Here population size : N = 7 N =7 N = 7 and we have to draw a sample of size 2
So, there are N C n ^{N}C_n N C n possible samples that is, 7 C 2 = 21 ^{7}C_2=21 7 C 2 = 21
Mean of population ( μ ) (\mu) ( μ ) = 3 + 6 + 7 + 9 + 10 + 4 + 8 7 = 47 7 ≈ 6.7143 \dfrac{3+6+7+9+10+4+8}{7}=\dfrac{47}{7}\approx6.7143 7 3 + 6 + 7 + 9 + 10 + 4 + 8 = 7 47 ≈ 6.7143
Variance of population
( σ 2 ) = Σ ( x i − x ˉ ) 2 n = 1 7 ( 676 49 + 25 49 + 4 49 (\sigma^2)=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{7}(\dfrac{676}{49}+\dfrac{25}{49}+\dfrac{4}{49} ( σ 2 ) = n Σ ( x i − x ˉ ) 2 = 7 1 ( 49 676 + 49 25 + 49 4
+ 256 49 + 529 49 + 361 49 + 81 49 ) +\dfrac{256}{49}+\dfrac{529}{49}+\dfrac{361}{49}+\dfrac{81}{49}) + 49 256 + 49 529 + 49 361 + 49 81 )
= 1932 343 =\dfrac{1932}{343} = 343 1932
n o S a m p l e S a m p l e m e a n ( x ˉ ) 1 3 , 4 3.5 2 3 , 6 4.5 3 3 , 7 5 4 3 , 8 5.5 5 3 , 9 6 6 3 , 10 6.5 7 4 , 6 5 8 4 , 7 5.5 9 4 , 8 6 10 4 , 9 6.5 11 4 , 10 7 12 6 , 7 6.5 13 6 , 8 7 14 6 , 9 7.5 15 6 , 10 8 16 7 , 8 7.5 17 7 , 9 8 18 7 , 10 8.5 19 8 , 9 8.5 20 8 , 10 9 21 9 , 10 9.5 \def\arraystretch{1.5}
\begin{array}{c:c:c:c:c}
no & Sample & Sample \\
& & mean\ (\bar{x})
\\ \hline
1 & 3,4 & 3.5 \\
\hdashline
2 & 3,6 & 4.5 \\
\hdashline
3 & 3,7 & 5 \\
\hdashline
4 & 3,8 & 5.5 \\
\hdashline
5 & 3,9 & 6 \\
\hdashline
6 & 3,10 & 6.5 \\
\hdashline
7 & 4,6 & 5 \\
\hdashline
8 & 4,7 & 5.5 \\
\hdashline
9 & 4,8 & 6 \\
\hdashline
10 & 4,9 & 6.5 \\
\hdashline
11 & 4,10 & 7 \\
\hdashline
12 & 6,7 & 6.5 \\
\hdashline
13 & 6,8 & 7 \\
\hdashline
14 & 6,9 & 7.5 \\
\hdashline
15 & 6,10 & 8 \\
\hdashline
16 & 7,8 & 7.5 \\
\hdashline
17 & 7,9 & 8 \\
\hdashline
18 & 7,10 & 8.5 \\
\hdashline
19 & 8,9 & 8.5 \\
\hdashline
20 & 8,10 & 9 \\
\hdashline
21 & 9,10 & 9.5 \\
\hdashline
\end{array} n o 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 S am pl e 3 , 4 3 , 6 3 , 7 3 , 8 3 , 9 3 , 10 4 , 6 4 , 7 4 , 8 4 , 9 4 , 10 6 , 7 6 , 8 6 , 9 6 , 10 7 , 8 7 , 9 7 , 10 8 , 9 8 , 10 9 , 10 S am pl e m e an ( x ˉ ) 3.5 4.5 5 5.5 6 6.5 5 5.5 6 6.5 7 6.5 7 7.5 8 7.5 8 8.5 8.5 9 9.5
Mean of sampling distribution
μ x ˉ = E ( x ˉ ) = ∑ x ˉ i f ( x ˉ i ) = 47 7 = μ \mu_{\bar{x}}=E(\bar{x})=\sum\bar{x}_if(\bar{x}_i)=\dfrac{47}{7}=\mu μ x ˉ = E ( x ˉ ) = ∑ x ˉ i f ( x ˉ i ) = 7 47 = μ
The variance of sampling distribution
V a r ( x ˉ ) = σ x ˉ 2 = ∑ x ˉ i 2 f ( x ˉ i ) − [ ∑ x ˉ i f ( x ˉ i ) ] 2 Var(\bar{x})=\sigma^2_{\bar{x}}=\sum\bar{x}_i^2f(\bar{x}_i)-\big[\sum\bar{x}_if(\bar{x}_i)\big]^2 Va r ( x ˉ ) = σ x ˉ 2 = ∑ x ˉ i 2 f ( x ˉ i ) − [ ∑ x ˉ i f ( x ˉ i ) ] 2 = 332 7 − 2209 49 = 115 49 = σ 2 n ( N − n N − 1 ) =\dfrac{332}{7}-\dfrac{2209}{49}=\dfrac{115}{49}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1}) = 7 332 − 49 2209 = 49 115 = n σ 2 ( N − 1 N − n )
σ x ˉ = 115 49 ≈ 1.5320 \sigma_{\bar{x}}=\sqrt{\dfrac{115}{49}}\approx1.5320 σ x ˉ = 49 115 ≈ 1.5320
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