Question

An engineer in an electronics company wants to estimate the proportion of electric circuit which are defective.
130 circuits were tested and 26 of them were found to be defective.
Give the 95% confidence interval for the true proportion. (3 decimal places)

Answer 1

The 95% confidence interval for proportion of defectives

$\hat{p}\pm zc*\sqrt{\hat{p}(1-\hat{p})/n}$

where

$\hat{p}$ = 26/130 = 0.2

For 95% confidence , zc =1.96

Therefore 95% confidence interva for true proportion is

$0.2\pm 1.96*\sqrt{0.2(1-0.2)/130}$

-02-00688

= ( 0.131 , 0.269 )