Question

According to a poll by the Gallup organization in December of 2012, among a random sample of 1,025 U.S. adults, 707 said they were optimistic about their finances in 2013. Calculate and interpret a 99% 4 confidence interval for the proportion of all U.S. adults in December of 2012 who were optimistic about their finances in 2013. Make sure to include all steps.

Answer 1

Answer with explanation:

As per given , we have

n= 1025

x=707

$\hat{p}=\dfrac{707}{1025}=0.689756097561\approx0.69$

Critical value for 99% confidence interval : $z_{\alpha/2}=2.576$

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

i.e. $0.69\pm (2.576)\sqrt{\dfrac{0.69(1-0.69)}{1025}}$

i.e. $0.69\pm 0.0372125403253$

i.e. $\approx 0.69\pm 0.037$

$=( 0.69- 0.037,\ 0.69+0.037)=(0.653,\ 0.727)$

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).