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 : [tex]p=\dfrac{531}{1005}\approx0.53[/tex]
Significance level : [tex]\alpha: 1-0.90=0.1[/tex]
Critical value : [tex]z_{\alpha/2}=1.645[/tex]
The confidence interval for population proportion is given by :-
[tex]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)[/tex]
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)
