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

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There are two general methods for obtaining information about a statistic’s

sampling distribution. One method involves calculations based on probability rules,

and the other involves carrying out a simulation experiment.

Deriving the Sampling Distribution of a Statistic

Probability rules can be used to obtain the distribution of a statistic provided that it

is a “fairly simple” function of the X

’s and either there are relatively few different

X values in the population or else the population distribution has a “nice” form. Our

next two examples illustrate such situations.

Example 6.2 A certain brand of MP3 player comes in three configurations: with 2 GB of

memory, costing $80, a 4 GB model priced at $100, and an 8 GB version with a

price tag of $120. If 20% of all purchasers choose the 2 GB model, 30% choose the

4 GB, and 50% choose the 8 GB model, then the probability distribution of the cost

of a single randomly selected MP3 player purchase is given by

x 80 100 120

p(x).2 .3 .5

with m = 106, s

= 244 (6.1)

Suppose only two MP3 players are sold today. Let X

¼ the cost of the first player

and X

¼ the cost of the second. Suppose that X

and X

are independent, each with

the probability distribution shown in (6.1), so that X

and X

constitute a random

sample from the distribution (6.1). Table 6.2 lists possible (x

, x

) pairs, the

probability of each computed using (6.1) and the assumption of independence,

and the resulting



x and s

values. (When n ¼ 2, s

¼ðx





xÞ

þðx





xÞ

Now to obtain the probability distribution of

X, the sample average cost per

MP3 player, we must consider each possible value



x and compute its probability.

For example,



x ¼ 100 occurs three times in the table with probabilities .10, .09,

and .10, so

Pð

X ¼ 100Þ¼:10 þ :09 þ :10 ¼ :29

Table 6.2 Outcomes, probabilities, and values



x and s

for Example 6.2

p(x

, x

)



80 80 (.2)(.2) ¼ .04 80 0

80 100 (.2)(.3) ¼ .06 90 200

80 120 (.2)(.5) ¼ .10 100 800

100 80 (.3)(.2) ¼ .06 90 200

100 100 (.3)(.3) ¼ .09 100 0

100 120 (.3)(.5) ¼ .15 110 200

120 80 (.5)(.2) ¼ .10 100 800

120 100 (.5)(.3) ¼ .15 110 200

120 120 (.5)(.5) ¼ .25 120 0

288

CHAPTER 6 Statistics and Sampling Distributions

Similarly, s

¼ 800 appears twice in the table with probability .10 each time, so

PðS

¼ 800Þ¼PðX

¼ 80; X

¼ 120ÞþPðX

¼ 120; X

¼ 80Þ

¼ :10 þ :10 ¼ :20

The complete sampling distributions of

X and S

appear in (6.2) and (6.3).



x 80 90 100 110 120



xÞ .2 .12 .29 .30 .5

(6.2)

0 200 800

ðs

.38 .42 .20

(6.3)

Figure 6.2 pictures a probability histogram for both the original distribution of

X (6.1) and the

X distribution (6.2). The figure suggests first that the mean

(i.e. expected value) of

X is equal to the mean $106 of the original distribution,

since both histograms appear to be centered at the same place. Indeed, from (6.2),

Eð

XÞ¼



xÞ¼80ð:04Þþþ120ð:25Þ¼106 ¼ m

Second, it appears that the

X distribution has smaller spread (variability) than the

original distribution, since the values of



x are more concentrated toward the mean.

Again from (6.2),

Vð

XÞ¼



x  mÞ



xÞ¼



x  106Þ



xÞ

¼ð80  106Þ

ð:04Þþþð120  106Þ

ð:25Þ¼122

Notice that the Vð

XÞ¼122 ¼ 244=2 ¼ s

=2, is exactly half the population vari-

ance; the division by 2 here is a consequence of the fact that n ¼ 2.

Finally, the mean value of S

EðS

Þ¼

ðs

Þ¼0ð:38Þþ200ð:42Þþ800ð:20Þ¼244 ¼ s

That is, the X sampling distribution is centered at the population mean m, and the S

sampling distribution (histogram not shown) is centered at the population variance s

If four MP3 players had been purch ased on the day of interest, the sample

average cost

X would be based on a random sample of four X

s, each having

the distribution (6.1). More calculation eventually y ields the distribution of

X for

n ¼ 4as



x 80 85 90 95 100 105 110 115 120



xÞ .0016 .0096 .0376 .0936 .1761 .2340 .2350 .1500 .0625

.04

.12

.29

.30

.25

10080

120 80 90 100 110 120

Figure 6.2 Probability histograms for (a) the underlying population distribution

and (b) the sampling distribution of

in Example 6.2

6.1 Statistics and Their Distributions 289

From this, EðXÞ¼106 ¼ m and VðXÞ¼61 ¼ s

=4. Figure 6.3 is a probability

histogram of this distribution.

