Question

A company wants to estimate the mean net weight of all 32-ounce packages of its yummy taste cookies at 95% confidence. it is known that the standard deviation of net weights is 0.1 ounce. the sample size that will yield the margin of error within 0.02 ounces of the population mean is

Answer 1

standard deviation= σ= 0.1
Confidence interval= 95%= 0.95
Significance level=α= 1-0.95= 0.05
α/2 = 0.025
At α/2 z( α/2 ) =1.960 [from statistic table]
Now,
the sample size at error of 0.02 is
0.02=z( α/2 )σ/√n
0.02=(1.960)*(0.1)/√n
5/49 = 1/√n
√n=49/5
n=96.04
n should be 96 or greater