin of error = 166.311

b) sample size ≥ 62

Step-by-step explanation:

Given:

Average cost = $1599

Sample size = 50 persons

Standard deviation = $600

Confidence level is 95%

a) Margin of error = $z\frac{\sigma}{\sqrt{n}}$

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

Thus,

Margin of error = $1.96\frac{600}{\sqrt{50}}$

or

Margin of error = 166.311

b) For Margin of error ≤ 150

$z\frac{\sigma}{\sqrt{n}}$ ≤ 150

or

$1.96\frac{600}{\sqrt{n}}$ ≤ 150

or

$1.96\frac{600}{150}$ ≤ √n

or

√n ≥ 7.84

or

n ≥ 61.4656

Therefore,

sample size ≥ 62