Example 6.2 should suggest first of all that the computation of p



xÞand p

ðs

can be tedious. If the original distribution (6.1) had allowed for more than three

possible values 80, 100, and 120, then even for n ¼ 2 the computations would have

been more involved. The example should also suggest, however, that there are some

general relationships between Eð

XÞ; VðXÞ; EðS

Þ,andthemeanm and variance s

the original distribution. These are stated in the next section. Now consider an

example in which the random sample is drawn from a continuous distribution.

Example 6.3 Th e time that it takes to serve a customer at the cash register in a minimarket is a

random variable having an exponential distribution wi th parameter l. Suppose X

and X

are service times for two different customers, assumed independent of each

other. Consider the total service time T

¼ X

þ X

for the two customers, also a

statistic. The cdf of T

is, for t  0,

ðtÞ¼PðX

þ X

 tÞ¼

ðð

fðx

Þ:x

þx

tg

f ð x

; x

Þ dx

tx

lx

 le

lx

ðle

lx

 le

lt

Þ dx

¼ 1  e

lt

 lte

lt

The region of integration is pictured in Figure 6.4.

Figure 6.3 Probability histogram for

based on

¼ 4 in Example 6.2 ■

t − x

)

+ x

= t

Figure 6.4 Region of integration to obtain cdf of

in Example 6.3

290

CHAPTER 6 Statistics and Sampling Distributions

The pdf of T

is obtained by differentiating F

ðtÞ:

ðtÞ¼

lt

t  0

0 t < 0



ð6:4Þ

This is a gamma pdf (a ¼ 2 and b ¼ 1/l). Th is distribution for T

can also be

derived by a moment generating function argument.

The pdf of

X ¼ T

=2 can be obtained by the method of Section 4.7 as



xÞ¼



2l



x  0



x < 0



ð6:5Þ

The mean and variance of the underlying exponential distribution are m ¼ 1/l and

¼ 1/l

. Using Expressions (6.4) and (6.5), it can be verified that EðXÞ¼

1=l; Vð

XÞ¼1=ð2l

Þ; EðT

Þ¼2=l, and VðT

Þ¼2=l

. These results again sug-

gest some general relationships between means and variances of

X, T

, and the

underlying distribution. ■

Simulation Experiments

The second method of obtaining information about a statistic’s sampling distribu-

tion is to perform a simulation experiment. This method is usually used when a

derivation via probability rules is too difficult or complicated to be carried out.

Such an experiment is virtually always done with the aid of a computer.

The following characteristics of an experiment must be specified:

1. Th e statistic of interest (

X, S, a particular trimmed mean, etc.)

2. Th e population distribution (normal with m ¼ 100 and s ¼ 15, uniform with

lower limit A ¼ 5 and upper limit B ¼ 10, etc.)

3. Th e sample size n (e.g., n ¼ 10 or n ¼ 50)

4. Th e number of replications k (e.g., k ¼ 1000)

Then use a computer to obtain k different random samples, each of size n, from the

designated population distribution. For each such sample, calculate the value of the

statistic and construct a histogram of the k calculated values. This histogram gives

the approximate sampling distribution of the statistic. The larger the value of k,

the better the approximation will tend to be (the actual sampling distributio n

emerges as k !1). In practice, k ¼ 1000 may be enough for a “fairly simple”

statistic and population distribution, but modern computers allow a much larger

number of replications.

Example 6.4 Th e population distribution for our first simul ation study is normal with m ¼ 8.25

and s ¼ .75, as pictured in Figure 6.5. [The article “Platelet Size in Myocardial

Infarction” (British Med. J., 1983: 449–451) sugges ts this distribution for platelet

volume in individuals wi th no history of serious heart problems.]

We actually performed four different experiments, with 500 replic ations for

each one. In the first experiment, 500 samples of n ¼ 5 observations each were

generated using MINITAB, and the sample sizes for the other three were n ¼ 10,

n ¼ 20, and n ¼ 30, respectively. The sample mean was calculated for each

sample, and the resulting histograms of



x values appear in Figure 6.6.

