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

Answer:

$n=(\frac{2.326(500)}{120})^2 =93.928 \approx 94$

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

Step-by-step 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:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$    (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:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (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:

$n=(\frac{2.326(500)}{120})^2 =93.928 \approx 94$

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