Answer to Question #241518 in Statistics and Probability for Vainqueur

Question

Suppose a production facility purchases a particular component part in large lots from a supplier. The production manager wants to estimate the proportion of defective is no more than 20 and want to be within 0.2 of the Trey proportion of defective parts with a 90% level of confidence. How large a sample should she take

Answer 1

QUETION

Suppose a production facility purchases a particular component part in large lots from a supplier. The production manager wants to estimate the proportion of defective parts received from this supplier. She believes the proportion defective is no more than 0.22 and wants to be within 0.02 of the true proportion of defective parts with a 90% level of confidence. How large a sample should she take?

SOLUTION

In a sample with a number n of people surveyed with a probability of a success of $\pi$ and a confidence level of $1-\alpha,$ we have the following:

confidence interval of proportions $\pi+z\sqrt\frac{\pi (1- \pi)}{n}$

In which z is the z score that has a p-value of $1-\frac{\alpha}{2}$

The margin of error for the interval is: $z=\sqrt\frac{\pi(1-\pi)}{n}$

We have $\pi=0.22$ and $\alpha=0.1$

90% confidence level $Z=1-\frac{0.1}2=0.95$

So, $Z=1.645$

How large a sample should she take?

We need a sample size of at least n.

n is found when $M=0.02$

So,

$M=z\sqrt\frac{\pi(1-\pi)}{n}$

$0.02=1.645\sqrt\frac{0.22*0.78}{n}$

$0.02\sqrt n=1.645\sqrt{0.22*0.78}$

$\sqrt n=\frac{1.645\sqrt{0.22*0.78}}{0.02}$

$(\sqrt n)^2=\Big(\frac{1.645\sqrt{0.22*0.78}}{0.02}\Big)^2\\n=1160.88$

Round off to the nearest whole numbers

Answer: $1161$