Question

During an independent research, 2500 randomly selected people were interviewed whether they visit their dentist for a regular dental checkup or not. Only 1750 gave a positive answer. Calculate a 90% two-sided confidence interval on the proportion of people who regularly have a dental checkup. Round your answers to three decimal places

Answer 1

Answer: $(0.685,0.715)$

Step-by-step explanation:

The confidence interval for population proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

Given : Sample size : n= 2500

Number of people gave a positive answer = 1750

Then , Proportion of people gave a positive answer =$p=\dfrac{1750}{2500}=0.7$

Significance level :$\alpha= 1-0.9=0.1$

Critical value : $z_{\alpha/2}=1.645$

Now, the 90% two-sided confidence interval on the proportion of people who regularly have a dental checkup will be :-

$0.7\pm (1.645)\sqrt{\dfrac{0.7(0.3)}{2500}}\\\\\approx0.7\pm0.015\\\\=(0.7-0.015,0.7+0.015)\\\\=(0.685,0.715)$