Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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If we think of the width of the interval as specifying its precision or accuracy,

then the confidence level (or reliability) of the interval is inversely related to its

precision. A highly reliable interval estimate may be imprecis e in that the endpoints

of the interval may be far apart, whereas a precise interval may entail relatively low

reliability. Thus it cannot be said unequivocally that a 99% interval is to be

preferred to a 95% interval; the gain in reliability entails a loss in precision.

An appealing strategy is to specify both the desired confidence level and

interval width and then determine the necessary sample size.

Example 8.4 Ex tensive monitoring of a computer time-sharing system has suggested that

response time to a particular editing command is normally distributed with standard

deviation 25 ms. A new operating system has been installed, and we wish to

estimate the true average response time m for the new environment. Assuming

that response times are still normally distributed with s ¼ 25, what sample size is

necessary to ensure that the resulting 95% CI has a width of (at most) 10? The

sample size n must satisfy

10 ¼ 2 ð1:96Þð25=

ﬃﬃﬃ

Rearranging this equation gives

ﬃﬃﬃ

¼ 2 ð1:96Þð25Þ=10 ¼ 9:80

n ¼ 9:80

¼ 96:04

Since n must be an integer, a sample size of 97 is required.

■

The general formula for the sample size n necessary to ensure an interval

width w is obtained from w ¼ 2  z

a=2

 s=

ﬃﬃﬃ

n ¼ 2z

a=2





ð8:6Þ

The smaller the desired width w, the larger n must be. In addition, n is an increasing

function of s (more population variability necessitates a larger sample size) and of

the confidence level 100(1  a) (as a decreases, z

a/2

increases).

The half-width 1:96  s=

ﬃﬃﬃ

of the 95% CI is sometimes called the bound on

the error of estimation associated with a 95% confidence level; that is, with 95%

confidence, the point estimate

x will be no farther than this from m. Before

obtaining data, an investigator may wish to determine a sample size for which a

particular value of the bound is achieved. For example, with m representing the

average fuel efficiency (mpg) for all cars of a certain type, the objective of an

investigation may be to estimate m to within 1 mpg with 95% confidence. More

generally, if we wish to estimat e m to within an amount B (the specified bound on

the error of estimation) with 100(1  a)% confidence, the necessary sample size

results from replacing 2/w by 1/B in (8.6).

388 CHAPTER 8 Statistical Intervals Based on a Single Sample

Deriving a Conﬁdence Interval

Let X

, X

, ... , X

denote the sample on which the CI for a parameter y is to be

based. Suppose a random variable satisfying the following two properties can be

found:

1. Th e variable depends functionally on both X

, ... , X

and y.

2. Th e probability distribution of the variable does not depend on y or on any other

unknown parameters.

Let h(X

, X

, ... , X

; y) denote this random variable. For example, if the

population distribution is normal with known s and y ¼ m, the variable

hðX

; ...; X

; yÞ¼ X  mðÞ= s=

ﬃﬃﬃ

ðÞsatisfies both properties; it clearly depends

functionally on m, yet has the standard normal probability distribution, which

does not depend on m. In general, the form of the h function is usually suggested

by examining the distribution of an appropriate estimator

For any a between 0 and 1, constants a and b can be found to satisfy

P½a < h ðX

; ...; X

; yÞ< b¼1  a ð8:7Þ

Because of the second property, a and b do not depend on y. In the normal example,

a ¼z

a/2

and b ¼ z

a/2

. Now suppose that the inequalities in (8.7) can be manipu-

lated to isolate y, giving the equivalent proba bility statement

P½lX

; ...; X

ðÞ< y < uX

; ...; X

ðÞ¼1  a

Then l(x

, x

, ... , x

) and u(x

, ... , x

) are the lower and upper confidence limits,

respectively, for a 1 00(1  a)% CI. In the normal example, we saw that

; ...; X

ðÞ¼X  z

a=2

 s=

ﬃﬃﬃ

and uX

; ...; X

ðÞ¼X þ z

a=2

 s=

ﬃﬃﬃ

Example 8.5 A theoretical model suggests that the time to breakdown of an insulating fluid

