Chapter 7
Sampling and Sampling Distributions
 Simple Random Sampling
 Point Estimation
 Introduction to Sampling Distributions
 Sampling Distribution of x
 Sampling Distribution of p
Statistical Inference
The purpose of statistical inference is to obtain
information about a population from information
contained in a sample.
A population is the set of all the elements of interest.
A sample is a subset of the population.
Statistical Inference
The sample results provide only estimates of the
values of the population characteristics.
A parameter is a numerical characteristic of a
population.
Simple Random Sampling:
Finite Population

Finite populations are often defined by lists such as:
• Organization membership roster
• Credit card account numbers
• Inventory product numbers

A simple random sample of size n from a finite
population of size N is a sample selected such
that each possible sample of size n has the same
probability of being selected.
Simple Random Sampling:
Finite Population
 Replacing each sampled element before selecting
subsequent elements is called sampling with
replacement.
 Computer-generated random numbers are often used
to automate the sample selection process.
Simple Random Sampling:
Infinite Population


Infinite populations are often defined by an ongoing
process whereby the elements of the population
consist of items generated as though the process would
operate indefinitely.
A simple random sample from an infinite population
is a sample selected such that the following conditions
are satisfied.
• Each element selected comes from the same
population.
• Each element is selected independently.
Point Estimation
We refer to
mean .
x as the point estimator of the population
s is the point estimator of the population standard
deviation .
p is the point estimator of the population proportion p.
Sampling Error
 When the expected value of a point estimator is equal
to the population parameter, the point estimator is said
to be unbiased.
 The difference between an unbiased point estimate and
the corresponding population parameter is called the
sampling error.
 Statistical methods can be used to make probability
statements about the size of the sampling error.
Example: St. Andrew’s
St. Andrew’s College receives
900 applications annually from
prospective students. The
application form contains
a variety of information
including the individual’s
scholastic aptitude test (SAT) score and whether or not
the individual desires on-campus housing.
Example: St. Andrew’s
The director of admissions
would like to know the
following information:
• the average SAT score for
the 900 applicants, and
• the proportion of
applicants that want to live on campus.
Conducting a Census

If the relevant data for the entire 900 applicants were
in the college’s database, the population parameters of
interest could be calculated using the formulas
presented in Chapter 3.

We will assume for the moment that conducting a
census is practical in this example.
Conducting a Census


Population Mean SAT Score
xi


 990
900
Population Standard Deviation for SAT Score


2
(
x


)
 i
 80
900
Population Proportion Wanting On-Campus Housing
648
p
 .72
900
Simple Random Sampling
 Now suppose that the necessary data on the
current year’s applicants were not yet entered in the
college’s database.
 Furthermore, the Director of Admissions must obtain
estimates of the population parameters of interest for
a meeting taking place in a few hours.
 She decides a sample of 30 applicants will be used.
 The applicants were numbered, from 1 to 900, as
their applications arrived.
Simple Random Sampling:
Using a Random Number Table

Use of Random Numbers for Sampling
3-Digit
Applicant
Random Number Included in Sample
744
No. 744
436
No. 436
865
No. 865
790
No. 790
835
No. 835
902
Number exceeds 900
190
No. 190
836
No. 836
. . . and so on
Simple Random Sampling:
Using a Random Number Table

Sample Data
Random
No. Number
1
744
2
436
3
865
4
790
5
835
.
.
.
.
30
498
Applicant
Conrad Harris
Enrique Romero
Fabian Avante
Lucila Cruz
Chan Chiang
.
.
SAT
Score
1025
950
1090
1120
930
.
.
Live OnCampus
Yes
Yes
No
Yes
No
.
.
Emily Morse
1010
No
Point Estimation

x as Point Estimator of 
x

x
30


29, 910
 997
30
s as Point Estimator of 
s

i
 (x
i
 x )2
29

163, 996
 75.2
29
p as Point Estimator of p
p  20 30  .68
Note: Different random numbers would have
identified a different sample which would have
resulted in different point estimates.
Summary of Point Estimates
Obtained from a Simple Random Sample
Population
Parameter
Parameter
Value
 = Population mean
990
x = Sample mean
997
 = Population std.
80
s = Sample std.
deviation for
SAT score
75.2
p = Population proportion wanting
campus housing
.72
p = Sample pro-
.68
SAT score
deviation for
SAT score
Point
Estimator
Point
Estimate
SAT score
portion wanting
campus housing
Sampling Distribution of x

Process of Statistical Inference
Population
with mean
=?
The value of x is used to
make inferences about
the value of .
A simple random sample
of n elements is selected
from the population.
The sample data
provide a value for
the sample mean x.
Sampling Distribution of x
The sampling distribution of x is the probability
distribution of all possible values of the sample
mean x.
Expected Value of
x
E( x ) = 
where:
 = the population mean
Sampling Distribution of x
Standard Deviation of x
Finite Population
 N n
x  ( )
n N 1
Infinite Population
x 

n
• A finite population is treated as being
infinite if n/N < .05.
• ( N  n) / ( N  1) is the finite correction factor.
•  x is referred to as the standard error of the
mean.
Form of the Sampling Distribution of x
If we use a large (n > 30) simple random sample, the
central limit theorem enables us to conclude that the
sampling distribution of x can be approximated by
a normal distribution.
When the simple random sample is small (n < 30),
the sampling distribution of x can be considered
normal only if we assume the population has a
normal distribution.
Sampling Distribution of x for SAT Scores
Sampling
Distribution
of x
E( x )  990
x 

