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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

Respuesta :

Answer:

[tex]n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663[/tex]

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:

[tex] ME = 50[/tex] margin of error desired

[tex]\sigma = 1000[/tex] the standard deviation for this case

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]    (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:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

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

[tex]n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663[/tex]

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

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