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

The number of the families that need to be surveyed will be 2663.

What is the margin of error?

Your results will deviate from the actual population value by how many percentage points, as indicated by the margin of error.

We have the following info:

ME = 50 margins of error desired

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

The margin of error is given by this formula:

(a) $ME=Z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

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:

(b) $n =(\dfrac{Z_{a\2}\sigma}{ME})^2$

The critical value for the 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:

$n=(\dfrac{2.58(1000)}{50})^2= 2663$

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

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