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, σ 2 = ( 1 − 4 ) 2 + ( 3 − 4 ) 2 + ( 8 − 4 ) 2 3 = 26 3 ,\sigma^2=\dfrac{(1-4)^2+(3-4)^2+(8-4)^2}{3}=\dfrac{26}3 , σ 2 = 3 ( 1 − 4 ) 2 + ( 3 − 4 ) 2 + ( 8 − 4 ) 2 = 3 26
(2): The variance of the sampling distribution of means, σ x 2 = σ 2 ( N − n ) n ( N − 1 ) ,\sigma_x^2=\dfrac{\sigma^2(N-n)}{n(N-1)} , σ x 2 = n ( N − 1 ) σ 2 ( N − n )
= 26 3 ( 3 − 2 ) 2 ( 3 − 1 ) = 13 6 =\dfrac{\dfrac{26}3(3-2)}{2(3-1)}=\dfrac{13}6 = 2 ( 3 − 1 ) 3 26 ( 3 − 2 ) = 6 13
(3): The standard deviation of the sampling distribution, σ x = 13 6 ,\sigma_x=\sqrt{\dfrac{13}6} , σ x = 6 13
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