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

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:

margin of error desired

the standard deviation for this case

The margin of error is given by this formula:

   (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:

  (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance is . And for this case would be , replacing into formula (b) we got:

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

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step-by-step explanation:

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