Consider a population consisting of the values (1, 3, 8), n= 2 with replacement.
Solution:
Possible sample of size 2 with replacement={(1,3)(3,8)(1,8),(1,1),(3,3),(8,8)}=\{(1,3)(3,8)(1,8),(1,1),(3,3),(8,8)\}={(1,3)(3,8)(1,8),(1,1),(3,3),(8,8)}
Sample means for these samples={2,5.5,4.5,1,3,8}=\{2,5.5,4.5,1,3,8\}={2,5.5,4.5,1,3,8}
(1): Mean of sample means=(2+5.5+4.5+1+3+8)/6=4=(2+5.5+4.5+1+3+8)/6=4=(2+5.5+4.5+1+3+8)/6=4
Population mean=(1+3+8)/3=4=(1+3+8)/3=4=(1+3+8)/3=4
Population variance,σ2=(1−4)2+(3−4)2+(8−4)23=263,\sigma^2=\dfrac{(1-4)^2+(3-4)^2+(8-4)^2}{3}=\dfrac{26}3,σ2=3(1−4)2+(3−4)2+(8−4)2=326
(2): The variance of the sampling distribution of means,σx2=σ2(N−n)n(N−1),\sigma_x^2=\dfrac{\sigma^2(N-n)}{n(N-1)},σx2=n(N−1)σ2(N−n)
=263(3−2)2(3−1)=136=\dfrac{\dfrac{26}3(3-2)}{2(3-1)}=\dfrac{13}6=2(3−1)326(3−2)=613
(3): The standard deviation of the sampling distribution,σx=136,\sigma_x=\sqrt{\dfrac{13}6},σx=613
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