Answer:
1. Compute the mean and the standard deviation of the population.
E[X]=(2+5+6+8+10+12+13)/7=8E[X]=(2+5+6+8+10+12+13)/7=8
E[X^2 ]=(2^2+5^2+6^2+8^2+10^2+12^2+13^2 )/7=542/7=77\frac{3}{7}E[X
2
]=(2
2
+5
2
+6
2
+8
2
+10
2
+12
2
+13
2
)/7=542/7=77
7
3
Var[X]=E[X^2 ]-E[X]^2=77 \frac{3}{7}-64=13\frac{3}{7}Var[X]=E[X
2
]βE[X]
2
=77
7
3
β64=13
7
3
\sigma(X)=\sqrt{Var[X] }=\sqrt{94/7}=3.6645Ο(X)=
Var[X]
=
94/7
=3.6645
2. List all samples of size 5 and compute the mean for each sample.
3. Construct the sampling distribution of the sample means.
P(\bar X_5=6.2)= P(\bar X_5=6.6)=P(\bar X_5=6.8)=P(\bar X_5=7.0)=P(\bar X_5=7.2)=P(\bar X_5=7.4)=P(\bar X_5=7.8)=P(\bar X_5=8.0)=P(\bar X_5=8.6)=P(\bar X_5=8.8)=P(\bar X_5=9.0)=P(\bar X_5=9.2)=P(\bar X_5=9.6)=P(\bar X_5=9.8)=1/21P(
X
Λ
5
=6.2)=P(
X
Λ
5
=6.6)=P(
X
Λ
5
=6.8)=P(
X
Λ
5
=7.0)=P(
X
Λ
5
=7.2)=P(
X
Λ
5
=7.4)=P(
X
Λ
5
=7.8)=P(
X
Λ
5
=8.0)=P(
X
Λ
5
=8.6)=P(
X
Λ
5
=8.8)=P(
X
Λ
5
=9.0)=P(
X
Λ
5
=9.2)=P(
X
Λ
5
=9.6)=P(
X
Λ
5
=9.8)=1/21
P(\bar X_5=8.2)=P(\bar X_5=8.4)=2/21P(
X
Λ
5
=8.2)=P(
X
Λ
5
=8.4)=2/21
P(\bar X_5=7.6)=3/21P(
X
Λ
5
=7.6)=3/21
4. Calculate the mean of the sampling distribution of the sample means. Compare this to mean of the population.
E[\bar X_5 ]=(6.2+6.6+6.8+7.0+7.2+7.4+7.8+8.0+8.6+8.8+9.0+9.2+9.6+9.8+2\cdot 8.2+2\cdot 8.4+3\cdot 7.6)/21=8.0=E[X]E[
X
Λ
5
]=(6.2+6.6+6.8+7.0+7.2+7.4+7.8+8.0+8.6+8.8+9.0+9.2+9.6+9.8+2β
8.2+2β
8.4+3β
7.6)/21=8.0=E[X]
5. Calculate the standard deviation of the sampling distribution of the sample means . Compare this to the standard deviation of the population.
E[\bar X_5^2] =(6.2^2+6.6^2+6.8^2+ 7.0^2+7.2^2+7.4^2+7.8^2+8.0^2+8.6^2+ 8.8^2+9.0^2+9.2^2+9.6^2 +9.8^2+2β
8.2^2+2β
8.4^2 +3β
7.6^2)/21=1362.8/21=64\frac{94}{105}E[
X
Λ
5
2
]=(6.2
2
+6.6
2
+6.8
2
+7.0
2
+7.2
2
+7.4
2
+7.8
2
+8.0
2
+8.6
2
+8.8
2
+9.0
2
+9.2
2
+9.6
2
+9.8
2
+2β
8.2
2
+2β
8.4
2
+3β
7.6
2
)/21=1362.8/21=64
105
94
Var[\bar X_5 ]=E[\bar X_5^2 ]-E[\bar X_5 ]^2=64\frac{94}{105}-64=\frac{94}{105}Var[
X
Λ
5
]=E[
X
Λ
5
2
]βE[
X
Λ
5
]
2
=64
105
94
β64=
105
94
\sigma(\bar X_5 )=\sqrt{Var[\bar X_5 ]}=\sqrt{94/105}=\sigma(X)/\sqrt{15}=0.946Ο(
X
Λ
5
)=
Var[
X
Λ
5
]
=
94/105
=Ο(X)/
15
=0.946