80

 14.6
n
30
x
Sampling Distribution of x for SAT Scores
What is the probability that a simple random sample
of 30 applicants will provide an estimate of the
population mean SAT score that is within +/10 of
the actual population mean  ?
In other words, what is the probability that x will be
between 980 and 1000?
Sampling Distribution of x for SAT Scores
Step 1: Calculate the z-value at the upper endpoint of
the interval.
z = (1000 - 990)/14.6= .68
Step 2: Find the area under the curve between µ and Z
P(µ ≤ z < .68) = .2517
Sampling Distribution of x for SAT Scores
Sampling
Distribution
of x
 x  14.6
Area = .2517
x
990 1000
Sampling Distribution of x for SAT Scores
Step 3: Calculate the z-value at the lower endpoint of
the interval.
z = (980 - 990)/14.6= - .68
Step 4: Find the area under the curve between µ and
lower endpoint.
P(µ ≥ z ≥ - .68) = .2517
Sampling Distribution of x for SAT Scores
Sampling
Distribution
of x
 x  14.6
Area = .2517
.2517
980
x
990 1000
Sampling Distribution of x for SAT Scores
Step 5: Calculate the area under the curve between
the lower and upper endpoints of the interval.
P(-.68 < z < .68) = .2517 + .2517 = .5034
The probability that the sample mean SAT score will
be between 980 and 1000 is:
P(980 <
x < 1000) = .5034
Sampling Distribution of x for SAT Scores
Sampling
Distribution
of x
 x  14.6
Area = .5034
980 990 1000
x
Relationship Between the Sample Size
and the Sampling Distribution of x
 Suppose we select a simple random sample of 100
applicants instead of the 30 originally considered.
 E(x ) =  regardless of the sample size. In our
example, E( x) remains at 990.
 Whenever the sample size is increased, the standard
error of the mean  x is decreased. With the increase
in the sample size to n = 100, the standard error of the
mean is decreased to:

80
x 

 8.0
n
100
Relationship Between the Sample Size
and the Sampling Distribution of x
With n = 100,
x  8
With n = 30,
 x  14.6
E( x )  990
x
Relationship Between the Sample Size
and the Sampling Distribution of x
 Recall that when n = 30, P(980 <
x < 1000) = .5034.
 We follow the same steps to solve for P(980 < x < 1000)
when n = 100 as we showed earlier when n = 30.
 Now, with n = 100, P(980 <
x < 1000) = .7888.
 Because the sampling distribution with n = 100 has a
smaller standard error, the values of x have less
variability and tend to be closer to the population
mean than the values of x with n = 30.
Relationship Between the Sample Size
and the Sampling Distribution of x
Sampling
Distribution
of x
x  8
Area = .7888
x
980 990 1000
Sampling Distribution of p

Making Inferences about a Population Proportion
Population
with proportion
p=?
The value of p is used
to make inferences
about the value of p.
A simple random sample
of n elements is selected
from the population.
The sample data
provide a value for the
sample proportion p.
Sampling Distribution of p
The sampling distribution of p is the probability
distribution of all possible values of the sample
proportion p.
Expected Value of p
E ( p)  p
where:
p = the population proportion
Sampling Distribution of p
Standard Deviation of p
Finite Population
p 
p(1  p) N  n
n
N 1
Infinite Population
p 
p(1  p)
n
 p is referred to as the standard error of the
proportion.
Form of the Sampling Distribution of p
The sampling distribution of p can be approximated
by a normal distribution whenever the sample size
is large.
The sample size is considered large whenever these
conditions are satisfied:
np > 5
and
n(1 – p) > 5
Sampling Distribution of p

Example: St. Andrew’s College
Recall that 72% of the
prospective students applying
to St. Andrew’s College desire
on-campus housing.
What is the probability that
a simple random sample of 30 applicants will provide
an estimate of the population proportion of applicant
desiring on-campus housing that is within plus or
minus .05 of the actual population proportion?
Sampling Distribution of p
Sampling
Distribution
of p
.72(1  .72)
p 
 .082
30
E( p )  .72
p
Sampling Distribution of p
Step 1: Calculate the z-value at the upper endpoint of
the interval.
z = (.77 - .72)/.082 = .61
Step 2: Find the area under the curve between µ and the
upper endpoint.
P(µ ≤ z < .61) = .2291
Sampling Distribution of p
Sampling
Distribution
of p
 p  .082
Area = .2291
p
.72 .77
Sampling Distribution of p
Step 3: Calculate the z-value at the lower endpoint of
the interval.
z = (.67 - .72)/.082 = - .61
Step 4: Find the area under the curve between µ and the
lower endpoint.
P(z < -.61) = P(µ ≥ z > -.61)
= .2291
Sampling Distribution of p
Sampling
Distribution
of p
Area = .2291
 p  .082
.2291
p
.67 .72 .77
Sampling Distribution of p
Step 5: Calculate the area under the curve between
the lower and upper endpoints of the interval.
P(-.61 < z < .61) = .2291 + .2291
= .4582
The probability that the sample proportion of applicants
wanting on-campus housing will be within +/-.05 of the
actual population proportion :
P(.67 < p < .77) = .4582
Sampling Distribution of p
Sampling
Distribution
of p
 p  .082
Area = .4582
p
.67
.72
.77