Question

(1 point) Suppose we want a 99% confidence interval for the average amount spent on books by freshmen in their first year at college. The amount spent has a normal distribution with standard deviation $30. How large should the sample be if the margin of error is to be less than $4? ANSWER:

Answer 1

We have the margin of error as,

$Z_{.005}\frac{\sigma}{\sqrt{n}}$

Also $\sigma$ = 30. Also for 99% CI, Z = 2.576

Hence want to determine n such that,

30 2.576- n <4

On rearranging and taking squares we have,

$n \geq (\frac{2.576*30}{4})^2$

Hence n>=373.26

Hence minimum sample size required is 374.