between electrodes at a particular voltage has an exponential distribution with

parameter l (see Section 4.4). A random sample of n ¼ 10 breakdown times yields

the following sample data (in min): x

¼ 41.53, x

¼ 18.73, x

¼ 2.99, x

¼ 30.34,

¼ 12.33, x

¼ 117.52, x

¼ 73.02, x

¼ 223.63, x

¼ 4.00, x

¼ 26.78.

A 95% CI for l and for the true average breakdown time are desired.

Let h(X

, X

, ... , X

; l ) ¼ 2lSX

. Using a moment generating function

argument, it can be shown that this random variable has a chi-squared distribution

with 2n degrees of freedom (df) (v ¼ 2n, as discussed in Sect ion 6.4). Appendix

Table A.6 pictures a typical chi-squared density curve and tabulates critical values

that capture specified tail areas. The relevant number of degrees of freedom here is

2(10) ¼ 20. The n ¼ 20 row of the table shows that 34.170 captures upper-tail area

.025 and 9.591 captures lower-tail area .025 (upper-tail area .975). Thus for n ¼ 10,

Pð9:591 < 2lSX

< 34:170Þ¼:95

Division by 2SX

isolates l, yielding

P 9:591= 2SX

ðÞ< l < 34:170= 2SX

ðÞ½¼:95

8.1 Basic Properties of Conﬁdence Intervals 389

The lower limit of the 95% CI for l is 9.591/(2Sx

), and the upper limit is 34.170/

(2Sx

). For the given data, Sx

¼ 550.87, giving the interval (.00871, .03101).

The expected value of an exponential rv is m ¼ 1/l. Since

P 2SX

=34:170 < 1=l < 2SX

=9:591ðÞ¼:95

the 95% CI for true average breakdown time is (2Sx

/34.170, 2Sx

/9.591) ¼

(32.24, 114.87). This interval is obviously quite wide, ref lecting substantial varia-

bility in breakdown times and a small sample size.

■

In general, the upper and lower confidence limits result from replacing each

< in (8.7) by ¼ and solving for y. In the insu lating fluid example just considered,

2lSx

¼ 34.170 gives l ¼ 34.170/(2Sx

) as the upper confidence limit, and the

lower limit is obtained from the other equation. Notice that the two interval limits

are not equidistant from the point estimate, since the interval is not of the form

y  c.

Exercises Section 8.1 (1–11)

1. Consider a normal population distribution with the

value of s known.

a. What is the confidence level for the interval

x  2:81s=

ﬃﬃﬃ

b. What is the confidence level for the interval

x  1:44s=

ﬃﬃﬃ

c. What value of z

a/2

in the CI formula (8.5)

results in a confidence level of 99.7%?

d. Answer the question posed in part (c) for a

confidence level of 75%.

2. Each of the following is a confidence interval for

m ¼ true average (i.e., population mean) reso-

nance frequency (Hz) for all tennis rackets of a

certain type:

(114.4, 115.6) (114.1, 115.9)

a. What is the value of the sample mean reso-

nance frequency?

b. Both intervals were calculated from the same

sample data. The confidence level for one of

these intervals is 90% and for the other is 99%.

Which of the intervals has the 90% confidence

level, and why?

3. Suppose that a random sample of 50 bottles of

a particular brand of cough syrup is selected and

the alcohol content of each bottle is determined.

Let m denote the average alcohol content for the

population of all bottles of the brand under study.

