stats help

kateeelllyyynn:

99% CI for the population proportion of voters opposing a new tax was calculated to be (.68,.76). What was the sample proportion of voters who oppose the new tax

The CI of a population proportion is

(p - z SE, p + z SE)

where p is the mean sample proportion, SE is its standard error, and z is the critical z-value of the confidence interval.

In other words, the requested sample proportion is the mean of the CI, which is:

p = (.68 +.76) / 2



Additionally, we can estimate the number of voters n.

From the CI we know that z SE = (.76 - .68) / 2 = 0.04.
The critical z-value of a 99% CI is z = 2.575.
So SE = 0.04 / 2.575 = 0.015534.

The SE of a proportion is:

SE = √ (p (1 - p) / n)

Therefore:

n = p (1 - p) / SE ² = .72 (1 - .72) / .015534 ² = 835