a.
We have population values 1 , 3 , 4 , 6 , 8 , 1,3,4,6,8, 1 , 3 , 4 , 6 , 8 , population size N = 5 N=5 N = 5
μ = 1 + 3 + 4 + 6 + 8 5 = 4.4 \mu=\dfrac{1+3+4+6+8}{5}=4.4 μ = 5 1 + 3 + 4 + 6 + 8 = 4.4 b.
σ 2 = 1 5 ( ( 1 − 4.4 ) 2 + ( 3 − 4.4 ) 2 + ( 4 − 4.4 ) 2 \sigma^2=\dfrac{1}{5}((1-4.4)^2+(3-4.4)^2+(4-4.4)^2 σ 2 = 5 1 (( 1 − 4.4 ) 2 + ( 3 − 4.4 ) 2 + ( 4 − 4.4 ) 2
+ ( 6 − 4.4 ) 2 ) + ( 8 − 4.4 ) 2 ) = 5.84 +(6-4.4)^2)+(8-4.4)^2)=5.84 + ( 6 − 4.4 ) 2 ) + ( 8 − 4.4 ) 2 ) = 5.84
c.
σ = σ 2 = 5.84 ≈ 2.4166 \sigma=\sqrt{\sigma^2}=\sqrt{5.84}\approx2.4166 σ = σ 2 = 5.84 ≈ 2.4166
d. We have population values 1 , 3 , 4 , 6 , 8 1,3,4,6,8 1 , 3 , 4 , 6 , 8 population size N = 5 N=5 N = 5 and sample size n = 3. n=3. n = 3. Thus, the number of possible samples which can be drawn without replacement is
( N n ) = ( 5 3 ) = 10 \dbinom{N}{n}=\dbinom{5}{3}=10 ( n N ) = ( 3 5 ) = 10 S a m p l e S a m p l e S a m p l e m e a n N o . v a l u e s ( X ˉ ) 1 1 , 3 , 4 8 / 3 2 1 , 3 , 6 10 / 3 3 1 , 3 , 8 4 4 1 , 4 , 6 11 / 3 5 1 , 4 , 8 13 / 3 6 1 , 6 , 8 5 7 3 , 4 , 6 13 / 3 8 3 , 4 , 8 5 9 3 , 6 , 8 17 / 3 10 4 , 6 , 8 6 \def\arraystretch{1.5}
\begin{array}{c:c:c}
Sample & Sample & Sample \ mean \\
No. & values & (\bar{X}) \\ \hline
1 & 1,3,4 & 8/3 \\
\hdashline
2 & 1,3,6 & 10/3 \\
\hdashline
3 & 1,3,8 & 4 \\
\hdashline
4 & 1,4,6 & 11/3 \\
\hdashline
5 & 1,4,8 & 13/3\\
\hdashline
6 & 1,6,8 & 5 \\
\hdashline
7 & 3,4,6 & 13/3 \\
\hdashline
8 & 3,4,8 & 5 \\
\hdashline
9 & 3,6,8 & 17/3 \\
\hdashline
10 & 4,6,8 & 6 \\
\hline
\end{array} S am pl e N o . 1 2 3 4 5 6 7 8 9 10 S am pl e v a l u es 1 , 3 , 4 1 , 3 , 6 1 , 3 , 8 1 , 4 , 6 1 , 4 , 8 1 , 6 , 8 3 , 4 , 6 3 , 4 , 8 3 , 6 , 8 4 , 6 , 8 S am pl e m e an ( X ˉ ) 8/3 10/3 4 11/3 13/3 5 13/3 5 17/3 6 The sampling distribution of the sample means.
X ˉ f f ( X ˉ ) X ˉ f ( X ˉ ) X ˉ 2 f ( X ˉ ) 8 / 3 1 1 / 10 8 / 30 64 / 90 10 / 3 1 1 / 10 10 / 30 100 / 90 11 / 3 1 1 / 10 11 / 10 121 / 90 4 1 1 / 10 12 / 30 144 / 90 13 / 3 2 2 / 10 26 / 30 338 / 90 5 2 2 / 10 30 / 30 450 / 90 6 1 1 / 10 18 / 30 324 / 90 17 / 3 1 1 / 10 17 / 30 289 / 90 T o t a l 10 1 4.4 61 / 3 \def\arraystretch{1.5}
\begin{array}{c:c:c:c:c:c}
& \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline
& 8/3 & 1 & 1/10 & 8/30 & 64/90 \\
\hdashline
& 10/3 & 1 & 1/10 & 10/30 & 100/90 \\
\hdashline
& 11/3 & 1 & 1/10 & 11/10 & 121/90 \\
\hdashline
& 4 & 1& 1/10 & 12/30 & 144/90 \\
\hdashline
& 13/3 & 2 & 2/10 & 26/30 & 338/90 \\
\hdashline
& 5 & 2 & 2/10 & 30/30 & 450/90 \\
\hdashline
& 6 & 1 & 1/10 & 18/30 & 324/90 \\
\hdashline
& 17/3 & 1 & 1/10 & 17/30 & 289/90 \\
\hdashline
Total & & 10 & 1 & 4.4 & 61/3 \\ \hline
\end{array} T o t a l X ˉ 8/3 10/3 11/3 4 13/3 5 6 17/3 f 1 1 1 1 2 2 1 1 10 f ( X ˉ ) 1/10 1/10 1/10 1/10 2/10 2/10 1/10 1/10 1 X ˉ f ( X ˉ ) 8/30 10/30 11/10 12/30 26/30 30/30 18/30 17/30 4.4 X ˉ 2 f ( X ˉ ) 64/90 100/90 121/90 144/90 338/90 450/90 324/90 289/90 61/3
E ( X ˉ ) = ∑ X ˉ f ( X ˉ ) = 4.4 E(\bar{X})=\sum\bar{X}f(\bar{X})=4.4 E ( X ˉ ) = ∑ X ˉ f ( X ˉ ) = 4.4
The mean of the sampling distribution of the sample means is equal to the the mean of the population.
E ( X ˉ ) = μ X ˉ = 4.4 = μ E(\bar{X})=\mu_{\bar{X}}=4.4=\mu E ( X ˉ ) = μ X ˉ = 4.4 = μ
e.
V a r ( X ˉ ) = ∑ X ˉ 2 f ( X ˉ ) − ( ∑ X ˉ f ( X ˉ ) ) 2 Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2 Va r ( X ˉ ) = ∑ X ˉ 2 f ( X ˉ ) − ( ∑ X ˉ f ( X ˉ ) ) 2
= 61 3 − ( 4.4 ) 2 = 2.92 3 =\dfrac{61}{3}-(4.4)^2=\dfrac{2.92}{3} = 3 61 − ( 4.4 ) 2 = 3 2.92
σ X ˉ = V a r ( X ˉ ) = 2.92 3 ≈ 0.9866 \sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{2.92}{3}}\approx0.9866 σ X ˉ = Va r ( X ˉ ) = 3 2.92 ≈ 0.9866 Verification:
V a r ( X ˉ ) = σ 2 n ( N − n N − 1 ) = 5.84 3 ( 5 − 3 5 − 1 ) Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{5.84}{3}(\dfrac{5-3}{5-1}) Va r ( X ˉ ) = n σ 2 ( N − 1 N − n ) = 3 5.84 ( 5 − 1 5 − 3 )
= 2.92 3 , T r u e =\dfrac{2.92}{3}, True = 3 2.92 , T r u e
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