ESci 117 - Engineering Data Analysis
1st Semester, AY 2022-2023
Module 6
Sampling Distributions
Name: Henry Gabriel A. Pradilla
Degree and Year: BSABE – 3
Date Submitted: January 5, 2023
Schedule: TTh (10:00 – 11:30)
Learning Task No. 6.1
Basic Concepts of Random Sampling
Perform as indicated given the following information.
There are 10 engineering students enrolled in an online course during the midyear. Their
ages are as follows:
Student, i
Age, Xi
1
19
2
21
3
20
4
19
5
20
6
20
7
21
8
19
9
20
10
20
1. Compute for the population mean, 𝜇, and variance, 𝜎 2 , of the ages of these students.
Solution:
𝜇=
∑𝑁
19 + 21 + 20 + 19 + 20 + 20 + 21 + 19 + 20 + 20 199
𝑖=1 𝑋𝑖
=
=
𝑁
10
10
𝜇 = 19.9
Interpretation: The average age of the 10 engineering students enrolled in an online
course during the midyear is 19.9 years.
2
∑𝑁
𝑖=1(𝑋𝑖 − 𝜇)
𝑁
0.81 + 1.21 + 0.01 + 0.81 + 0.01 + 0.01 + 1.21 + 0.81 + 0.01 + 0.01 4.90
𝜎2 =
=
10
10
𝜎2 =
𝜎 2 = 0.49
Interpretation: The variance of the ages of the 10 engineering students enrolled in an
online course during the midyear is 0.49 years. This means that the
individual ages of the students do not vary greatly.
2. Suppose we select a sample consisting of 2 students from this online class using
SRSWOR. Present all possible samples of size 2 and the corresponding values of the
sample mean following the table below.
Physical
Sample
{1, 2}
{1,3}
{1,4}
{1,5}
{1,6}
{1,7}
{1,8}
{1,9}
{1,10}
{2,3}
{2,4}
{2,5}
{2,6}
{2,7}
{2,8}
{𝑋1 , 𝑋2 }
𝑥̅
{19, 21}
{19, 20}
{19, 19}
{19, 20}
{19, 20}
{19, 21}
{19, 19}
{19,20}
{19, 20}
{21, 20}
{21, 19}
{21, 20}
{21, 20}
{21, 21}
{21, 19}
20.0
19.5
19.0
19.5
19.5
20.0
19.0
19.5
19.5
20.5
20.0
20.5
20.5
21.0
20.0
Physical
Sample
{2,9}
{2,10}
{3,4}
{3,5}
{3,6}
{3,7}
{3,8}
{3,9}
{3,10}
{4,5}
{4,6}
{4,7}
{4,8}
{4,9}
{4,10}
{𝑋1 , 𝑋2 }
𝑥̅
{21, 20}
{21, 20}
{20, 19}
{20, 20}
{20, 20}
{20, 21}
{20, 19}
{20, 20}
{20, 20}
{19, 20}
{19, 20}
{19, 21}
{19, 19}
{19, 20}
{19, 20}
20.5
20.5
19.5
20.0
20.0
20.5
19.5
20.0
20.0
19.5
19.5
20.0
19.0
19.5
19.5
Physical
Sample
{5,6}
{5,7}
{5,8}
{5,9}
{5,10}
{6,7}
{6,8}
{6,9}
{6,10}
{7,8}
{7,9}
{7,10}
{8,9}
{8,10}
{9,10}
{𝑋1 , 𝑋2 }
𝑥̅
{20, 20}
{20, 21}
{20, 19}
{20, 20}
{20, 20}
{20, 21}
{20, 19}
{20, 20}
{20, 20}
{21, 19}
{21, 20}
{21, 20}
{19, 20}
{19, 20}
{20, 20}
20.0
20.5
19.5
20.0
20.0
20.5
19.5
20.0
20.0
20.0
20.5
20.5
19.5
19.5
20.0
3. Construct the sampling distribution of the sample mean, 𝑋̅. Give two statements about the
behavior of the sample mean.
