Computing Standard Deviations for Proportions

You already learned about the standard error for the sampling distribution of means, s.emean =

My lecture notes for yesterday gave the formula for computing the standard error for proportions, which is simply a mean computed for data scored 1 (for p) or 0 (for q). It so happens that the variance for data in proportions is simply

variance = pq
So the standard deviation =

In case you don't believe this, here is a computed example for these data inspired by the CBS/New York Times poll reported on October 29, 2001.

Sixty-one percent think the war in Afghanistan would be worth it even if it meant several thousand American troops would lose their lives; 27 percent say the war there would not be worth that cost.

Let's round off the 61% to 60% for easier computation and consider only a sub-sample of ten cases:

 Case Worth It? Score (X) Mean (X-mean) (X-mean)2 1 yes 1 0.6 0.4 0.16 2 no 0 0.6 -0.6 0.36 3 no 0 0.6 -0.6 0.36 4 yes 1 0.6 0.4 0.16 5 yes 1 0.6 0.4 0.16 6 yes 1 0.6 0.4 0.16 7 yes 1 0.6 0.4 0.16 8 no 0 0.6 -0.6 0.36 9 yes 1 0.6 0.4 0.16 10 no 0 0.6 -0.6 0.36 6/10 =.6 (mean of proportion) · = 2.4 (sum of squares)
• Given a sum of squares of 2.4 for ten cases, the variance is .24.
• Now let's multiply the p (.6) by the q (.4): .6 * .4 = .24 -- so pq = variance.
• We can compute the s.e. of the proportion for the CBS/New York Times poll of 1,024 respondents, using yesterday's formula:
• This result is one standard error of a proportion; we multiply by 100 to make it a percentage: 1.5%
• But remember we need to double the 1.5% to produce an estimate of +/- 3%--such that it will embrace 95% of the possible samples.