Chapter 3 sampling and sampling distribution

Chapter 3
Sampling and Sampling
Distribution
Mr. Anthony F. Balatar Jr.
Subject Instructor

Sampling Distribution of Sample Means
Steps in Constructing the Sampling Distribution of
the Means
1. Determine the number of possible samples that can be
drawn from the population using the formula:
NCn
where N = size of the population
n = size of the sample size
2. List all the possible samples and compute the mean of
each sample.
3. Construct a frequency distribution of the sample means
obtained in Step 2.

Illustrative Example. A population consists of the
numbers 2, 4, 9, 10 and 5. List all possible sample of
size 3 from this population and compute the mean of
each sample.
Solution:
Using the formula NCn to determine the number of
possible samples that can be drawn from the
population. (N = 5 and n = 3)
NCn = 5C3 = 10

Sample Mean
2, 4, 9 5
2, 4, 10 5.33
2, 4, 5 3.67
2, 9, 10 7
2, 9, 5 5.33
2, 10, 5 5.67
4, 9, 10 7.67
4, 9, 5 6
4, 10, 5 6.33
9, 10, 5 8
There are 10
possible
samples of size
3 that can be
drawn from the
given
population.

Sample Mean Frequency
3.67 1
5 1
5.33 2
5.67 1
6 1
6.33 1
7 1
7.67 1
8 1
Total n = 10
Make a
frequency
distribution of
sample means or
the sampling
distribution of
the sample
means.

A sampling distribution of sample means is a
frequency distribution using the means computed
from all possible random samples of a specific size
taken from a population.

Sample Mean
(X)
Frequency
Probability
P(X)
3.67 1 1/10 or 0.1
5 1 1/10 or 0.1
5.33 2 2/10 or 0.2
5.67 1 1/10 or 0.1
6 1 1/10 or 0.1
6.33 1 1/10 or 0.1
7 1 1/10 or 0.1
7.67 1 1/10 or 0.1
8 1 1/10 or 0.1
Total n = 10 ΣP(X) = 1
Make a
probability
distribution of
sample means
or the
complete
sampling
distribution of
the sample
means.

Individual Activity #1
Construct a Complete Sampling Distribution of the
Means using the problem below.
Sample of three cards are drawn at random from a
population of eight cards numbered from 1 to 8.

QUIZ #1
How many different samples of size n = 3 can be
selected from a population with the following sizes?
1. N = 4
2. N = 8
3. N = 15
4. N = 20
5. N = 50

QUIZ #2
Construct a Sampling Distribution of the Means
using the problem below.
1.A population consists of the five numbers 2, 3, 6,
8, and 11. Consider samples of size 2 that can be
drawn from this population.
2.A group of students who got the following scores in
a test: 6, 9, 12, 15, 18, and 21. Consider samples
of size 3 that can be drawn from this population.

QUIZ #2
Construct a Sampling Distribution of the Means
using the problem below.
3.A finite population consists of 8 elements:
10 10 10 10 10 12 18 40
a. How many samples of size n = 2 can be drawn from
this population?
b. List all the possible samples and the corresponding
means.
c. Construct a complete sampling distribution of the
sample means.

Finding the Mean and Variance of the
Sampling Distribution of Means
Consider a population consisting 1, 2, 3, 4, 5.
Suppose samples of size (a) 2 and (3) are drawn from
this population. Describe the sampling distribution of
the sample means.
1. What is the mean and variance of the sampling
distribution of the sample means?
2. Compare these values to the mean and variance of
the population.

Solutions for (a) n = 2:
1. Compute the mean of the population (μ).
𝜇 =
Σ𝑋
𝑁
=
1+2+3+4+5
5
= 3
2. Compute for the variance of the population (σ).
X X - μ (X – μ)2
1 -2 4
2 -1 1
3 0 0
4 1 1
5 2 4
Σ(X – μ)2 = 10
Σ(X – μ)2

3. Determine the number of
possible samples of size n = 2.
Use the formula NCn. Here N = 5
and n = 2. 5C2 = 10
So, there are 10 possible samples
of size 2 that can be drawn.
4. List all possible samples and
their corresponding means.
Samples Mean
1, 2 1.5
1, 3 2
1, 4 2.5
1, 5 3
2, 3 2.5
2, 4 3
2, 5 3.5
3, 4 3.5
3, 5 4
4, 5 4.5

5. Construct the sampling
distribution of the sample
means.
Sample
Mean (X)
Frequency
Probability
P(X)
1.5 1 0.1
2 1 0.1
2.5 2 0.2
3 2 0.2
3.5 2 0.2
4 1 0.1
4.5 1 0.1
Total N = 10 1

