Question

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 14 defectives. Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool. Round the answers to 4 decimal places.

Answer 1

Answer: $(0.0228\ ,0.0706)$

Step-by-step explanation:

Given : Sample size : n= 300

The sample proportion of defectives : $\hat{p}=\dfrac{14}{300}=0.0467$

Significance level for 95% confidence level =$\alpha=1-0.95=0.05$

Critical z-value:$z_{\alpha/2}=\pm1.96$

Confidence interval for population proportion :

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$= 0.0467\pm (1.96)\sqrt{\dfrac{0.0467(1-0.0467)}{300}}$

$\approx\ 0.0467\pm 0.0239\\\\=(0.0467-0.0239\ , \ 0.0467-0.0239)\\\\=(0.0228\ ,0.0706)$

Hence, a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool= $(0.0228\ ,0.0706)$