Respuesta :
Solution :
From the table,
Number of countries = [tex]$N_i$[/tex]
Mean = [tex]$\overline Y_i$[/tex]
Standard deviation = [tex]$\sigma_i$[/tex]
Margin of error of B = 50,000 acres
So we determine the [tex]$\text{sample size n}$[/tex] and then allocate the four regions of [tex]$n_1, n_2, n_3$[/tex] and [tex]$n_4$[/tex].
Under the [tex]$\text{proportional allocation}$[/tex], the [tex]$\text{equation}$[/tex] for the value of [tex]$n_1$[/tex] which yields [tex]$V(\overline Y_{st}) = D = \frac{B^2}{H}$[/tex] is given by :
[tex]$n=\frac{\sum_{K=1}^L N_K \sigma^2_K}{ND+\frac{1}{N}\sum_{K=1}^L N_K \sigma^2_K}$[/tex]
Let us complete the numerator
[tex]$\sum_{K=1}^L N_K \sigma^2_K = N_1\sigma_1^2 + N_2\sigma_2^2 + N_3\sigma_3^2 + N_4\sigma_4^2$[/tex]
[tex]$= 1052 \times (271)^2 + 210 \times (79)^2 + 1376 \times (244)^2 + 418 \times (837)^2 $[/tex]
= 453329920
The bound on the error of estimation is B = 50,000 acres
Hence we get
[tex]$D=\frac{B^2}{H}$[/tex]
[tex]$=\frac{50000^2}{4}$[/tex]
= 625,000,000
[tex]$n=\frac{\sum_{K=1}^L N_K \sigma^2_K}{ND+\frac{1}{N}\sum_{K=1}^L N_K \sigma^2_K}$[/tex]
[tex]$=\frac{453329920}{3056 \times 625 + \frac{1}{3056} \times 453329920}$[/tex]
= 220.24044
≈ 221 (approximately)
Therefore, the sample size under the proportional allocation from each stratum are given by :
[tex]$n_1=n\left[\frac{N_1}{\sum_{K=1}^\mu N_K}\right]$[/tex]
[tex]$=221\left[\frac{1052}{3056}\right]$[/tex]
≈ 76
[tex]$n_2=n\left[\frac{N_2}{\sum_{K=1}^\mu N_K}\right]$[/tex]
[tex]$=221\left[\frac{210}{3056}\right]$[/tex]
≈ 15
[tex]$n_3=n\left[\frac{N_3}{\sum_{K=1}^\mu N_K}\right]$[/tex]
[tex]$=221\left[\frac{1376}{3056}\right]$[/tex]
≈ 100
[tex]$n_4=n\left[\frac{N_4}{\sum_{K=1}^\mu N_K}\right]$[/tex]
[tex]$=221\left[\frac{418}{3056}\right]$[/tex]
≈ 30
Thus, we should select 76 counties from North central region, 15 counties from the north east region, 100 from south region and 30 from west region.
We can also estimate mean acreage of each county across all the county with a margin of error of 50,000 acres.
