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 = 100
Sample 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 = 1.96
Then, 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)
