Answer to Question #238995 in Statistics and Probability for Matthew

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 parts

received from this supplier. She believes the proportion defective is no more than .20 and

wants to be within .02 of the true proportion of defective parts with a 90% level of

confidence. How large a sample should she take?

Answer 1

Proportion p has approximately normal distribution with parameters $\mu_p=\hat p. \space \sigma_p=\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$. Thus for p analisis.we use normal distribution$N(\hat p,\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}})$. So C- confidence interval for mean M(p) will be $\hat p \pm z^*_\frac{1-C}{2}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$

where $z^*_\frac{1+C}{2}$ is such level that P(N(0,1)$> z^*_\frac{1-C}{2}=\frac{1-C}{2}$ Therfore margin error will be E=$z^*_\frac{1-C}{2}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}$ . Fron this equation we define size of sample $n=\frac{\hat p\cdot (1-\hat p)\cdot(z^*_\frac{1-C}{2})^2 }{E^2}$ .We are given with E=0.02, $\hat p=0.2$ we use the worst case . C=90%=0.9 and calculate: $z^*_\frac{1-C}{2}=z^*_{0.05}=$

=qnorm(0.95,0,1)=1.645 where qnorm is statistical function from Mathcad therefore we have $n=\frac{0.2\cdot(1-0.2)\cdot 1.645^2}{0.02^2}=956.046\approx 957$

So size of sample should be 957, this value is enough for all $\hat p\le 0.2$