A manufacturer turns out a product item that is labeled either defective or not defective. a....

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?

Solutions

Expert Solution

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

orchestra answered 2 years ago

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