Answer to Question #173598 in Statistics and Probability for ANJU JAYACHANDRAN

Question

8b) Suppose from a total of 120 guava trees, 5 clusters of 4 trees each selected and the

yield (in kg) is recorded below:

Cluster Trees

1 2 3 4

1 5 4 2 15

2 11 1 4 7

3 26 10 19 11

4 7 15 12 10

5 2 22 8 6

Estimate the average yield per tree and its standard error.

Answer 1

Solution:

Let $Y_i$ denotes the yield in $i^{th}$ tree.

So, $Y_1=\dfrac{51}{5}=10.2$

$Y_2=\dfrac{52}{5}=10.4 \\ Y_3=\dfrac{45}{5}=9 \\ Y_4=\dfrac{49}{5}=9.8$

Now, standard deviation for each tree, $\sigma_i=\sqrt{\dfrac{\Sigma_{j=1}^5(Y_j-\bar{Y_i})^2}{n-1}}$

So, $\sigma_1=\sqrt{\dfrac{\Sigma_{j=1}^5(Y_j-10.2)^2}{5-1}}$

$=\sqrt{\dfrac{(5-10.2)^2+(11-10.2)^2+(26-10.2)^2+(7-10.2)^2+(2-10.2)^2}{4}}$

$=9.418$

Similarly, $\sigma_2=8.444,\sigma_3=6.782,\sigma_4=3.564$

Now, standard error, $\sigma_{\bar{y_i}}=\dfrac{\sigma_i}{n}$

So,

$\sigma_{\bar{y_1}}=\dfrac{\sigma_1}{n}=\frac{9.418}{5}=1.8836 \\ \sigma_{\bar{y_2}}=\dfrac{\sigma_2}{n}=\frac{8.444}{5}=1.6888 \\ \sigma_{\bar{y_3}}=\dfrac{\sigma_3}{n}=\frac{6.782}{5}=1.3564 \\ \sigma_{\bar{y_4}}=\dfrac{\sigma_4}{n}=\frac{3.564}{5}=0.7128$