6.1 Statistics and Their Distributions 291

m = 8.25

s = .75

⎧

⎨

⎩

6.00 6.75 7.50 9.00 9.75 10.50

Figure 6.5 Normal distribution, with m ¼ 8.25 and s ¼ .75

7.35 7.65 7.95 8.25 8.55 8.85 9.15

7.50 7.80 8.10 8.40 8.70 9.00 9.30

.05

.10

.15

.20

.25

.05

.10

.15

.20

.25

.05

.10

.15

.20

.25

.05

.10

.15

.20

.25

Relative

frequency

Relative

frequency

7.65 7.95 8.25 8.55 8.85

7.50 7.80 8.10 8.40 8.70

7.80 8.10 8.40 8.70

7.95 8.25 8.55

7.80 8.10 8.40 8.70

7.95 8.25 8.55

Relative

frequency

Relative

frequency

Figure 6.6 Sample histograms for

based on 500 samples, each consisting of

observations: (a)

¼ 5; (b)

¼ 10; (c)

¼ 20; (d)

¼ 30

292

CHAPTER 6 Statistics and Sampling Distributions

The first thing to notice about the histograms is their shape. To a reasonable

approximation, each of the four looks like a normal curve. The resemblance would

be even more striking if each histogram had been based on many more than 500



values. Second, each histogram is centered approximately at 8.25, the mean of the

population being sampled. Had the histograms been based on an unending sequence



x values, their centers would have been exactly the population mean, 8.25.

The final aspect of the histograms to note is their spread relative to each

other. The smaller the value of n, the greater the exte nt to which the sampling

distribution spreads out about the mean value. This is why the histograms for

n ¼ 20 and n ¼ 30 are based on narrow er class intervals than those for the two

smaller sample sizes. For the larger sample sizes, most of the



x values are quite

close to 8.25. This is the effect of aver aging. When n is small, a single unusual x

value can result in an



x valu e far from the center. With a larger sample size, any

unusual x values, when averaged in with the other sample values, still tend to yield



x value close to m. Combining these insights yields a result that should appeal

to your intuition:

X based on a large n tends to be closer to m than does X based on a

small n.

■

Example 6.5 Con sider a simulation experiment in which the population distribution is quite

skewed. Figure 6.7 shows the density curve for lifetimes of a certain type of

electronic control (This is actually a lognormal distribution with E[ln(X)] ¼ 3

and V[ln(X)] ¼ .16; that is, ln(X) is normal with mean 3 and variance .16.).

Again the statistic of interest is the sample mean

X. The experiment utilized

500 replications and considered the same four sample sizes as in Example 6.4.

The resulting histograms along with a normal probability plot from MINITAB for

the 500



x values based on n ¼ 30 are shown in Figure 6.8.

Unlike the normal case, these histograms all differ in shape. In particular,

they become progressively less skewed as the sample size n increases. The averages

of the 500



x values for the four different sample sizes are all quite close to the mean

value of the population distribution. If each histogram had been based on an

0 25 50 75

.01

.02

.03

.04

.05

f (x)

Figure 6.7 Density curve for the simulation experiment of Example 6.5 [

(

) ¼

m ¼ 21.7584,

(

) ¼ s

¼ 82.1449]

6.1 Statistics and Their Distributions 293

unending sequence of



x values rather than just 500, all four would have been

centered at exactly 21.7584. Thus different values of n change the shape but not

the center of the sampling distribution of

X. Comparison of the four histograms in

Figure 6.8 also shows that as n increases, the spread of the histograms decreases.

Increasing n results in a greater degree of concentration about the population mean

value and makes the histogram look more like a normal curve. The histogram of

Figure 6.8(d) and the normal probability plot in Figure 6.8(e) provide convincing

evidence that a sample size of n ¼ 30 is sufficient to overcome the skewness of the

population distribut ion and give an approximately normal

X sampling distribution.

