A.We have population values 1,2,3,4, population size N=5 and sample size n=2.
The number of possible samples which can be drawn with replacement is
Nn=42=16.
S={(1,1),(1,2),(1,3),(1,4),
(2,1),(2,2),(2,3),(2,4),
(3,1),(3,2),(3,3),(3,4),
(4,1),(4,2),(4,3),(4,4)}
B.
no12345678910111213141516Sample1,11,21,31,42,12,22,32,43,13,23,33,44,14,24,34,4Samplemean (xˉ)2/23/24/25/23/24/25/26/24/25/26/27/25/26/27/28/2
C.
Xˉ2/23/24/25/26/27/28/2f(Xˉ)1/162/163/164/163/162/161/16Xˉf(Xˉ)2/326/3212/3220/3218/3214/328/32Xˉ2f(Xˉ)4/6418/6448/64100/64108/6498/6464/64
D.
Mean of population
μ=41+2+3+4=2.5
Mean of sampling distribution
μXˉ=E(Xˉ)=∑Xˉif(Xˉi)=2.5=μ
Variance of population
σ2=nΣ(xi−xˉ)2=41(2.25+0.25+0.25
+2.25)=1.25
The variance of sampling distribution
Var(Xˉ)=σXˉ2=∑Xˉi2f(Xˉi)−[∑Xˉif(Xˉi)]2=64440−(2.5)2=0.625=nσ2
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