Chapter 8
Sampling Variability and Sampling Distributions
In inferential statistics, we want to use information contained in a sample to reach conclusions
about one or more characteristics of the population from which the sample was selected. In the
subsequent chapters, we will introduce some inferential methods.
8.1 Statistics and Sampling Variability
A statistic  any quantity computed from values in a sample (for example, x , s, the sample
median, the sample interquartile range and so on).
A population parameter  any quantity computed from values in a population (for example, ,
, the population median, the population interquartile range and so on).

The difference between a statistic and a population parameter.
(1) A statistic is a sample characteristic, whereas a population parameter is a population
characteristic.
(2) The observed value of a statistic depends on the particular sample selected from the
population; typically, it varies from sample to sample. This variability is called sampling
variability. However, a population parameter is a fixed number, which is generally unknown.
A population is an entire collection of individuals or objects about which information is desired.
Generally we are interested in a variable of a population. A variable associates a value with each
individual or object in a population. For example,
Population = {all students at NAU}
= {student 1, student 2, , student k}
The variable of interest for this population may be x = height of a student.
Now we consider a hypothetical “population”:
Population of samples = {all possible samples of a given size n}
= {sample 1, sample 2,, sample k}
For a population of samples, we may be interested in a statistic (for example, x ).
Just as a variable associates a value with every individual or object in a population and can be
described by its distribution, a statistic associates a value with each individual sample in the
population of samples, and can also be described by a distribution.
Definition 8.1: The distribution of a statistic is called its sampling distribution.
Example 8.1 Population = {6 students}. The variable of interest is x = the amount of money
(dollars) each student spent on textbooks for the current semester, which is given in the
following table.
Student Student 1 Student 2 Student 3 Student 4 Student 5 Student 6
x
267
258
261
275
252
288
The distribution of x can be summarized by a density histogram.
Class Interval Relative Frequency Density
250 to < 260
0.333
0.0333
260 to < 270
0.333
0.0333
270 to < 280
0.167
0.0167
280 to < 290
0.167
0.0167
Density
Table 8.1 Frequency distribution for x
0.035
0.03
0.025
0.02
0.015
0.01
0.005
0
255
265
275
285
x
Figure 8.1 Density histogram for x
Now we consider all possible samples of size 2 and calculate their means.
Sample
{student1, student 2}
{student1, student 3}
{student1, student 4}
{student1, student 5}
{student1, student 6}
{student2, student 3}
{student2, student 4}
{student2, student 5}
x
262.5
264
271
259.5
277.5
259.5
266.5
255
Sample
{student2, student 6}
{student3, student 4}
{student3, student 5}
{student3, student 6}
{student4, student 5}
{student4, student 6}
{student5, student 6}
x
273
268
256.5
274.5
263.5
281.5
270
Table 8.2 All possible samples of size 2 and their means
The sampling distribution of x can also be summarized by a density histogram.
Class Interval Relative Frequency Density
250 to < 260
0.267
0.0267
260 to < 270
0.333
0.0333
270 to < 280
0.333
0.0333
280 to < 290
0.067
0.0067
Table 8.3 Frequency distribution for x
0.035
0.03
Density
0.025
0.02
0.015
0.01
0.005
0
255
265
275
285
Figure 8.2 Density histogram for x
8.2 The sampling Distribution of a Sample Mean
When we want to make an inference about the population mean, , it is natural to consider to use
the sample mean, x . The behavior of x is described by its sampling distribution. The sample
size n and characteristics of the population (for example, mean value , and standard deviation ,
and its shape) are important in determining the sampling distribution of x .
Generally, it is hard to obtain the true sampling distribution of x . We approximate the sampling
distribution by some sampling experiments. Let us recall the experiment of tossing a coin.
Population = {head, tail}. We can obtain an approximate distribution of the population as
follows: toss the coin 500 times; construct a relative frequency histogram for the outcomes of the
500 tosses. Since 500 is reasonably large, the histogram should rather closely resemble the
population distribution.
Relative Frequency
Head
Tail
Figure 8.3 An approximate distribution of the population = {head, tail} based on 500 tosses
An approximate sampling distribution of x can be obtained in the same way: (1) select 500
different random samples of a given size n; (2) compute x for each sample; (3) construct a
sample histogram of these 500 x values. The sample histogram should rather closely resemble
the true sampling distribution of x .