Suppose that the resulting 95% confidence inter-

val is (7.8, 9.4).

a. Would a 90% confidence interval calculated

from this same sample have been narrower or

wider than the given interval? Explain your

reasoning.

b. Consider the following statement: There is a

95% chance that m is between 7.8 and 9.4. Is

this statement correct? Why or why not?

c. Consider the following statement: We can be

highly confident that 95% of all bottles of this

type of cough syrup have an alcohol content

that is between 7.8 and 9.4. Is this statement

correct? Why or why not?

d. Consider the following statement: If the pro-

cess of selecting a sample of size 50 and then

computing the corresponding 95% interval is

repeated 100 times, 95 of the resulting intervals

will include m. Is this statement correct? Why or

why not?

4. A CI is desired for the true average stray-load loss

m (watts) for a certain type of induction motor

when the line current is held at 10 amps for a

speed of 1,500 rpm. Assume that stray-load loss

is normally distributed with s ¼ 3.0.

a. Compute a 95% CI for m when n ¼ 25 and

x ¼ 58:3.

b. Compute a 95% CI for m when n ¼ 100 and

x ¼ 58:3.

c. Compute a 99% CI for m when n ¼ 100 and

x ¼ 58:3.

d. Compute an 82% CI for m when n ¼ 100 and

x ¼ 58:3.

e. How large must n be if the width of the 99%

interval for m is to be 1.0?

5. Assume that the helium porosity (in percentage)

of coal samples taken from any particular seam is

normally distributed with true standard devia-

tion .75.

390

CHAPTER 8 Statistical Intervals Based on a Single Sample

a. Compute a 95% CI for the true average porosity

of a certain seam if the average porosity for 20

specimens from the seam was 4.85.

b. Compute a 98% CI for true average porosity of

another seam based on 16 specimens with a

sample average porosity of 4.56.

c. How large a sample size is necessary if the

width of the 95% interval is to be .40?

d. What sample size is necessary to estimate true

average porosity to within .2 with 99% confi-

dence?

6. On the basis of extensive tests, the yield point of a

particular type of mild steel reinforcing bar is

known to be normally distributed with s ¼ 100.

The composition of the bar has been slightly mod-

ified, but the modification is not believed to have

affected either the normality or the value of s.

a. Assuming this to be the case, if a sample of 25

modified bars resulted in a sample average

yield point of 8439 lb, compute a 90% CI for

the true average yield point of the modified bar.

b. How would you modify the interval in part (a)

to obtain a confidence level of 92%?

7. By how much must the sample size n be increased

if the width of the CI (8.5) is to be halved? If the

sample size is increased by a factor of 25, what

effect will this have on the width of the interval?

Justify your assertions.

8. Let a

> 0, a

> 0, with a

+ a

¼ a. Then

P z

X  m

ﬃﬃﬃ

< z



¼ 1  a

a. Use this equation to derive a more general

expression for a 100(1  a)% CI for m of

which the interval (8.5) is a special case.

b. Let a ¼ .05 and a

¼ a/4, a

¼ 3a/4. Does

this result in a narrower or wider interval than

the interval (8.5)?

9. a. Under the same conditions as those leading to

the CI (8.5), P

X mðÞ= s=

ﬃﬃﬃ

ðÞ<1:645½¼:95.

Use this to derive a one-sided interval for m

that has infinite width and provides a lower

confidence bound on m. What is this interval

for the data in Exercise 5(a)?

b. Generalize the result of part (a) to obtain

a lower bound with a confidence level of

100(1  a)%.

c. What is an analogous interval to that of part (b)

that provides an upper bound on m? Compute

this 99% interval for the data of Exercise 4(a).

10. A random sample of n ¼ 15 heat pumps of a

certain type yielded the following observations

on lifetime (in years):

2.0 1.3 6.0 1.9 5.1 .4 1.0 5.3

15.7 .7 4.8 .9 12.2 5.3 .6

a. Assume that the lifetime distribution is expo-

nential and use an argument parallel to that of

Example 8.5 to obtain a 95% CI for expected

(true average) lifetime.

b. How should the interval of part (a) be altered to

achieve a confidence level of 99%?

c. What is a 95% CI for the standard deviation of

the lifetime distribution? [Hint: What is the

standard deviation of an exponential random

variable?]