𝑥̅
𝑓(𝑥̅ ) = 𝑃(𝑋̅ = 𝑥̅ )
19
3/45
19.5
15/45
20
16/45
20.5
10/45
21
1/45
Interpretation: We see from the sampling distribution of the sample mean, 𝑋̅, of sample
size, 2, that the most likely estimate of the population mean is 20 years of
age, which is not equal to the population mean. However, 20 years of age is
only 0.1 years larger than the population mean which is 19.9 years of age.
4. Find the mean of the sample mean, 𝐸(𝑋̅), and the standard error of the sample mean,
𝑆𝐸(𝑋̅) using two ways (i.e., by definition and by the general behavior of the sample mean
under the sampling scheme used). Interpret your computed values.
Solution:
(By definition)
𝐸(𝑋̅) = ∑ 𝑥̅ 𝑓(𝑥̅ )
3
15
16
10
1
𝐸(𝑋̅) = (19) ( ) + (19.5) ( ) + (20) ( ) + (20.5) ( ) + (21) ( )
45
45
45
45
45
𝐸(𝑋̅) = 19.9
Interpretation: Solving for the mean (or expected value) of the sample mean by its
definition, we can observe that it is just equal to the population mean which
is 19.9 years of age. This shows that 𝑋̅ is an unbiased estimator of 𝜇.
𝑆𝐸(𝑋̅) = √𝑉𝑎𝑟(𝑋̅) = √𝐸(𝑋̅ − 𝜇)2
3
15
16
10
1
𝑆𝐸(𝑋̅) = √(19 − 19.9)2 ( ) + (19.5 − 19.9)2 ( ) + (20 − 19.9)2 ( ) + (20.5 − 19.9)2 ( ) + (21 − 19.9)2 ( )
45
45
45
45
45
49
𝑆𝐸(𝑋̅) = √
225
𝑆𝐸(𝑋̅) ≈ 0.467
Interpretation: Solving for the standard error (or standard deviation) of the sample mean by
its definition, we first find the variance of the sample mean equal to 49/225
years of age which is approximately equal to 0.218 years of age. Computing
for the square root of the variance of the sample mean, we now have the
computed value of its standard error which tells us that, on average, the
sample mean will differ from the population mean by 7/15 years of age,
which is approximately equal to 0.467 years of age.
(By the general behavior of the sample mean under the sampling scheme used)
𝐸(𝑋̅) = 𝜇 =
∑𝑁
19 + 21 + 20 + 19 + 20 + 20 + 21 + 19 + 20 + 20 199
𝑖=1 𝑋𝑖
=
=
𝑁
10
10
𝐸(𝑋̅) = 19.9
Interpretation: Solving for the mean (or expected value) of the sample mean by the general
behavior of the sample mean under the sampling scheme used, we arrived
with the same value when we solved by its definition. We can observe that it
is still just equal to the population mean which is 19.9 years of age. Still, this
shows that 𝑋̅ is an unbiased estimator of 𝜇.
𝜎2 𝑁 − 𝑛
0.49 10 − 2
0.49 8
7
𝑆𝐸(𝑋̅) = √𝑉𝑎𝑟(𝑋̅ ) = √ (
)=√
(
)=√
( )=
𝑛 𝑁−1
2 10 − 1
2 9
15
𝑆𝐸(𝑋̅) ≈ 0.467
Interpretation: Solving for the standard error (or standard deviation) of the sample mean by
the general behavior of the sample mean under the sampling scheme used,
we first find the population variance, 𝜎 2 , which is equal to 0.49 (we solved
this earlier). Evaluating equation further, we arrived with the value of the
variance of the sample mean, which is the same as the one we computed
by definition, 49/225 years of age which is approximately equal to 0.218
years of age. Computing for the square root of the variance of the sample
mean, we arrived with the same value of the standard error as we
previously computed, telling us that, on average, the sample mean will differ
from the population mean by 7/15 years of age, which is approximately
equal to 0.467 years of age.
5. Find the 𝑃(|𝑋̅ − 𝜇| ≤ 1) and interpret.