Let’s Apply
A. Direction: Identify the terms that will complete each description.
1. The Mean of Sampling Distribution is always equal to population mean.
2. We write Finite Population Correction Factor as: FPC factor.
3. The sample mean is another label for Mean of Sampling Distribution.
4. The Standard Deviation of Sampling Distribution can be represented as SEM.
5. The Population Variance and Standard Deviation is square root of the variance than their
counter parts in Sampling Distribution of sample means.
B. Direction: Tell what will happen to each standard error of the mean when the sample
size:
1. Increased from 50 to 200? Explain: If the sample size is increased from 50 to 200 then the
standard error of the mean will decrease. It is because an inverse relationship exists between
the standard error of the mean and the sample size. A larger sample size leads to a smaller
value of the standard deviation.
2. Decreased from 250 to 225? Explain: If the sample size is decreased from 250 to 225 then
the value of standard error will increase. In case of a decrease in sample size, smaller data
values are clustered to the mean which leads to an increase in the standard error of the mean.
3. Increased from 200 to 400? Explain: If the sample size is increased from 200 to 400 then the
standard error of the sampling distribution of the mean will be declined. Sample means cluster
more closely around the population mean in case of a large sample size that has the potential to
smaller the spread.
4. Increased from 40 to 50? Explain: If the sample size is increased from 40 to 50 then standard
error will decrease due to the inverse relationship between two variables.
5. Decreased from 600 to 40? Explain: In case of a decrease in sample size from 600 to 40, the
standard error of the mean will be declined. Standard error equals the standard deviation
divided by the square root of the sample size, so an increase or decrease in sample size leads
to a fall or rise in the standard error.
C. Direction: Solve for the value of the population correction factor. Round your answer
to 2-decimal place, if necessary.
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
�−� �−� �−� �−� �−�
FPC = �−1
FPC = �−1
FPC = �−1
FPC = �−1
FPC = �−1
200−10 2000−10 400−40 500−10 200−20
FPC = 200−1
FPC = 2000−1
FPC = 400−1
FPC = 500−1
FPC = 200−1
FPC = 0.97 FPC = 0.99 FPC = 0.95 FPC = 0.99 FPC = 0.95
Let’s Analyze
Direction: Solve the problem below.
1. A population consists of three numbers (3, 6, 9). Consider all possible samples of size 2
which can be drawn without replacement from the population. Find the following:
a. Population mean b. Population variance c. Populations standard deviation
∑�� ∑(�� −�)2 ∑(�� −�)2
μ= σ2= σ=
� � �
μ σ2 =6 σ = �
3+6+9
= 3 σ = 2.45
d. Mean of the sampling distribution of the means.
= Possible samples of size 2 mean
; (3, 6) = 4.5
; (3, 9) = 6
; (6, 9) = 7.5
= Mean of the sampling distribution of the
means:
�.�+�+�.�
�
=6
e. Variance of the sampling distribution of the means
(�.�−�)� +(�−�)� +(�.�−�)�
�
= 1. 5
f. Standard deviation of the sampling distributions of the
means.
(�.�−�)� +(�−�)� +(�.�−�)�
3
= �. � = 1.22
Let’s Evaluate
Direction: Solve the following problems
1. Consider all samples of size 4 from this population without replacement: 6 8 10 12 13. What is the
mean and standard deviation of the Sampling Distribution?
ANSWER:
SAMPLE VALUES Mean X � − �� (x − �� )2
6,8,10,12 9 0.92 0.85
6,8,10,13 9.25 0.55 0.30
6,8,12,13 9.75 - 0.05 0.0025
6,10,12,13 10.25 0.45 0.20
8,10,12,13 10.75 0.95 0.90
�� = 9.8 Σ(x − �� )2 = 2.25
Mean of Sampling Distribution Standard Deviation of Sampling Distribution
Σx̅
�� = � σ� =
∑(� −�� )2
�
9+9.25+9.75+10.25+10.75
�� = 5 σ� =
2.25
5
49
�� = 5
= 9.8 σ� = 0.45
�� = 0.67
2. Given the population 1, 3, 4, 6, and 8. Suppose samples of size 3 with replacement were drawn from
this population. What is the mean and standard deviation of the Sampling Distribution?
ANSWER:
SAMPLE VALUES Mean X � − �� (x − �� )2
1,3,4 2.67 -1.73 2.99
1,3,6 3.33 -1.07 1.14
1,3,8 4 -0.4 0.16
1,4,6 3.67 -0.73 0.53
1,4,8 4.33 -0.07 0.0049
1,6,8 5 0.6 0.36
3,4,6 4.33 -0.07 0.0049
3,4,8 5 0.6 0.36
3,6,8 5.67 1.27 1.61
4,6,8 6 1.6 2.56
�� = 4.4 Σ(x − �� )2 = 9.72
Mean of Sampling Distribution Standard Deviation of Sampling Distribution
Σx̅
�� = � σ� =
∑(� −�� )2
�
�� 9.72
2.67+3.33+4+3.67+4.33+5+4.33+5+5.67+6 σ� = 10
= 10
σ� = 0.972
44
�� = 10
= 4.4
�� = 0.99
3. A random sample of n=100 measurements is obtained from a population of N=120, with �=55 and
σ=20. Describe the sampling distribution of sample means by computing the ��̅ and ��̅ .
Σx̅ σ N−n
�� = �
σ� =
�
· �−1
100+120+55+20
�� = 4 σ� =
20
·
120 −100
100 120 −1
295
�� = 4
= 73.75 20 20
σ� = ·
100 119
σ� = (2) (0.41)
�� = 0.82
4. Mang Renato’s ice cream machine serves about 220 ice creams daily, with an average of 315.45 ml
each serving. The machine has some variability so the standard deviation of the serving is 3.67 ml. A
sample of 10 ice creams is inspected. What are the mean and standard deviation of this sampling
distribution?
Σx̅ σ N−n
�� = �
σ� = ·
� �−1
220+315.45+3.67+10
�� = 3.67 315.45 −220
4 σ� = 220
· 315.45 −1
549.12
�� = 4
= 137.28
3.67 95.45
σ� = · 314.45
220
σ� = (0.25) (0.55)
�� = 0.121
5. A random sample of n=100 measurements is obtained from a population with �=55 and σ=20.
Describe the sampling distribution of sample means by computing the ��̅ and ��̅.
��̅ = μ = 55
σ 20
σ� = = =2
� 100
= The sampling distribution of the sample means has ��̅ = 55 and ��̅ = 2.