11. Consider the next 1,000 95% CIs for m that a

statistical consultant will obtain for various

clients. Suppose the data sets on which the inter-

vals are based are selected independently of one

another. How many of these 1,000 intervals do

you expect to capture the corresponding value of

m? What is the probability that between 940 and

960 of these intervals contain the corresponding

value of m?[Hint: Let Y ¼ the number among

the 1,000 intervals that contain m. What kind of

random variable is Y?]

8.2

Large-Sample Confidence Intervals

for a Population Mean and Proportion

The CI for m given in the previous section assumed that the population distribution

is normal and that the value of s is known. We now present a large-sample CI

whose validity does not require these assumptions. After showing how the argu-

ment leading to this interval generalizes to yield other large-sample intervals, we

focus on an interval for a population proportion p.

8.2 Large-Sample Conﬁdence Intervals for a Population Mean and Proportion 391

A Large-Sample Interval for m

Let X

, X

, ... , X

be a random sample from a population having a mean m and

standard deviation s. Provided that n is large, the Central Limit Theorem (CLT)

implies that

X has approximately a normal distribution whatever the nature of the

population distribution. It then follows that Z ¼ð

X  mÞ=ðs=

ﬃﬃﬃ

Þ has approxi-

mately a standard normal distribution, so that

P z

a=2

X  m

ﬃﬃﬃ

< z

a=2



 1  a

An argument parallel with that given in Section 8.1 yields

x  z

a=2

 s=

ﬃﬃﬃ

as a

large-sample CI for m with a confidence level of approximately 100(1  a)%. That

is, when n is large, the CI for m given previously remains valid whatever the

population distribution, provided that the qualifier “approximately” is inserted in

front of the confidence level.

One practical difficulty with this development is that computation of the

interval requires the value of s, which will almost never be known. Consider the

standardized variable

Z ¼

X  m

ﬃﬃﬃ

in which the sample standard deviation S replaces s. Previously there was random-

ness only in the numerator of Z (by virtue of

X). Now there is randomness in both

the numerator and the denominator—the values of both

X and S vary from sample

to sample. However, when n is large, the use of S rather than s adds very little extra

variability to Z. More specifically, in this case the new Z also has approximatel y a

standard normal distribution. Manipulation of the inequalities in a probability

statement involving this new Z yields a gener al large-sample interval for m.

PROPOSITION

If n is sufficiently large, the standardized variable

Z ¼

X  m

ﬃﬃﬃ

has approximately a standard normal distribution. This implies that

x  z

a=2



ﬃﬃﬃ

ð8:8Þ

is a large-sample confidence interval for m with confidence level approxi-

mately 100(1  a)%. This formula is valid regardless of the shape of the

population distribution.

Generally speaking, n > 40 will be sufficient to justify the use of this interval. This

is somewhat more conservative than the rule of thumb for the CLT because of the

additional variability introduced by using S in place of s.

392 CHAPTER 8 Statistical Intervals Based on a Single Sample

Example 8.6 Haven’t you always wanted to own a Porsche? One of the authors thought maybe he

could afford a Boxster, the cheapest model. So he went to www.cars.com on Nov.

18, 2009 and found a total of 1,113 such cars listed. Asking prices ranged from

$3,499 to $130,000 (the latter price was one of only two exceeding $70,000). The

prices depressed him, so he focused instead on odometer readings (miles). Here are

reported readings for a sample of 50 of these Boxsters:

2948 2996 7197 8338 8500 8759 12710 12925

15767 20000 23247 24863 26000 26210 30552 30600

35700 36466 40316 40596 41021 41234 43000 44607

45000 45027 45442 46963 47978 49518 52000 53334

54208 56062 57000 57365 60020 60265 60803 62851

64404 72140 74594 79308 79500 80000 80000 84000

113000 118634

A boxplot of the data (Figure 8.5) shows that, except for the two mild outliers at the

upper end, the distribution of values is reasonably symmetric (in fact, a normal

probability plot exhibits a reasonably linear pattern, though the points

corresponding to the two smallest and two largest observations are somewhat

removed from a line fit through the remaining points).

