Answer to Question #325465 in Statistics and Probability for Imran

Question

For a population 0,4,8,12, construct the sampling distribution of mean for samples of size 2 taken with replacement and the find its mean and standard error

Answer 1

m(0,4)=(0+4)/2=2

m(0,8)=(0+8)/2=4

m(0,12)=(0+12)/2=6

m(4,8)=(4+8)/2=6

m(4,12)=(4+12)/2=8

m(8,12)=(8+12)/2=10

m(0,0=(0+0)/2=0

m(4,4)=(4+4)/2=4

m(8,8)=(8+8)/2=8

m(12,12)=(12+12)/2=12

Frequency

F(2)=F(12)=F(10)=F(0)=1

F(6)=F(8)=F(4)=2

Probability$P(x)=F(x)/\sum F(x)$

P(2)=P(12)=P(10)=P(0)=1/10

P(6)=F(8)=F(4)=2/10=1/5

$E(x)=\sum Px=1/10(2+12+0+10)+2/10(6+8+4)=2.4+3.6=6$

$\sigma^2=\sum Px^2-(\sum Px)^2=1/10(4+144+0+100)+2/10(36+64+16)-36=24.8+23.2-36=12$

$\mu=E(x)$

$\sigma=\sqrt{\frac{\sum(x-\mu)^2}{n-1}}=\sqrt{\frac{36+4+4+36}{3}}=5.16$

$SE=\sigma/\sqrt{n}=5.16/\sqrt{4}=2.58$