Introduction to Probability and Statistics for Engineers and Scientists

264 Chapter 7: Parameter Estimation

approximately .05 would require an approximate sample of size

n=

4(z.005)^2

(.05)^2

26

30

(

1 −

26

30

)

=

4(2.58)^2

(.05)^2

26

30

4

30

=1,231

Hence, we should now sample an additional 1,201 chips and if, for instance, 1,040 of
them are acceptable, then the final 99 percent confidence interval forpis

(

1,066

1,231

−

√

1,066

(

1 −

1,066

1,231

)

z.005

1,231

,

1,066

1,231

+

√

1,066

(

1 −

1,066

1,231

)

z.005

1,231

)

or

p∈(.84091, .89101) ■

REMARK

As shown, a 100(1−α) percent confidence interval forpwill be of approximate lengthb
when the sample size is

n=

(2zα/2)^2

b^2

p(1−p)

Now it is easily shown that the functiong(p)=p(1−p) attains its maximum value of^14 ,
in the interval 0≤p≤1, whenp=^12. Thus an upper bound onnis

n≤

(zα/2)^2

b^2

and so by choosing a sample whose size is at least as large as (zα/2)^2 /b^2 , one can be
assured of obtaining a confidence interval of length no greater thanbwithout need of any
additional sampling. ■

One-sided approximate confidence intervals forpare also easily obtained; Table 7.3
gives the results.

TABLE 7.3 Approximate100(1−α)Percent Confidence Intervals for p

XIs a Binomial (n,p) Random Variable

pˆ=X/n

Type of Interval Confidence Interval

Two-sided pˆ±zα/2

√

pˆ(1−pˆ)/n

One-sided lower

(

−∞,pˆ+zα

√

pˆ(1−pˆ)/n

)

One-sided upper

(

pˆ−zα

√

pˆ(1−pˆ)/n,∞

)