6. Compute the mean of the sampling
distribution of the sample means (μX).
Follow these steps:
a. Multiply the sample mean by the
corresponding probability.
b. Add the results.
μX = ΣX · P(X) = 3
So, the mean of the sampling
distribution of the sample means is 3.
Sample
Mean (X)
Probability
P(X)
X · P(X)
1.5 0.1 0.15
2 0.1 0.2
2.5 0.2 0.5
3 0.2 0.6
3.5 0.2 0.7
4 0.1 0.4
4.5 0.1 0.45
Total 1 ΣX · P(X) = 3

7. Compute for the variance (σ2
X)
of the sampling distribution of
the sample means. Follow these
steps:
a. Subtract the population
mean (μ) from each sample mean
(x). Label this as X – μ.
b. Square the difference. Label
this as (X – μ)2.
c. Multiply the results by the
corresponding probability. Label
this as P(X) · (X – μ)2
d. Add the results.
Sample
Mean (X)
Probability
P(X)
X – μ (X – μ)2 P(X) · (X – μ)2
1.5 0.1 -1.5 2.25 0.225
2 0.1 -1 1 0.1
2.5 0.2 -0.5 0.25 0.050
3 0.2 0 0 0
3.5 0.2 0.5 0.25 0.050
4 0.1 1 1 0.1
4.5 0.1 1.5 2.25 0.225
Total 1 0.750
σ2 = ΣP(X) · (X – μ)2 = 0.75
So, the variance of the sampling
distribution is 0.75.

1. Compute the mean of the population (μ).
𝜇 =
Σ𝑋
𝑁
=
1+2+3+4+5
5
= 3
2. Compute for the variance of the population (σ).
X X - μ (X – μ)2
1 -2 4
2 -1 1
3 0 0
4 1 1
5 2 4
Σ(X – μ)2 = 10
Σ(X – μ)2

3. Determine the number of
possible samples of size n = 3.
Use the formula NCn. Here N = 5
and n = 3. 5C3 = 10
So, there are 10 possible samples
of size 3 that can be drawn.
4. List all possible samples and
their corresponding means.
Samples Mean
1, 2, 3 2
1, 2, 4 2.33
1, 2, 5 2.67
1, 3, 4 2.67
1, 3, 5 3
1, 4, 5 3.33
2, 3, 4 3
2, 3, 5 3.33
2, 4, 5 3.67
3, 4, 5 4

5. Construct the sampling
distribution of the sample
means.
Sample
Mean (X)
Frequency
Probability
P(X)
2 1 0.1
2.33 1 0.1
2.67 2 0.2
3 2 0.2
3.33 2 0.2
3.67 1 0.1
4 1 0.1
Total N = 10 1

6. Compute the mean of the sampling
distribution of the sample means (μX).
Follow these steps:
a. Multiply the sample mean by the
corresponding probability.
b. Add the results.
μX = ΣX · P(X) = 3
So, the mean of the sampling
distribution of the sample means is 3.
Sample
Mean (X)
Probability
P(X)
X · P(X)
2 0.1 0.2
2.33 0.1 0.23
2.67 0.2 0.53
3 0.2 0.6
3.33 0.2 0.67
3.67 0.1 0.37
4 0.1 0.4
Total 1 ΣX · P(X) = 3

7. Compute for the variance (σ2
X)
of the sampling distribution of
the sample means. Follow these
steps:
a. Subtract the population
mean (μ) from each sample mean
(x). Label this as X – μ.
b. Square the difference. Label
this as (X – μ)2.
c. Multiply the results by the
corresponding probability. Label
this as P(X) · (X – μ)2
d. Add the results.
Sample
Mean (X)
Probability
P(X)
X – μ (X – μ)2 P(X) · (X – μ)2
2 0.1 -1 1 1/10
2.33 0.1 -0.67 4/9 2/45
2.67 0.2 -0.33 1/9 1/45
3 0.2 0 0 0
3.33 0.2 0.33 1/9 1/45
3.67 0.1 0.67 4/9 2/45
4 0.1 1 1 1/10
Total 1 1/3 = 0.33
σ2 = ΣP(X) · (X – μ)2 = 0.33
So, the variance of the sampling
distribution is 0.33.

Summary of comparison of the means and variances of the
population and the sampling distribution of the means.
The mean of the sampling distribution of the sample means
is always equal to the mean of the population. μx = μ
For n = 2 For n = 3
Population
(N=5)
Sampling
Distribution of
the Sample
Means
Population
(N=5)
Sampling
Distribution of
the Sample
Means
Mean μ = 3 μx = 3 μ = 3 μx = 3
Variance σ2 = 2 σ2
x = 0.75 σ2 = 2 σ2
x = 0.33
Standard
Variation
σ = 1.41 σx = 0.87 σ = 1.41 σx = 0.57

The variance of the sampling distribution is obtained
by using the formula 𝝈 𝟐
𝒙 =
𝝈 𝟐
𝒏
∙
𝑵 − 𝒏
𝑵 −𝟏
. This formula
holds when the population is finite. The examples
shown are all finite population. A finite population
is one that consists of a finite or fixed number of
elements, measurements, or observations; while an
infinite population contains, hypothetically at least,
infinitely elements.

