Question

If the manager of a bottled water distributor wants to​ estimate, with 95​% ​confidence, the mean amount of water in a​ 1-gallon bottle to within ± 0.004 gallons and also assumes that the standard deviation is 0.04 ​gallons, what sample size is​ needed? (Round up to the nearest​ integer)

If the inspection division of a county weights and measures department wants to estimate the mean amount of​ soft-drink fill in​ 2-liter bottles to within ± 0.02 liter with 90​% confidence and also assumes that the standard deviation is .09 ​liters, what sample size is​ needed? (Round up to the nearest​ integer.)  

Answer 1

Solution: We can use the formula for finding sample size(n)

$n=\left [ \frac{Z_{\alpha/2 }*\sigma }{E} \right ]^{2}$

where, $\sigma$ is a standard deviation, E is margin of error and $Z_{\alpha /2}$ is critical value.

(a) Given that

$E=\pm 0.004$ , $S.D(\sigma )=0.04$ and critical value at 5% level of significance (using Z-table) will be 1.96

Therefore,

$n=\left [ \frac{1.96*0.04}{0.004} \right ]^{2}=384.16$

rounding up you get n = 384

(b) Given that

$E=\pm 0.02$ , $S.D(\sigma )=0.09$ and critical value at 10% level of significance (using Z-table) will be 1.64

Therefore,

$n=\left [ \frac{1.64*0.09}{0.02} \right ]^{2}=54.46$

rounding up you get n = 54