Solution :
From the table,
Number of countries = 
Mean = 
Standard deviation = 
Margin of error of B = 50,000 acres
So we determine the
and then allocate the four regions of
and
.
Under the
, the
for the value of
which yields
is given by :

Let us complete the numerator


= 453329920
The bound on the error of estimation is B = 50,000 acres
Hence we get


= 625,000,000


= 220.24044
≈ 221 (approximately)
Therefore, the sample size under the proportional allocation from each stratum are given by :
![$n_1=n\left[\frac{N_1}{\sum_{K=1}^\mu N_K}\right]$](https://tex.z-dn.net/?f=%24n_1%3Dn%5Cleft%5B%5Cfrac%7BN_1%7D%7B%5Csum_%7BK%3D1%7D%5E%5Cmu%20N_K%7D%5Cright%5D%24)
![$=221\left[\frac{1052}{3056}\right]$](https://tex.z-dn.net/?f=%24%3D221%5Cleft%5B%5Cfrac%7B1052%7D%7B3056%7D%5Cright%5D%24)
≈ 76
![$n_2=n\left[\frac{N_2}{\sum_{K=1}^\mu N_K}\right]$](https://tex.z-dn.net/?f=%24n_2%3Dn%5Cleft%5B%5Cfrac%7BN_2%7D%7B%5Csum_%7BK%3D1%7D%5E%5Cmu%20N_K%7D%5Cright%5D%24)
![$=221\left[\frac{210}{3056}\right]$](https://tex.z-dn.net/?f=%24%3D221%5Cleft%5B%5Cfrac%7B210%7D%7B3056%7D%5Cright%5D%24)
≈ 15
![$n_3=n\left[\frac{N_3}{\sum_{K=1}^\mu N_K}\right]$](https://tex.z-dn.net/?f=%24n_3%3Dn%5Cleft%5B%5Cfrac%7BN_3%7D%7B%5Csum_%7BK%3D1%7D%5E%5Cmu%20N_K%7D%5Cright%5D%24)
![$=221\left[\frac{1376}{3056}\right]$](https://tex.z-dn.net/?f=%24%3D221%5Cleft%5B%5Cfrac%7B1376%7D%7B3056%7D%5Cright%5D%24)
≈ 100
![$n_4=n\left[\frac{N_4}{\sum_{K=1}^\mu N_K}\right]$](https://tex.z-dn.net/?f=%24n_4%3Dn%5Cleft%5B%5Cfrac%7BN_4%7D%7B%5Csum_%7BK%3D1%7D%5E%5Cmu%20N_K%7D%5Cright%5D%24)
![$=221\left[\frac{418}{3056}\right]$](https://tex.z-dn.net/?f=%24%3D221%5Cleft%5B%5Cfrac%7B418%7D%7B3056%7D%5Cright%5D%24)
≈ 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.