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}=(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{\frac{\hat{p}(1-\hat{p})}{n}}`

i.e.
`0.69\pm (2.576)\sqrt{(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).