Question

You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the amount a family has spent on food during a year has been approximately $1000. If you want to be 99% sure that you have estimated average family food expenditures within (error) $50, how many families do you need to survey? Round your answer to a whole number

Answer 1

Answer:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer

Step-by-step explanation:

We have the following info:

$ME = 50$ margin of error desired

$\sigma = 1000$ the standard deviation for this case

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =50 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 critical value for 99% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=0.01$. And for this case would be $z_{\alpha/2}=2.58$, replacing into formula (b) we got:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer