In: Statistics and Probability
A manufacturer turns out a product item that is labeled either defective or not defective.
a. if the manufacturer wants to estimate the percentage of defective products produced with 99% confidence to within 0.03 of the true value, what is the sample size they should take?
Solution :
Given that,
= 0.5 ( assume 0.5)
1 - = 1 - 0.5 = 0.5
margin of error = E = 0.03
At 99% confidence level the z is,
= 1 - 99%
= 1 - 0.99 = 0.01
/2 = 0.005
Z/2 = 2.576 ( Using z table )
Sample size = n = (Z/2 / E)2 * * (1 - )
= (2.576 / 0.03)2 * 0.5 * 0.5
=1843.27
Sample size = 1844