Question

In a survey of 1005 ​adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not. Of the 1005 ​surveyed, 531 stated that they were worried about having enough money to live comfortably in retirement. Construct a 90​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement

Answer 1

Answer: (0.504, 0.556)

Step-by-step explanation:

Given : Sample size : n= 1005

Number of adults stated that they were worried about having enough money to live comfortably in retirement = 531

Then the proportion of adults stated that they were worried about having enough money to live comfortably in retirement : $p=\dfrac{531}{1005}\approx0.53$

Significance level : $\alpha: 1-0.90=0.1$

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

The confidence interval for population proportion is given by :-

$p\pm\ z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}\\\\=0.53\pm(1.645)\sqrt{\dfrac{0.53(1-0.53)}{1005}}\\\\\approx0.53\pm0.026\\\\=(0.504,\ 0.556)$

Hence, the 90​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement = (0.504, 0.556)