Introduction to Probability and Statistics for Engineers and Scientists

7.5Approximate Confidence Interval for the Mean of a Bernoulli Random Variable 263

wherenis the size of the sample. Since the “margin of error” is±4 percent, it follows that

1.96

√

.52(.48)/n=.04

or

n=

(1.96)^2 (.52)(.48)

(.04)^2

=599.29

Thatis, approximately599peopleweresampled, and52percentofthemreportedfavorably
on President Bush’s job performance. ■

We often want to specify an approximate 100(1−α) percent confidence interval for
pthat is no greater than some given length, sayb. The problem is to determine the
appropriate sample sizento obtain such an interval. To do so, note that the length of
the approximate 100(1−α) percent confidence interval forpfrom a sample of sizenis

2 zα/2

√

pˆ(1−pˆ)/n

which is approximately equal to 2zα/2

√
p(1−p)/n. Unfortunately,pis not known in
advance, and so we cannot just set 2zα/2

√
p(1−p)/nequal tobto determine the necessary
sample sizen. What we can do, however, is to first take a preliminary sample to obtain
a rough estimate ofp, and then use this estimate to determinen. That is, we usep∗, the
proportion of the preliminary sample that meets the standards, as a preliminary estimate
ofp; we then determine the total sample sizenby solving the equation

2 zα/2

√

p∗(1−p∗)/n=b

Squaring both sides of the preceding yields that

(2zα/2)^2 p∗(1−p∗)/n=b^2

or

n=

(2zα/2)^2 p∗(1−p∗)

b^2

That is, ifkitems were initially sampled to obtain the preliminary estimate ofp, then an
additionaln−k(or0ifn≤k) items should be sampled.

EXAMPLE 7.5c A certain manufacturer produces computer chips; each chip is indepen-
dently acceptable with some unknown probabilityp. To obtain an approximate 99 percent
confidence interval forp, whose length is approximately .05, an initial sample of 30
chips has been taken. If 26 of these chips are of acceptable quality, then the prelimi-
nary estimate ofpis 26/30. Using this value, a 99 percent confidence interval of length