0.00

0.05

0.10

0.15

0.20

0.25

16 20 24 28 32 36

n=5

12 16 20 24 28 32 36

n=20

12 16 20 24 28 32 36

n=10

12 16 20 24 28 32 36

n=30

Density

0.00

0.05

0.10

0.15

0.20

0.25

Density

0.00

0.05

0.10

0.15

0.20

0.25

Density

0.00

0.05

0.10

0.15

0.20

0.25

Density

Figure 6.8 Results of the simulation experiment of Example 6.5: (a)

histogram for

¼ 5;

(b)

histogram for

¼ 10; (c)

histogram for

¼ 20; (d)

histogram for

¼ 30; (e) normal

probability plot for

¼ 30 (from MINITAB) ■

294

CHAPTER 6 Statistics and Sampling Distributions

Exercises Section 6.1 (1–10)

1. A particular brand of dishwasher soap is sold in

three sizes: 25 oz, 40 oz, and 65 oz. Twenty

percent of all purchasers select a 25-oz box, 50%

select a 40-oz box, and the remaining 30% choose

a 65-oz box. Let X

and X

denote the package

sizes selected by two independently selected pur-

chasers.

a. Determine the sampling distribution of

X, cal-

culate E(

X), and compare to m.

b. Determine the sampling distribution of the

sample variance S

, calculate E(S

), and com-

pare to s

2. There are two traffic lights on the way to work.

Let X

be the number of lights that are red, requir-

ing a stop, and suppose that the distribution of X

is as follows:

012

p(x

).2 .5 .3

m ¼ 1.1, s

¼ .49

Let X

be the number of lights that are red on

the way home; X

is independent of X

. Assume

that X

has the same distribution as X

, so that X

is a random sample of size n ¼ 2.

a. Let T

¼ X

þ X

, and determine the probabil-

ity distribution of T

b. Calculate m

. How does it relate to m, the

population mean?

c. Calculate s

. How does it relate to s

, the

population variance?

3. It is known that 80% of all brand A DVD players

work in a satisfactory manner throughout the

warranty period (are “successes”). Suppose that

n ¼ 10 players are randomly selected. Let X ¼

the number of successes in the sample. The statis-

tic X/ n is the sample proportion (fraction) of suc-

cesses. Obtain the sampling distribution of this

statistic. [Hint: One possible value of X/n is .3,

corresponding to X ¼ 3. What is the probability of

this value (what kind of random variable is X)?]

4. A box contains ten sealed envelopes numbered 1,

..., 10. The first five contain no money, the next

three each contain $5, and there is a $10 bill in

each of the last two. A sample of size 3 is selected

with replacement (so we have a random sample),

and you get the largest amount in any of the

envelopes selected. If X

, X

, and X

denote the

amounts in the selected envelopes, the statistic of

interest is M ¼ the maximum of X

, X

, and X

a. Obtain the probability distribution of this

statistic.

b. Describe how you would carry out a simulation

experiment to compare the distributions of M

for various sample sizes. How would you guess

the distribution would change as n increases?

5. Let X be the number of packages being mailed by

a randomly selected customer at a shipping facil-

ity. Suppose the distribution of X is as follows:

x 1234

p(x).4 .3 .2 .1

a. Consider a random sample of size n ¼ 2 (two

customers), and let

X be the sample mean num-

ber of packages shipped. Obtain the probability

distribution of

b. Refer to part (a) and calculate Pð

X  2:5Þ.

c. Again consider a random sample of size n ¼ 2,

but now focus on the statistic R ¼ the sample

range (difference between the largest and smal-

lest values in the sample). Obtain the distribu-

tion of R.[Hint: Calculate the value of R for

each outcome and use the probabilities from

part (a).]

d. If a random sample of size n ¼ 4 is selected,

what is Pð

X  1:5Þ?[Hint: You should not

have to list all possible outcomes, only those

for which



x  1:5.]

6. A company maintains three offices in a region,

each staffed by two employees. Information con-

cerning yearly salaries (1000’s of dollars) is as

follows:

Office 112233

Employee 123456

Salary 29.7 33.6 30.2 33.6 25.8 29.7

a. Suppose two of these employees are randomly

selected from among the six (without replace-

ment). Determine the sampling distribution of

the sample mean salary

b. Suppose one of the three offices is randomly

selected. Let X

and X

denote the salaries of

the two employees. Determine the sampling

distribution of

c. How does Eð

XÞ from parts (a) and (b) compare

to the population mean salary m?