General Properties of the Sampling Distribution of x
Let x denote the mean of the observations in a random sample of size n from a population
having mean  and standard deviation . Denote the mean value of the x distribution by  x and
the standard deviation of x distribution by  x . Then the following rules hold.
Rule 1:  x =
Rule 2:  x = n .
Rule 3: When the population distribution is normal, the sampling distribution of x is also normal
for any sample size n. Thus, the standardized variable
z
x x
x

x 
/ n
has the standard normal (z) distribution.
Rule 4: (Central Limit Theorem) When n is sufficiently large, the sampling distribution of x is
well approximated by a normal curve, even when the population distribution is not itself
normal. So, the standardized variable
z
x x
x

x 
/ n
has approximately the standard normal (z) distribution.
Note: The Central Limit Theorem can safely be applied if n exceeds 30.
Example 8.2 A patient visits her doctor with concerns about her blood pressure. If the systolic
blood pressure exceeds 150, the patient is considered to have high blood pressure and medication
may be prescribed. The problem is that there is a considerable variation in a patient’s systolic
blood pressure readings during a given day. Suppose that the patient’s systolic readings during a
given day have a normal distribution with a mean of  = 160 mm mercury and a standard
deviation of  = 20 mm. If five measurements are taken at various times during the day, what is
the probability that the average blood pressure reading will be less than 150 and hence fail to
indicate that the patient has a high blood pressure problem?
Let x denote the average blood pressure reading. Because the patient’s systolic readings during a
given day have a normal distribution, by rule 3, the sampling distribution of x is also normal,
and by rules 1 and 2, the sampling distribution of x has mean value
 x =  = 160
and standard deviation
x =

n

20
5
= 8.9443.
P( x < 150) = P( z < (? - ?) / ?) = P(z < ?) = ?.
Example 8.3 An anthropologist wishes to estimate the average height  of men for a certain race
of people. Suppose that the population standard deviation  is 2.5 inches and that she randomly
samples 100 men.
(1) Find the probability that the difference between the sample mean x and the true population
mean  will not exceed .5 inch.
(2) If the true population mean  = 68 inches, find the probability that x will exceed 68.4 inches.
(1) By rule 2,  x =

n

2.5
100
= .25. Since n = 100, we can invoke the Central Limit Theorem and
regard the x distribution as approximately normal. Then,
P( x    0.5 ) = P ( x0.25 
0.5
0.25
)  P( z  2 ) = P(  2  z  2 )
= P( z  2 ) - P(z < -2) = 0.9772 -0.0228 = 0.9544.
(2) By the Central Limit Theorem,
P( x > 68.4) ≈ P( z > (? - ?) / ?) = P(z > ?) = 1 - P(z  ?) = ?.
8.3 The Sampling Distribution of a Sample Proportion
When we investigate the proportion of individuals or objects in a population that possess a
specified property, traditionally, we label the individual or object that possesses the property of
interest S (for success), and the one that does not possess the property F (for failure). Let 
denote the proportion of S’s in the population. For example, population = {Students at NAU}. 
= the proportion of female students.
The value of  is usually unknown. When a random sample of size n is selected from this type of
population, some of the individuals in the sample are S’s, and the remaining individuals in the
sample are F’s. The statistic that will provide a basis for making inferences about  is p, the
sample proportion of S’s:
p = (number of S’s in the sample) / n.
Just as making inference about  requires knowing something about the sampling distribution of
x , making inferences about  requires knowing properties of the sampling distribution of the
statistic p.
Let
1
if it is a success (S),
0
if it is a failure (F).
x=
Then, x has the Bernoulli( ) distribution and the sample becomes
Sample F S S  F
S
x
0 1 1  0
1
x1 x2 x3  xn-1 xn
x
Now x  n = (number of S’s in the sample) / n = p. Thus we have following rules:
Note: For the Bernoulli( ) distribution, mean  =  and standard deviation  =

 (1   ) .
General Properties of the Sampling Distribution of p
Let p be the proportion of S’s in a random sample of size n from a population whose proportion
of S’s is . Denote the mean value of p by p and the standard deviation of p by p. Then the
following rules hold.
Rule 1: p = 
Rule 2:  p 
 (1   )
n
Rule 3: (Central Limit Theorem) When n is large and  is not too near 0 or 1, the sampling
distribution of p is approximately normal. Thus, the standardized variable
z
p p
p

p 
 (1 ) / n
has approximately the standard normal (z) distribution.
Note: If both n  10 and n(1- ) 10 , it is safe to use a normal approximation.
Example 8.4 Population = {All blood recipients}.  = The proportion of all blood recipients
stricken with viral hepatitis = .07. A new treatment is given to n = 200 blood recipients. Only 6
of the 200 patients contract hepatitis. The question of interest to medical researchers is: Is the
new treatment effective?
If the new treatment is ineffective, then
p =  = .07,  p 
 (1   )
n

(.07 )(1  .07 )
200
= .018.
Here, p = 6 / 200 = .03. Since n = 200(.07) = 14 > 10 and n(1-) = 200(1-.07) = 186 > 10, the
sampling distribution of p is approximately normal. Then
P( p  .03 )  P( z  (.03 - .07) / .018 )= P( z  -2.22 ) = .0132
Since the probability is very small, as means that it is unlikely that a sample proportion .03 or
smaller would be observed if the new treatment really were ineffective, the new treatment is
effective.
Exercise in class: Suppose that the sample size n = 100 and the population proportion  = 0.95.
a) Does the sample proportion p have approximately a normal distribution? Explain.
b) What is the smallest value of n for which the sampling distribution of p is approximately
normal?