Answer: [tex](0.0445,\ 0.0755)[/tex]
Step-by-step explanation:
The confidence interval for the population proportion is given by :-
[tex]p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}[/tex]
Given : A Bernoulli random variable X has unknown success probability p.
Sample size : [tex]n=100[/tex]
Unknown success probability : [tex]p=0.06[/tex]
Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Critical value : [tex]z_{\alpha/2}=2.576[/tex]
Now, the 99% confidence interval for true proportion will be :-
[tex]0.06\pm(2.576)\sqrt{\dfrac{0.06(0.06)}{100}}\\\\\approx0.06\pm(0.0155)\\\\=(0.06-0.0155,\ 0.06+0.0155)\\\\=(0.0445,\ 0.0755)[/tex]
Hence, the 99% confidence interval for true proportion= [tex](0.0445,\ 0.0755)[/tex]