You already learned about
the standard error for the sampling distribution of means,
s.e_{mean} =
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.
Sixtyone
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 subsample of
ten cases:
Case

Worth
It?

Score
(X)

Mean

(Xmean)

(Xmean)^{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.
