Solution :
Let Y i Y_i Y i denotes the yield in i t h i^{th} i t h tree.
So, Y 1 = 51 5 = 10.2 Y_1=\dfrac{51}{5}=10.2 Y 1 = 5 51 = 10.2
Y 2 = 52 5 = 10.4 Y 3 = 45 5 = 9 Y 4 = 49 5 = 9.8 Y_2=\dfrac{52}{5}=10.4
\\ Y_3=\dfrac{45}{5}=9
\\ Y_4=\dfrac{49}{5}=9.8 Y 2 = 5 52 = 10.4 Y 3 = 5 45 = 9 Y 4 = 5 49 = 9.8
Now, standard deviation for each tree, σ i = Σ j = 1 5 ( Y j − Y i ˉ ) 2 n − 1 \sigma_i=\sqrt{\dfrac{\Sigma_{j=1}^5(Y_j-\bar{Y_i})^2}{n-1}} σ i = n − 1 Σ j = 1 5 ( Y j − Y i ˉ ) 2
So, σ 1 = Σ j = 1 5 ( Y j − 10.2 ) 2 5 − 1 \sigma_1=\sqrt{\dfrac{\Sigma_{j=1}^5(Y_j-10.2)^2}{5-1}} σ 1 = 5 − 1 Σ j = 1 5 ( Y j − 10.2 ) 2
= ( 5 − 10.2 ) 2 + ( 11 − 10.2 ) 2 + ( 26 − 10.2 ) 2 + ( 7 − 10.2 ) 2 + ( 2 − 10.2 ) 2 4 =\sqrt{\dfrac{(5-10.2)^2+(11-10.2)^2+(26-10.2)^2+(7-10.2)^2+(2-10.2)^2}{4}} = 4 ( 5 − 10.2 ) 2 + ( 11 − 10.2 ) 2 + ( 26 − 10.2 ) 2 + ( 7 − 10.2 ) 2 + ( 2 − 10.2 ) 2
= 9.418 =9.418 = 9.418
Similarly, σ 2 = 8.444 , σ 3 = 6.782 , σ 4 = 3.564 \sigma_2=8.444,\sigma_3=6.782,\sigma_4=3.564 σ 2 = 8.444 , σ 3 = 6.782 , σ 4 = 3.564
Now, standard error, σ y i ˉ = σ i n \sigma_{\bar{y_i}}=\dfrac{\sigma_i}{n} σ y i ˉ = n σ i
So,
σ y 1 ˉ = σ 1 n = 9.418 5 = 1.8836 σ y 2 ˉ = σ 2 n = 8.444 5 = 1.6888 σ y 3 ˉ = σ 3 n = 6.782 5 = 1.3564 σ y 4 ˉ = σ 4 n = 3.564 5 = 0.7128 \sigma_{\bar{y_1}}=\dfrac{\sigma_1}{n}=\frac{9.418}{5}=1.8836
\\ \sigma_{\bar{y_2}}=\dfrac{\sigma_2}{n}=\frac{8.444}{5}=1.6888
\\ \sigma_{\bar{y_3}}=\dfrac{\sigma_3}{n}=\frac{6.782}{5}=1.3564
\\ \sigma_{\bar{y_4}}=\dfrac{\sigma_4}{n}=\frac{3.564}{5}=0.7128 σ y 1 ˉ = n σ 1 = 5 9.418 = 1.8836 σ y 2 ˉ = n σ 2 = 5 8.444 = 1.6888 σ y 3 ˉ = n σ 3 = 5 6.782 = 1.3564 σ y 4 ˉ = n σ 4 = 5 3.564 = 0.7128
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