2. Consider a population of four values (3, 5, 7, 9). a. Find all possible samples of size 2 which can be drawn without replacement from this population. b. Find the mean of the sampling distribution of means. c. Find the variance of the sampling distribution of the means.

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Please answer the activity below
Let's have another example.
Example 2: Consider a population with values (2, 3, 8).
a. Find all possible samples of size 2 which can be drawn without replacement from this
population.
b.
Find the mean of the sampling distribution of means.
c. Find the variance of the sampling distribution of means.
Solution:
a.
Again, we must make a table with the list of all possible samples.
Observation
Sample
1
(2, 3)
7
(2,8)
3
(3, 2)
4
(3,8)
5
(8, 2)
6
(8,3)
Notice that the observation was lessen because there was not repletion of the values in each sample.
b. Add another column for the average/ mean of each sample. Then compute for the mean of the
sampling distribution of the mean.
Observation Sample
X
Ex
1
(2,3)
2.5
Mx = n
2
(2,8)
5
3
(3, 2)
2.5
H₂ = 26
6
4
(3, 8)
5.5
5
(8,2)
5
H = 4.33
6
(8,3)
5.5
x=26
So the mean of the sampling distribution of the mean is
4.33.
c. Computing for the variance of the sampling distribution of the means, we have
Observation
X
x-μe
Sample
(2, 3)
(x-µ₂) ²
3.3489
02
Σ(x-μχ)2
1
2.5
-1.83
72
2
2
(2,8)
5
0.67
0.4489
3
(3,2)
2.5
-1.83
3.3489
0 =
10.3334
6
4
(3,8)
5.5
1.17
1.3689
5
(8, 2)
5
0.67
0.4489
o=1.7222
6
(8,3)
5.5
1.17
1.3689
So the variance is 1.72.
Σ*
Σ(8-4₂)² =
= 26
10.3334
Now it's your turn to practice.
Practice Exercise
of four values /
3. Tiu air POSSIDIO
wit Witteplacement
"ation.
2. Consider a population of four values (3, 5, 7, 9).
a. Find all possible samples of size 2 which can be drawn without replacement from this
population.
b. Find the mean of the sampling distribution of means.
c. Find the variance of the sampling distribution of the means.
2
5
=
Transcribed Image Text:Let's have another example. Example 2: Consider a population with values (2, 3, 8). a. Find all possible samples of size 2 which can be drawn without replacement from this population. b. Find the mean of the sampling distribution of means. c. Find the variance of the sampling distribution of means. Solution: a. Again, we must make a table with the list of all possible samples. Observation Sample 1 (2, 3) 7 (2,8) 3 (3, 2) 4 (3,8) 5 (8, 2) 6 (8,3) Notice that the observation was lessen because there was not repletion of the values in each sample. b. Add another column for the average/ mean of each sample. Then compute for the mean of the sampling distribution of the mean. Observation Sample X Ex 1 (2,3) 2.5 Mx = n 2 (2,8) 5 3 (3, 2) 2.5 H₂ = 26 6 4 (3, 8) 5.5 5 (8,2) 5 H = 4.33 6 (8,3) 5.5 x=26 So the mean of the sampling distribution of the mean is 4.33. c. Computing for the variance of the sampling distribution of the means, we have Observation X x-μe Sample (2, 3) (x-µ₂) ² 3.3489 02 Σ(x-μχ)2 1 2.5 -1.83 72 2 2 (2,8) 5 0.67 0.4489 3 (3,2) 2.5 -1.83 3.3489 0 = 10.3334 6 4 (3,8) 5.5 1.17 1.3689 5 (8, 2) 5 0.67 0.4489 o=1.7222 6 (8,3) 5.5 1.17 1.3689 So the variance is 1.72. Σ* Σ(8-4₂)² = = 26 10.3334 Now it's your turn to practice. Practice Exercise of four values / 3. Tiu air POSSIDIO wit Witteplacement "ation. 2. Consider a population of four values (3, 5, 7, 9). a. Find all possible samples of size 2 which can be drawn without replacement from this population. b. Find the mean of the sampling distribution of means. c. Find the variance of the sampling distribution of the means. 2 5 =
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