Suppose that in sampling for the population proportion, it is found that 20 out of 100 items are defective. Construct a 95% confidence interval for the proportion of defective items in the population

Answer: (0.1216,0.2784)Step-by-step explanation:The confidence interval for population proportion is given by :-[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex], where [tex]\hat{p}[/tex] = Sample proportion.n= sample size.z* = critical z-value.Given : sample size = 100Sample proportion of defective items = [tex]\hat{p}=\dfrac{20}{100}=0.2[/tex]Confidence level =95%We know that , Critical z-value for 95% confidence interval  = 1.96Then, the confidence interval for the proportion of defective items in the population will be :-[tex]0.2\pm (1.96)\sqrt{\dfrac{0.2(1-0.2)}{100}}[/tex][tex]0.2\pm (1.96)\sqrt{0.0016}[/tex][tex]0.2\pm (1.96)(0.04)[/tex][tex]0.2\pm0.0784[/tex][tex](0.2-0.0784,\ 0.2+ 0.0784)= (0.1216,\ 0.2784)[/tex]Hence, the 95% confidence interval for the proportion of defective items in the population =  (0.1216,0.2784)