We have population values 2 , 4 , 6 , 8 2,4,6,8 2 , 4 , 6 , 8 population size N = 4 N=4 N = 4 and sample size n = 2. n=2. n = 2.
Thus, the number of possible samples which can be drawn with replacement is 4 2 = 16. 4^2=16. 4 2 = 16.
a.
S a m p l e v a l u e s S a m p l e m e a n ( X ˉ ) 2 , 2 2 2 , 4 3 2 , 6 4 2 , 8 5 4 , 2 3 4 , 4 4 4 , 6 5 4 , 8 6 6 , 2 4 6 , 4 5 6 , 6 6 6 , 8 7 8 , 2 5 8 , 4 6 8 , 6 7 8 , 8 8 \def\arraystretch{1.5}
\begin{array}{c:c}
Sample\ values & Sample\ mean(\bar{X}) \\ \hline
2,2 & 2\\
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2,4 & 3\\
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2,6 & 4\\
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2,8 & 5\\
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4,2 & 3\\
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4,4 & 4\\
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4,6 & 5\\
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4,8 & 6\\
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6,2 & 4\\
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6,4 & 5\\
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6,6 & 6\\
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6,8 & 7\\
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8,2 & 5\\
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8,4 & 6\\
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8,6 & 7\\
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8,8 & 8\\
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\end{array} S am pl e v a l u es 2 , 2 2 , 4 2 , 6 2 , 8 4 , 2 4 , 4 4 , 6 4 , 8 6 , 2 6 , 4 6 , 6 6 , 8 8 , 2 8 , 4 8 , 6 8 , 8 S am pl e m e an ( X ˉ ) 2 3 4 5 3 4 5 6 4 5 6 7 5 6 7 8
b.The sampling distribution of the sample mean X ˉ \bar{X} X ˉ is
X ˉ f f ( X ˉ ) X f ( X ˉ ) X 2 f ( X ˉ ) 2 1 1 / 16 2 / 16 4 / 16 3 2 2 / 16 6 / 16 18 / 16 4 3 3 / 16 12 / 16 48 / 16 5 4 4 / 16 20 / 16 100 / 16 6 3 3 / 16 18 / 16 108 / 16 7 2 2 / 16 14 / 16 98 / 16 8 1 1 / 16 8 / 16 64 / 16 S u m = 16 1 5 55 / 2 \def\arraystretch{1.5}
\begin{array}{c:c:c:c:c:c:c}
& \bar{X} & f & f(\bar{X}) & Xf(\bar{X})& X^2f(\bar{X}) \\ \hline
& 2 & 1 & 1/16 & 2/16 & 4/16\\
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& 3 & 2 & 2/16 & 6/16 & 18/16 \\
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& 4 & 3 & 3/16 & 12/16 & 48/16\\
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& 5 & 4 & 4/16 & 20/16 & 100/16 \\
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& 6 & 3 & 3/16 & 18/16 & 108/16 \\
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& 7 & 2 & 2/16 & 14/16 & 98/16 \\
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& 8 & 1 & 1/16 & 8/16 & 64/16\\
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Sum= & & 16 & 1 & 5 & 55/2\\
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\end{array} S u m = X ˉ 2 3 4 5 6 7 8 f 1 2 3 4 3 2 1 16 f ( X ˉ ) 1/16 2/16 3/16 4/16 3/16 2/16 1/16 1 X f ( X ˉ ) 2/16 6/16 12/16 20/16 18/16 14/16 8/16 5 X 2 f ( X ˉ ) 4/16 18/16 48/16 100/16 108/16 98/16 64/16 55/2
c.
μ = 2 + 4 + 6 + 8 4 = 5 \mu=\dfrac{2+4+6+8}{4}=5 μ = 4 2 + 4 + 6 + 8 = 5
σ 2 = 1 4 ( ( 2 − 5 ) 2 + ( 4 − 5 ) 2 + ( 6 − 5 ) 2 + ( 8 − 5 ) 2 ) \sigma^2=\dfrac{1}{4}((2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2) σ 2 = 4 1 (( 2 − 5 ) 2 + ( 4 − 5 ) 2 + ( 6 − 5 ) 2 + ( 8 − 5 ) 2 )
= 5 =5 = 5
Check
The mean of the sample means is
μ X ˉ = E ( X ˉ ) = 5 = μ \mu_{\bar{X}}=E(\bar{X})=5=\mu μ X ˉ = E ( X ˉ ) = 5 = μ
V a r ( X ˉ ) = σ X ˉ 2 = E ( X ˉ 2 ) − ( E ( X ˉ ) ) 2 Var(\bar{X})=\sigma^2_{\bar{X}}=E(\bar{X}^2)-(E(\bar{X}))^2 Va r ( X ˉ ) = σ X ˉ 2 = E ( X ˉ 2 ) − ( E ( X ˉ ) ) 2
= 55 2 − ( 5 ) 2 = 5 2 =\dfrac{55}{2}-(5)^2=\dfrac{5}{2} = 2 55 − ( 5 ) 2 = 2 5 The standard deviation of the sample means is
σ X ˉ = σ X ˉ 2 = 5 / 2 \sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\sqrt{5/2} σ X ˉ = σ X ˉ 2 = 5/2
σ 2 n = 5 2 = σ X ˉ 2 \dfrac{\sigma^2}{n}=\dfrac{5}{2}=\sigma^2_{\bar{X}} n σ 2 = 2 5 = σ X ˉ 2
μ X ˉ = μ \mu_{\bar{X}}=\mu μ X ˉ = μ
σ X ˉ = σ n \sigma_{\bar{X}}=\dfrac{\sigma}{\sqrt{n}} σ X ˉ = n σ
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