Question

A machine that produces a major part for an airplane engine is monitored closely. In the past, 10% of the parts produced would be defective. With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is

Answer 1

Answer: 216

Step-by-step explanation:

The formula to find the sample size , if prior population proportion is known :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Given : The prior proportion of defective parts : p= 0.10

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

Margin of error : $E=0.04$

Now, the required sample size will be :-

$n=0.1(0.9)(\dfrac{(1.96)}{0.04})^2=216.09\approx216$

Hence, the minimum required sample size = 216