We have population values 1,2,3,4,5 and 6 population size N=6 and sample size n=3.
Mean of population (μ) =
61+2+3+4+5+6=3.5
Variance of population
σ2=nΣ(xi−xˉ)2=617.5
σ=σ2=617.5≈1.7078 The number of possible samples which can be drawn without replacement is NCn=6C3=20.
no1234567891011121314151617181920Sample1,2,31,2,41,2,51,2,61,3,41,3,51,3,61,4,51,4,61,5,62,3,42,3,52,3,62,4,52,4,62,5,63,4,53,4,63,5,64,5,6Samplemean (xˉ)6/37/38/39/38/39/310/310/311/312/39/310/311/311/312/313/312/313/314/315/3
Xˉ6/37/38/39/310/311/312/313/314/315/3f(Xˉ)1/201/202/203/203/203/203/202/201/201/20Xˉf(Xˉ)6/607/6016/6027/6030/6033/6036/6026/6014/6015/60Xˉ2f(Xˉ)36/18049/180128/180243/180300/180363/180432/180338/180196/180225/180
Mean of sampling distribution
μXˉ=E(Xˉ)=∑Xˉif(Xˉi)=3.5=μ
The variance of sampling distribution
Var(Xˉ)=σXˉ2=∑Xˉi2f(Xˉi)−[∑Xˉif(Xˉi)]2=1802310−(3.5)2=127=nσ2(N−1N−n)
σXˉ=127≈0.7638
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