A population consist of five number 246810. Consider all possible sample of size two which can be taken with replacement from this population
Answers
Answer:
Mean
\mu=\dfrac{2+6+8+0+1}{5}=3.4μ=
5
2+6+8+0+1
=3.4
Variance
\sigma^2=\dfrac{1}{5}\big((2-3.4)^2+(6-2.4)^2+(8-3.4)^2σ
2
=
5
1
((2−3.4)
2
+(6−2.4)
2
+(8−3.4)
2
(0-3.4)^2+(1-3.4)^2\big)=9.44(0−3.4)
2
+(1−3.4)
2
)=9.44
Standard deviation
\sigma=\sqrt{\sigma^2}=\sqrt{9.44}\approx3.0725σ=
σ
2
=
9.44
≈3.0725
2. We have population values 2,6,8,0,12,6,8,0,1 population size N=5N=5 and sample size n=2.n=2. Thus, the number of possible samples which can be drawn without replacement is
\dbinom{N}{n}=\dbinom{5}{2}=10(
n
N
)=(
2
5
)=10
\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,6 & 4 \\ \hdashline 2 & 2,8 & 5 \\ \hdashline 3 & 2,0 & 1 \\ \hdashline 4 & 2,1 & 1.5 \\ \hdashline 5 & 6,8 & 7 \\ \hdashline 6 & 6,0 & 3 \\ \hdashline 7 & 6,1 & 3.5 \\ \hdashline 8 & 8,0 & 4 \\ \hdashline 9 & 8,1 & 4.5 \\ \hdashline 10 & 0,1 & 0.5 \\ \hline \end{array}
Sample
No.
1
2
3
4
5
6
7
8
9
10
Sample
values
2,6
2,8
2,0
2,1
6,8
6,0
6,1
8,0
8,1
0,1
Sample mean
(
X
ˉ
)
4
5
1
1.5
7
3
3.5
4
4.5
0.5
3. The sampling distribution of the sample means.
\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 1/2 & 1& 1/10 & 1/20 & 1/40 \\ \hdashline 1 & 1 & 1/10 & 2/20 & 4/40 \\ \hdashline 3/2 & 1 & 1/10 & 3/20 & 9/40 \\ \hdashline 3 & 1 & 1/10 & 6/20 & 36/40 \\ \hdashline 7/2 & 1 & 1/10 & 7/20 & 49/40 \\ \hdashline 4 & 2 & 2/10 & 16/20 & 128/40 \\ \hdashline 9/2 & 1& 1/10 & 9/20 & 81/40 \\ \hdashline 5 & 1& 1/10 & 10/20 & 100/40 \\ \hdashline 7 & 1 & 1/10 & 14/20 & 196/40 \\ \hdashline Total & 10 & 1 & 68/20 & 604/40 \\ \hline \end{array}
X
ˉ
1/2
1
3/2
3
7/2
4
9/2
5
7
Total
f
1
1
1
1
1
2
1
1
1
10
f(
X
ˉ
)
1/10
1/10
1/10
1/10
1/10
2/10
1/10
1/10
1/10
1
X
ˉ
f(
X
ˉ
)
1/20
2/20
3/20
6/20
7/20
16/20
9/20
10/20
14/20
68/20
X
ˉ
2
f(
X
ˉ
)
1/40
4/40
9/40
36/40
49/40
128/40
81/40
100/40
196/40
604/40
(
X
ˉ
)=∑
X
ˉ
f(
X
ˉ
)=
20
68
=3.4
The mean of the sampling distribution of the sample means is equal to the
the mean of the population.
E(\bar{X})=3.4=\muE(
X
ˉ
)=3.4=μ
5.
Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2Var(
X
ˉ
)=∑
X
ˉ
2
f(
X
ˉ
)−(∑
X
ˉ
f(
X
ˉ
))
2
=\dfrac{604}{40}-(\dfrac{68}{20})^2=\dfrac{1416}{400}=\dfrac{354}{100}=3.54=
40
604
−(
20
68
)
2
=
400
1416
=
100
354
=3.54
\sqrt{Var(\bar{X})}=\sqrt{\dfrac{354}{100}}\approx1.8815
Var(
X
ˉ
)
=
100
354
≈1.8815
Verification:
Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{9.44}{2}(\dfrac{5-2}{5-1})Var(
X
ˉ
)=
n
σ
2
(
N−1
N−n
)=
2
9.44
(
5−1
5−2
3.54,