Answer :
Answer: b) [tex](0.035,\ 0.065)[/tex]
Step-by-step explanation:
The confidence interval for proportion (p) 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.
Let p be the true proportion of defective manhole covers, based on this sample.
Given : The workers at Sandbachian, Inc. took a random sample of 800 manhole covers and found that 40 of them were defective.
Then , n= 800
[tex]\hat{p}=\dfrac{40}{800}=0.05[/tex]
Confidence interval = 95%
We know that the critical value for 95% Confidence interval : z*=1.96
Then, the 95% CI for p, the true proportion of defective manhole covers will be :-
[tex]0.05\pm (1.96)\sqrt{\dfrac{0.05(1-0.05)}{800}}\\\\=0.05\pm (1.96)(0.0077055)\\\\=0.05\pm0.01510278\\\\=(0.05-0.01510278,\ 0.05+0.01510278)\\\\=(0.03489722,\ 0.06510278)\approx(0.035,\ 0.065) [/tex]
Hence, the required confidence interval : b) [tex](0.035,\ 0.065)[/tex]