Summary quantitie s include n ¼ 50,

x ¼ 45;679: 4;

x ¼ 45;013: 5;

s ¼ 26;641:675; f

¼ 34;265. The mean and median are reasonably close (if the

two largest values were each reduced by 30,000, the mean would fall to 44,479.4

while the median would be unaffected). The boxplot and the magnitudes of s and f

relative to the mean and median both indicate a substantial amount of variability.

A confidence level of about 95% requires z

.025

¼ 1.96, and the interval is

45;679:4 ð1:96Þ

26;641:675

ﬃﬃﬃﬃﬃ



¼ 45;679:4  7384:7 ¼ð38;294:7; 53;064:1Þ

That is, 38,294.7 < m < 53,064.1 with 95% confidence. This interval is rather

wide because a sample size of 50, even though large by our rule of thumb, is not

large enough to overcome the substantial variability in the sample. We do not have

a very precise estimat e of the population mean odometer reading.

Is the interval we’ve calculated one of the 95% that in the long run includes

the parameter being estimated, or is it one of the “bad” 5% that does not do so?

Without knowing the value of m, we cannot tell. Remember that the confidence

0 20000 40000 60000 80000 100000 120000

mileage

Figure 8.5 A boxplot of the odometer reading data from Example 8.6

8.2 Large-Sample Conﬁdence Intervals for a Population Mean and Proportion 393

level refers to the long run capture percentage when the formula is used repeatedly

on various samples; it cannot be interpreted for a single sample and the resulting

interval.

■

Unfortunately, the choice of sample size to yield a desired interval width is

not as straightforward here as it was for the case of known s. This is because the

width of (8.8)is2z

ﬃﬃﬃ

. Since the value of s is not available before data

collection, the width of the interval cannot be determined solely by the choice of

n. The only option for an investigator who wishes to specify a desired width is to

make an educated guess as to what the value of s might be. By being conservative

and guessing a larger value of s,ann larger than necessary will be chosen. The

investigator may be able to specify a reasonably accurate value of the population

range (the difference between the largest and smallest values). Then if the popula-

tion distribution is not too skewed, dividing the range by four gives a ballpark value

of what s might be. The idea is that roughly 95% of the data lie within 2s of the

mean, so the range is roughly 4 s (range/6 might be too optimistic).

Example 8.7 An investigator wishes to estimate the true average score on an algebra placement

test. Suppose she believes that virtually all values in the p opulation are between 10

and 30. Then (30  10)/4 ¼ 5 gives a reasonable value for s. The appropriate

sample size for estimating the true average mileage to within one with confidence

level 95%—that is, for the 95% CI to have a width of 2—is

n ¼ 1:96ðÞ5ðÞ=1½

 96

■

A General Large-Sample Conﬁdence Interval

The large-sample intervals x  z

a=2

 s=

ﬃﬃﬃ

and x  z

a=2

 s=

ﬃﬃﬃ

are special cases of

a general large-sample CI for a parameter y. Suppose that

y is an estimator satisfying

the following properties: (1) It has approximat ely a normal distribution; (2) it is (at

