Advanced High-School Mathematics

SECTION 6.4 Confidence Intervals 391

Ifb 1 , b 2 ,...,bnare the observed outcomes, i.e.,

bi =







1 if type A is observed;

0 if type B is observed,

then the relevant test statistic is

pˆ =

b 1 +b 2 +···+bn

n

.

Notice that since we don’t know p(we’re tying to estimate it), we
knowneither the mean nor the variance of the test statistic. With
a large enough sample,P̂will be approximately normally distributed

with mean p and variance p(1−p). Therefore
P̂−p
»
p(1−p)/n

will be

approximately normal with mean 0 and variance 1. The problem with
the above is all of the occurrences of the unknownp. The remedy is to

approximate the variance

p(1−p)

n

by the sample variance based on ˆp:

pˆ(1−pˆ)

n

. Therefore, we may regard

Z =

P̂−p

√̂

P(1−P̂)/n

as being approximately normally distributed with mean 0 and variance

1. Having this we now build our (1−α)×100% confidence intervals
   based on the valueszα/ 2 taken from normal distribution with mean 0
   and variance 1. That is to say, the (1−α)×100% confidence interval
   for the population proportionpis



pˆ−z

α/ 2

Ã

pˆ(1−pˆ)

n

, x+zα/ 2

Ã

pˆ(1−pˆ)

n



.

Caution: If we are trying to estimate a population parameter which
we know to be either very close to 0 or very close to 1, the method
above performs rather poorly unless the sample size is very large.