Question

A quality control process for boxes of cereal production measures the weight of a box of cereal. The population standard deviation is known to be .0911 ounces. In order to achieve a 96% confidence with a margin of error of .02 ounces, how many boxes of cereal should be sampled? Report your answer as a whole number.

Answer:  n =  boxes

Answer 1

Solution :

Given that,

standard deviation = $\sigma$ = 0.911

margin of error = E = 0.02

At 96% confidence level the z is ,

  $\alpha$ = 1 - 96% = 1 - 0.96 = 0.04

$\alpha$ / 2 = 0.04 / 2 = 0.02

Z_$\alpha$/2 = Z_0.02 = 2.054

Sample size = n = ((Z_$\alpha$/2 * $\sigma$ ) / E)²

= ((2.054 * 0.911 ) / 0.02)²

= 8753

Answer: Sample size = 8753 boxes