Question

The Wall Street Journal reported that automobile crashes cost the United States $162 billion annually (2008 data). 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 100 persons who had been involved in car crashes and that the population standard deviation is σ=$500. What the sample size would you recommend if the study required a 98% confidence level and a margin of error of $120 or less?

Answer 1

`n=((2.326(500))/(120))^2 =93.928 \approx 94`

So the answer for this case would be n=94 rounded up to the nearest integer

Explanation:

Information given

`\bar X` represent the sample mean

`\mu` population mean (variable of interest)

`\sigma= 500` represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

`ME=z_(\alpha/2)(s)/(√(n))` (a)

And on this case we have that ME =120 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

`n=((z_(\alpha/2) \sigma)/(ME))^2` (b)

The confidence2 level is 98% or 0.98 then the significance level would be
`\alpha=1-0.98=0.02` and
`\alpha/2=0.01`, the critical value for this case would be
`z_(\alpha/2)=2.326`, replacing into formula (b) we got:

`n=((2.326(500))/(120))^2 =93.928 \approx 94`

So the answer for this case would be n=94 rounded up to the nearest integer