least approximately) unbiased; and (3) an expression for s

, the standard deviation

y, is available. For example, in the case y ¼ m,

m ¼

X is an unbiased estimator

whose distribution is approximately normal when n is large and s

¼ s

¼ s=

ﬃﬃﬃ

Standardizing

y yields the rv Z ¼ð

y  yÞ=s

, which has approximately a standar d

normal distribution. This justifies the probability statement

P z

a=2

y  y

< z

a=2

 1  a ð8:9Þ

Suppose, first, that s

does not involve any unknown parameters (e.g., known

s in the case y ¼ m). Then replacing each < in (8.9)by¼ results in

y ¼

y  z

a=2

 s

, so the lower and upper confidence limits are

y  z

a=2

 s

and

y þ z

a=2

 s

, respectively. Now suppose that s

does not involve y but does involve

at least one other unknown parameter. Let s

be the estimate of s

obtained by using

estimates in place of the unknown parameters (e.g., s=

ﬃﬃﬃ

estimates s=

ﬃﬃﬃ

). Under

general conditions (essentially that s

be close to s

for most samples), a valid CI is

y  z

a=2

 s

. The interval x  z

a=2

 s=

ﬃﬃﬃ

is an example.

394 CHAPTER 8 Statistical Intervals Based on a Single Sample

Finally, suppose that s

does involve the unknown y. This is the case, for

example, when y ¼ p, a population proportion. Then ð

y  yÞ=s

¼ z

a=2

can be

difficult to solve. An approximate solution can often be obtained by replacing y in

by its estimate

y. This results in an estimated standard deviation s

, and the

corresponding interval is again

y  z

a=2

 s

A Conﬁdence Interval for a Population Proportion

Let p denote the proportion of “successes” in a population, where success identifies

an individual or object that has a specified property . A random sample of n

individuals is to be selected, and X is the number of successes in the sample.

Provided that n is small compared to the population size, X can be regarded as a

binomial rv with E(X) ¼ np and s

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

npð1  pÞ

. Furthermore, if n is large

(np  10 and nq  10), X has approximately a normal distribution.

The natural estimator of p is

p ¼ X=n, the sample fraction of successes. Since

p is just X multiplied by a constant 1/n,

p also has approximately a normal

distribution. As shown in Section 7.1, Eð

pÞ¼p (unbiasedness) and

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

pð1  pÞ=n

. The standard deviation s

involves the unknown parameter

p. Standardizi ng

p by subtracting p and dividing by s

then implies that

P z

a=2

p  p

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

pð1  pÞ=n

< z

a=2

 1  a

Proceeding as suggested in the subsection “Deriving a Confidence Interval”

(Section 8.1), the confidence limit s result from replacing each < by ¼ and solving

the resulting quadratic equation for p. With

q ¼ 1 

p, this gives the two roots

p ¼

p þ z

a=2

=2n

1 þ z

a=2

 z

a=z

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

pð1 

pÞ=n þ z

a=2

=4n

1 þ z

a=2

p  z

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

pð1 

pÞ=n þ z

a=2

=4n

1 þ z

a=2

PROPOSITION

Let

p ¼

p þ z

a=2

=2n

1 þ z

a=2

. Then a confidence interval for a pop ulation propor-

tion p with confidence level approximately 100(1  a)% is

p  z

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

q=n þ z

a=2

=4n

1 þ z

a=2

ð8:10Þ

where

q ¼ 1 

p and, as before, the  in (8.10) corr esponds to the lower

confidence limit and the + to the upper confidence limit .

This is often referred to as the “score CI” for p.

8.2 Large-Sample Conﬁdence Intervals for a Population Mean and Proportion 395

If the sample size n is very large, then z

/2n is generally quite negligible

(small) compared to

p and z

/n is quite negligible compared to 1, from which

p 

In this case z

/4n

is also negligible compared to

q=n (n

is a much larger divisor

than is n); as a result, the dominant term in the  expression is z

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

q=n

and the

score interval is approximately

p  z

a=2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

q=n

ð8:11Þ

This latter interval has the general form

y  z

a=2

of a large-sample interval

suggested in the last subsection. The approximate CI (8.11) is the one that for

decades has appeared in introductory statistics textbooks. It clearly has a much

simpler and more appealing form than the score CI. So why bother with the latter?

First of all, suppos e we use z

.025

¼ 1.96 in the traditional formula (8.11).

