Answer: [tex]0.791\leq p\leq0.909[/tex]
Step-by-step explanation:
The confidence interval for the population proportion is given by:-
[tex]\hat{p}\pm z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}[/tex], where [tex]\hat{p}[/tex]= sample proportion, n =sample space , z* = critical z-value.
As per given, we have
[tex]n = 247[/tex]
[tex]\hat{p}=0.85[/tex]
Critical value for 99.9% confidence = 2.576
The required confidence interval will be:
[tex]0.85-2.576\sqrt{\frac{0.85\times0.15}{247}}\leq p\leq 0.85+2.576\sqrt{\frac{0.85\times0.15}{247}}\\\\\Rightarrow\ 0.85-(0.058526)\leq p \leq0.85+(0.058526)\\\\\Rightarrow\ 0.791474\leq p\leq0.908526\approx0.791\leq p\leq0.909[/tex]
The required confidence interval: [tex]0.791\leq p\leq0.909[/tex]
