Question

Ou 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 σ = $1200. If you want to be 99% sure that you have estimated average family food expenditures within $60, how many families do you need to survey? Place your answer, a whole number, do not use any decimals, in the blank___.

Answer 1

Answer:

The answer is "2653".

Step-by-step explanation:

$Sample \ Size = 99\% \ Cl\\\\ E =60 \\\\\sigma= 1200$

Minimum sample size for Cl level $= 99\%$ and Desired Margin of Error, $E = 60$ is:  

$\to P(|\bar{x}-\mu| < E) \geq 1-\alpha \\\\\to P(\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$

$Margin \ Error = \frac{\text{length of the CI}}{2}\\\\\sigma=1200\\\\\alpha=1 - Confidence =1-0.99=0.01\\\\Z_{\frac{\alpha}{2}}=z_{0.005}=2.58\\\\n\geq (\frac{z_{0.005} \times \sigma }{E})^2 \\\\= (\frac{2.58 \times 1200 }{60})^2\\\\=2662.560$

The minimum n has to be integer, we take the ceiling Of above number and get n = 2663

The exact z-value.

$n \geq (\frac{2.5758293035489 \times 1200}{60})^2 \\\\= 2653.95864$

using critical value of 2,575, which gives 2652.25  

$\to n =2653$