Answer to Question #170540 in Statistics and Probability for Sean

Question

A population consists of the four numbers: 2, 3, 6, and 9. You have to consider all possible samples of size 2 that can be drawn from this population. Find the following:

a. the mean of the population;

b. the standard deviation of the mean of the population;

c. the mean of the sampling distribution of the sample means; and

d. the standard deviation of the sampling distribution of the sample means.

Answer 1

(a),(b). Compute the mean and the standard deviation of the population. 

$E[X]=(2+3+6+9)/4=5$

$E[X^2 ]=(2^2+3^2+6^2+9^2)/4=32.5$

$Var[X]=E[X^2 ]-E[X]^2=32.5-25=7.5$

$\sigma(X)=\sqrt{Var[X] }=\sqrt{7.5}=2.7386$

Now we list all samples of size 2 and compute the mean for each sample. 

The sampling distribution of the sample means:

$P(\bar X_2=2.0)= P(\bar X_2=3.0)=P(\bar X_2=9.0)=1/16$

$P(\bar X_2=2.5)=P(\bar X_2=4.0)=P(\bar X_2=4.5)=P(\bar X_2=5.5)=P(\bar X_2=7.5)=1/8$

$P(\bar X_2=6/0)=3/16$

(c). Calculate the mean of the sampling distribution of the sample means. Compare this to mean of the population. 

$E[\bar X_2]=(2.0+3.0+9.0+2.5\cdot 2+4.0\cdot 2+4.5\cdot 2+5.5\cdot 2+7.5\cdot 2+6.0\cdot 3)/16=80/16=5.0=E[X]$

(d). Calculate the standard deviation of the sampling distribution of the sample means.

$E[\bar X_2^2] =(2.0^2+3.0^2+9.0^2+2.5^2\cdot 2+4.0^2\cdot 2+4.5^2\cdot 2+5.5^2\cdot 2+7.5^2\cdot 2+6.0^2\cdot 3)/16= 460/16=28.75$

$Var[\bar X_2 ]=E[\bar X_2^2 ]-E[\bar X_2 ]^2=28.75-5^2=3.75$

$\sigma(\bar X_2 )=\sqrt{Var[\bar X_2 ]}=\sqrt{3.75}=1.9365$