Then our nominal confidence level (the one we think we’re buying by using that z

critical value) is approximately 95%. So before a sample is selected, the probability

that the random interval includes the actual value of p (i.e., the coverage probabil-

ity) should be about .95. But as Figure 8.6 shows for the case n ¼ 100, the actual

coverage probability for this interval can differ considerably from the nominal

probability .95, particularly when p is not close to .5 (the graph of coverage

probability versus p is very jagged because the underlying binomial probability

distribution is discrete rather than continuous). This is generally speaking a defi-

ciency of the traditional interval – the actual confidence level can be quite different

from the nominal level even for reasonably large sample sizes. Recent research has

shown that the score interval rectifies this behavior – for virtually all sample sizes

and values of p, its actual confiden ce level will be quite close to the nominal level

specified by the choice of z

a/2

. This is due largely to the fact that the score interval is

shifted a bit toward .5 compared to the traditional interval. In particular, the

midpoint

p of the score interval is always a bit closer to .5 than is the midpoint

of the traditional interval. This is especially important when p is close to 0 or 1.

In addition, the score interval can be used with nearly all sample sizes and

parameter values. It is thus not necessary to check the condi tions n

p  10 and

nð1 

pÞ10 which would be required were the traditional interval employed. So

rather than asking when n is large enough for (8.11) to yield a good approximation

Figure 8.6 Actual coverage probability for the interval (8.11) for varying values of

p when n ¼ 100

396

CHAPTER 8 Statistical Intervals Based on a Single Sample

to (8.10), our recommendation is that the score CI should always be used. The slight

additional tediousness of the computation is outweighed by the desirable properties

of the interval.

Example 8.8 Th e article “Repeatability and Reproducibility for Pass/Fail Data” (J. Testing Eval.,

1997: 151–153) reported that in n ¼ 48 trials in a particular laboratory, 16 resulted

in ignition of a particular type of substrate by a lighted cigarette. Let p denote the

long-run proportion of all such trials that would result in ignition. A point estimate

for p is

p ¼ 16=48 ¼ :333. A confidence interval for p with a confidence level of

approximately 95% is

:333 þ 1:96

=96

1 þ 1:96

=48

 1:96

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð:333Þð:667Þ=48 þ 1:96

=ð4  48

1 þ 1:96

=48

¼ :346  :129 ¼ð:217;:475Þ

The traditional interval is

:333  1:96

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

ð:333Þð:667Þ=48

¼ :333  :133 ¼ð:200;:466Þ

These two intervals would be in much closer agreement were the sample size

substantially larger.

■

Equating the width of the CI for p to a prespecified width w gives a quadratic

equation for the sample size n necessary to give an interval with a desired degree of

precision. Suppressing the subscript in z

a/2

, the solution is

n ¼

q  z



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

qð

q  w

Þþw

ð8:12Þ

Neglecting the terms in the numerator involving w

gives

n ¼

This latter expression is what results from equating the width of the traditional

interval to w.

These formulas unfortunately involve the unknown p. The most conservative

approach is to take advantage of the fact that

q½¼

pð1 

pÞ is a maximum when

p ¼ :5. Thus if

p ¼

q ¼ :5 is used in (8.12), the width will be at most w regardless

of what value of

p results from the sample. Alternatively, if the investigator

believes strongly, based on prior information, that p  p

 .5, then p

can be

used in place of

p. A similar comment applies when p  p

 .5.

Example 8.9 Th e width of the 95% CI in Example 8.8 is .258. The value of n neces sary to ensure

a width of .10 irrespective of the value of p is

n ¼

2ð1:96Þ

ð:25Þð1:96 Þ

ð:01Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

4ð1:96Þ

ð:25Þð:25  :01Þþð:01Þð1:96Þ

:01

¼ 380:3

8.2 Large-Sample Conﬁdence Intervals for a Population Mean and Proportion 397