As per given , we have
n= 1025
x=707
[tex]\hat{p}=\dfrac{707}{1025}=0.689756097561\approx0.69[/tex]
Critical value for 99% confidence interval : [tex]z_{\alpha/2}=2.576[/tex]
Confidence interval : [tex]\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
i.e. [tex]0.69\pm (2.576)\sqrt{\dfrac{0.69(1-0.69)}{1025}}[/tex]
i.e. [tex]0.69\pm 0.0372125403253[/tex]
i.e. [tex]\approx 0.69\pm 0.037[/tex]
[tex]=( 0.69- 0.037,\ 0.69+0.037)=(0.653,\ 0.727) [/tex]
99% confidence interval for the proportion of all U.S. adults in December of 2012 who were optimistic about their finances in 2013= (0.653, 0.727)
Interpretation : We are 99% confident that the true population of all U.S. adults in December of 2012 who were optimistic about their finances in 2013 lies in interval (0.653, 0.727).
