Introduction to Probability and Statistics for Engineers and Scientists

262 Chapter 7: Parameter Estimation

pˆ is the maximum likelihood estimator ofp, and so should be approximately equal
top. As a result,

√

npˆ(1−pˆ) will be approximately equal to

√
np(1−p) and so from
Equation 7.5.1 we see that

X−np

√

npˆ(1−pˆ)

∼·N(0, 1)

Hence, for anyα∈(0, 1) we have that

P

{

−zα/2<

X−np

√

npˆ(1−pˆ)

<zα/2

}

≈ 1 −α

or, equivalently,

P{−zα/2

√

npˆ(1−pˆ)<np−X<zα/2

√

npˆ(1−pˆ)}≈ 1 −α

Sincepˆ=X/n, the preceding can be written as

P{pˆ−zα/2

√

pˆ(1−pˆ)/n<p<pˆ+zα/2

√

pˆ(1−pˆ)/n}≈ 1 −α

which yields an approximate 100(1−α) percent confidence interval forp.

EXAMPLE 7.5a A sample of 100 transistors is randomly chosen from a large batch and
tested to determine if they meet the current standards. If 80 of them meet the standards,
then an approximate 95 percent confidence interval forp, the fraction of all the transistors
that meet the standards, is given by

(.8−1.96

√

.8(.2)/100, .8+1.96

√

.8(.2)/100)=(.7216, .8784)

That is, with “95 percent confidence,” between 72.16 and 87.84 percent of all transistors
meet the standards. ■

EXAMPLE 7.5b On October 14, 2003, theNew York Timesreported that a recent poll
indicated that 52 percent of the population was in favor of the job performance of
President Bush, with a margin of error of±4 percent. What does this mean? Can we
infer how many people were questioned?

SOLUTION It has become common practice for the news media to present 95 percent
confidence intervals. Sincez.025 = 1.96, a 95 percent confidence interval forp, the
percentage of the population that is in favor of President Bush’s job performance, is
given by

pˆ±1.96

√

pˆ(1−pˆ)/n=.52±1.96

√

.52(.48)/n