Answer: [tex](0.04,\ 0.21)[/tex]
Step-by-step explanation:
The confidence interval for population proportion is given by :-
[tex]p_1-p_2\pm z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}[/tex]
Given : Significance level : [tex]\alpha=1-0.99=0.01[/tex]
Critical value : [tex]z_{\alpha/2}=2.576[/tex]
[tex]n_1=400 ;\ p_1=\dfrac{160}{400}=0.4[/tex]
[tex]n_2=400 ;\ p_2=\dfrac{110}{400}=0.275[/tex]
Then a 99% confidence interval is constructed to estimate the difference in population proportions which possess the given characteristic will be :-
[tex]0.4-0.275\pm (2.576)\sqrt{\dfrac{0.4(1-0.4)}{400}+\dfrac{0.275(1-0.275)}{400}}\\\\\approx0.125\pm0.085=(0.125-0.085,0.125+0.085)=(0.04,\ 0.21)[/tex]
Hence, the resulting confidence interval is [tex](0.04,\ 0.21)[/tex] .
