Estimation of a Population Proportion, p
A 100(1-)% confidence interval for p is
pˆ  (z/2 )
pˆ (1  pˆ )
n

Assumptions:
1. A random sample is selected from a population
2. The sample size is sufficiently large ( npˆ (1  pˆ )  10 ) such that the sampling
distribution of the sample proportion p̂ is approximately normal.
Properties of the sampling distribution of p̂ :
1. The sampling distribution of p̂ is centered at p ; that is,  p̂ = p .
2. The standard deviation of p̂ is  pˆ 
p(1  p)
n
3. As long as n is large ( np(1  p)  10 ), the sampling distribution of p̂ is well
approximated by a normal curve.
EXAMPLE In a recent survey of 1068 households in the United States, 673 stated that they had
answering machines. Using these sample results, construct a 95% CI for p , the proportion of all
households in the United States that have answering machines.
pˆ 
673
 .630
1068
.630  1.96 (.630)(.370)
1068
.630  .0290
95% CI FOR p
(.601, .659)
WE CAN BE 95% CONFIDENT THAT
p
IS BETWEEN .601 and .659.
IN NEWSPAPERS AND ON TV, THE RESULTS ARE OFTEN REPORTED AS FOLLOWS:
AMONG U.S. HOUSEHOLDS, THE PERCENTAGE THAT HAVE ANSWERING MACHINES
IS ESTIMATED TO BE 63%, WITH A MARGIN OF ERROR PLUS OR MINUS 2.9
PERCENTAGE POINTS.
Example Studies are performed to estimate the percentage of the nation’s 10 million asthmatics who
are allergic to sulfites. In one survey, 38 of 500 randomly selected U.S. asthmatics were found to be
allergic to sulfites. Find a 95% confidence interval for the proportion, p , of all U.S. asthmatics who
are allergic to sulfites. Interpret the interval!