The expression
𝑵 − 𝒏
𝑵 −𝟏
is called the finite population
correction factor. In general, when the population
is large and the sample size is small, the correlation
factor is not used since it will be very close to 1.

Properties of the Sampling Distribution of Sample Mean
If all possible samples of size n are drawn from a population
of size N with mean μ and the variance σ2, then the
sampling distribution of the sample means has the
following properties:
1. The mean of the sampling distribution of the sample
means is equal to the population mean μ. That is, μx = μ.
2. The variance of the sampling distribution of the sample
means σ is given by:
𝝈 𝟐
𝒙 =
𝝈 𝟐
𝒏
𝝈 𝟐
𝒙 =
𝝈 𝟐
𝒏
∙
𝑵 − 𝒏
𝑵 − 𝟏
for finite population for infinite population

Properties of the Sampling Distribution of Sample Mean
3. The standard deviation of the sampling distribution of
the sample means σ is given by:
𝝈 𝒙 =
𝝈
𝒏
𝝈 𝒙 =
𝝈 𝟐
𝒏
∙
𝑵 − 𝒏
𝑵 − 𝟏
for finite population
for infinite population
where
𝑵 − 𝒏
𝑵 −𝟏
is the finite
population correction factor

The standard deviation (σX) of the sampling distribution of the
sample means is also known as the standard error of the
mean. It measures the degree of accuracy of the sample mean
(σX) as an estimate of the population mean (μ).
A good estimate of the mean is obtained if the standard error of
the mean (σX) is small or close to zero, while a poor estimate, if
the standard error of the mean (σX) is large. Observe that the
value of σX depends on the size of the sample size.
Thus, if we want to get a good estimate of the population mean,
we have to make n sufficiently large. This fact is stated as a
theorem, which is known as the Central Limit Theorem (CLT).

The Central Limit Theorem (CLT)
If random samples of size n are drawn from a
population, then as n becomes larger, the sampling
distribution of the mean approaches the normal
distribution, regardless of the shape of the
population distribution.

Illustrative Example.
A population has a mean of 60 and a standard
deviation of 5. A random sample of 16 measurements
is drawn from this population. Describe the sampling
distribution of the sample means by computing its
mean and standard deviation.
We shall assume that the population is infinite.

Steps Solution
1. Identify the given information. μ = 60, σ = 5, and n = 16
2. Find the mean of the sampling
distribution. Use the property
that μx = μ.
μx = μ = 60
3. Find the standard deviation of
the sampling distribution. Use
the property that 𝜎𝑥 =
𝜎
𝑛
𝜎𝑥 =
𝜎
𝑛
=
5
16
=
5
4
= 1.25

Illustrative Example.
The heights of male college students are normally
distributed with mean of 68 inches and standard
deviation of 3 inches. If 80 samples consisting of 25
students each are drawn from the population, what
would be the expected mean and standard deviation
of the resulting sampling distribution of the means?
We shall assume that the population is infinite.

Steps Solution
1. Identify the given information. μ = 68, σ = 3, and n = 25
2. Find the mean of the sampling
distribution. Use the property
that μx = μ.
μx = μ = 68
3. Find the standard deviation of
the sampling distribution. Use
the property that 𝜎𝑥 =
𝜎
𝑛
𝜎𝑥 =
𝜎
𝑛
=
3
25
=
3
5
= 0.6

QUIZ #3
A.Find the finite population correction factor given
the following:
1.N = 200, n = 10
2.N = 2000, n = 10
3.N = 400, n = 40
4.N = 500, n = 10
5.N = 200, n = 20

QUIZ #3
B.Solve the following problems. Assume that the
population is infinite in each case.
1.The scores of individual students on a national
test have a normal distribution with mean 18.6
and standard deviation 5.9. at Federico Ramos
Rural High School, 76 students took the test. If
the scores at this school have the same
distribution as national scores, what are the
mean and standard deviation of the sample
mean for 76 students?

2.A molding machine prepares a certain kind of car
spare part with a target diameter of μ = 40.265
mm. The machine has some variability, so the
standard deviation of the diameters is σ = 0.004
mm. A sample of 6 spare parts is inspected each
hour for process control purposes and records
are kept of the sample mean diameter. What will
be the mean and standard deviation of the
numbers recorded?