6.1 Statistics and Their Distributions 295

7. The number of dirt specks on a randomly selected

square yard of polyethylene film of a certain type

has a Poisson distribution with a mean value of 2

specks per square yard. Consider a random sample

of n ¼ 5 film specimens, each having area 1

square yard, and let

X be the resulting sample

mean number of dirt specks. Obtain the first 21

probabilities in the

X sampling distribution. [Hint:

What does a moment generating function argument

say about the distribution of X

þ ···þ X

8. Suppose the amount of liquid dispensed by a

machine is uniformly distributed with lower limit

A ¼ 8 oz and upper limit B ¼ 10 oz. Describe

how you would carry out simulation experiments

to compare the sampling distribution of the

(sample) fourth spread for sample sizes n ¼ 5,

10, 20, and 30.

9. Carry out a simulation experiment using a statistical

computer package or other software to study the

sampling distribution of

X when the population dis-

tribution is Weibull with a ¼ 2andb ¼ 5, as in

Example 6.1. Consider the four sample sizes n ¼ 5,

10, 20, and 30, and in each case use 500 replications.

For which of these sample sizes does the

X sampling

distribution appear to be approximately normal?

10. Carry out a simulation experiment using a statisti-

cal computer package or other software to study

the sampling distribution of

X when the popula-

tion distribution is lognormal with E[ln(X)] ¼ 3

and V[ln(X)] ¼ 1. Consider the four sample sizes

n ¼ 10, 20, 30, and 50, and in each case use 500

replications. For which of these sample sizes does

the

X sampling distribution appear to be approxi-

mately normal?

6.2

The Distribution of the Sample Mean

The importance of the sam ple mean X springs from its use in drawing concl usions

about the population mean m. Some of the mos t frequently used inferential proce-

dures are based on properties of the sampling distribution of

X. A preview of these

properties appeared in the calculations and simulation experiments of the previous

section, where we noted relationships between Eð

XÞ and m and also among VðXÞ,

, and n .

PROPOSITION

Let X

, X

, ..., X

be a random sample from a distribution with mean value m

and standard deviation s.Then

1. Eð

XÞ¼m

¼ m

2. Vð

XÞ¼s

¼ s

=n and s

¼ s=

ﬃﬃﬃ

In addition, with T

¼ X

þ ···þ X

(the sample total), E(T

) ¼ nm,

V(T

) ¼ ns

, and s

ﬃﬃﬃ

Proofs of these results are deferred to the next section. According to Result 1, the

sampling (i.e., probability) distribution of

X is centered precisely at the mean of

the population from which the sample has been selected. Result 2 shows that the

distribution becomes more concentrated about m as the sample size n increases. In

marked contrast, the distribution of T

becomes more spread out as n increases.

Averaging moves probability in toward the middle, whereas totaling spreads

probability out over a wider and wider range of values.

296 CHAPTER 6 Statistics and Sampling Distributions

Example 6.6 Th e amount of time that a patient spends in a certain outpatient surgery center

is a random variable with a mean value of 4.5 h. and a standard deviation of 2 h.

Let X

, ..., X

be the times for a random sample of 25 patients. Then the expected

value of the sample mean amount of time is Eð

XÞ¼m ¼ 4:5, and the expected total

time for the 25 patients is E(T

) ¼ nm ¼ 25(4.5) ¼ 112.5. The standard deviations

X and T

are

¼ s=

ﬃﬃﬃ

ﬃﬃﬃﬃﬃ

¼ :4

ﬃﬃﬃ

s ¼

ﬃﬃﬃﬃﬃ

ð2Þ¼10

If the sample size increases to n ¼ 100, Eð

XÞ is unchanged, but s

¼ :2, half of its

previous value (the sample size must be quadrupled to halve the standard deviation

X) . ■

The Case of a Normal Population Distribution

Looking back to the simulation experiment of Example 6.4, we see that when

the population distribution is normal, each histogram of



x values is well approxi-

mated by a normal curve. The precise result follows (see the next section for a

derivation).

PROPOSITION

Let X

, X

, ..., X

be a random sample from a normal distribution with mean

m and standard deviation s. Then for any n,

X is normally distributed (with

mean m and standard deviation s=

ﬃﬃﬃ

), as is T

(with mean nm and standard

deviation

ﬃﬃﬃ

s).

We know everything there is to know about the

X and T

distributions when the

population distribution is normal. In particular, probabilities such as Pð a 

X  bÞ

and Pc T

 dðÞcan be obtained simply by standardizing. Figure 6.9 illustrates

the proposition.

X distribution

when n = 10

X distribution

when n = 4

Population

distribution

Figure 6.9 A normal population distribution and

sampling distributions

6.2 The Distribution of the Sample Mean 297