In survey conducted by Quinnipiac University from October 25-31, 2011, 47% of a sample of 2,294 registered voters approved of the job Barack Obama was doing as president.

What is the 99% confidence interval for the proportion of all registered voters who approved of the job Barack Obama was doing as president?

A) (0.460, 0.480)
B) (0.453, 0.487)
C) (0.450, 0.490)
D) (0.443, 0.497)

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for a proportion is given by this formula

[tex]\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]

For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.

[tex]z_{\alpha/2}=2.58[/tex]

And replacing into the confidence interval formula we got:

[tex]0.47 - 2.58 \sqrt{\frac{0.47(1-0.47)}{2294}}=0.443[/tex]

[tex]0.47 + 2.58 \sqrt{\frac{0.47(1-0.47)}{2294}}=0.497[/tex]

And the 99% confidence interval would be given (0.443;0.497).

We are confident that about 44.3% to 49.7% of registered voters approved of the job Barack Obama was doing as president.