Solving Problems Involving Sampling
Distribution of the Sample Means
The Central Limit Theorem is of fundamental
importance in statistics because it justifies the use of
normal curve methods for a wide range of problems.
This theorem applies automatically to sampling from
infinite population. It also assures us that no matter
what the shape of the population distribution of the
mean is, the sampling distribution of the sample
means is closely normally distributed whenever n is
large.

Consequently, it justifies the use of the formula 𝐳 =
𝑿 −μ
σ
𝒏
when computing for the probability that X will take on a
value within a given range in the sampling distribution of
X.
𝐳 =
𝑿 − μ
σ
𝒏
where X = sample mean
μ = population mean
σ = population standard deviation
n = sample size

Illustrative Example:
The average time it takes a group of college students to
complete a certain examination is 46.2 minutes. The
standard deviation is 8 minutes. Assume that the variable
is normally distributed.
a. What is the probability that a randomly selected college
student will complete the examination in less than 43
minutes?
b. If 50 randomly selected college students take the
examination, what is the probability that the mean time it
takes the group to complete the test will be less than 43
minutes?

Solution to (a):
Steps Solution
Identify the given information. μ = 46.2, σ = 8, and X = 43
Identify what is asked for. P(X < 43)
Identify the formula to be used.
Here we are dealing with an individual data
obtained from the population. So, we will be
use the formula z =
𝑥 − 𝜇
𝜎
to standardize 43.
Solve the problem
z =
𝑥 − 𝜇
𝜎
=
43 − 46.2
8
= −0.4
We shall find P(X < 43) by getting the area
under the normal curve.
P(X < 43) = P(z < -0.4)
= 0.5 – 0.1554
= 0.3446
= 34.46%

Solution to (b):
Steps Solution
Identify the given information. μ = 46.2, σ = 8, X = 43, n = 50
Identify what is asked for. P(< 43)
Here we are dealing with data from the sample
means. So, we will be use the formula 𝐳 =
𝑿 −μ
σ
𝒏
to
standardize 43.
Solve the problem
𝐳 =
𝑿 − μ
σ
𝒏
=
43 − 46.2
8
50
= −0.2.83
We shall find P(< 43) by getting the area under the
normal curve.
P(X < 43) = P(z < -2.83)
= 0.5 – 0.4997
= 0.0023
= 0.23%

Illustrative Example:
The average number of milligrams (mg) of cholesterol in a
cup of a certain brand of ice cream is 660 mg, and the
standard deviation is 35 mg. Assume the variable is
normally distributed.
a. If a cup of ice cream is selected, what is the probability
that the cholesterol content will be more than 670 mg?
b. If a sample of 10 cups of ice cream is selected, what is the
probability that the mean of the sample will be larger
than 670 mg?

Solution to (a):
Steps Solution
Identify the given information. μ = 660, σ = 35, and X = 670
Identify what is asked for. P(X > 670)
Here we are dealing with an individual data
obtained from the population. So, we will be
use the formula z =
𝑥 − 𝜇
𝜎
to standardize 670.
Solve the problem
z =
𝑥 − 𝜇
𝜎
=
670 − 660
35
= 0.29
We shall find P(X > 670) by getting the area
under the normal curve.
P(X > 670) = P(z > 0.29)
= 0.5 – 0.1141
= 0.3859
= 38.59%

Solution to (b):
Steps Solution
Identify the given information. μ = 660, σ = 35, X = 670 and n = 10
Identify what is asked for. P(X > 670)
Here we are dealing with data from the sample
means. So, we will be use the formula 𝐳 =
𝑿 −μ
σ
𝒏
to
standardize 670.
Solve the problem
𝐳 =
𝑿 − μ
σ
𝒏
=
670 − 660
35
10
= 0.9
We shall find P(< 43) by getting the area under the
normal curve.
P(X > 670) = P(z < 0.9)
= 0.5 – 0.3159
= 0.1841
= 18.41%

QUIZ #4
Solve the following problems.
1.The average cholesterol content of a certain canned
goods is 215 mg and the standard deviation is 15
mg. Assume the variable is normally distributed.
a.If a a canned good is selected, what is the
probability that the cholesterol content will be
greater than 220 mg?
b.if a sample of 25 canned goods is selected, what is
the probability that the mean of the sample will be
larger than 220 mg?

QUIZ #4
Solve the following problems.
2.The average public high school has 468 students
with a standard deviation of 87.
a.If a public school is selected, what is the
probability that the number of students enrolled is
greater than 400?
b.if a random sample of 38 public elementary school
is selected, what is the probability that the
number of students enrolled is between 445 and
485?

Chapter 3 sampling and sampling distribution

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