The U.S. Department of Agriculture records data on farm acreage and number of farms by county for every county in the country. This takes considerable time and energy, and the results for each update (usually occurring approximately every five years) are a long time in process. A quicker way to update the data is by sampling. The table below shows the number of counties in each geographic region and a summary of farm acres (in thousands) for independent random samples of 22 counties from each region.

Region Number of counties
North Central 1052
North East 210
South 1376
West 418

Required:
a. Estimate the mean acreage for each region, with a margin of error.
b. Estimate the total acreage for each region, with a margin of error.
c. Estimate the difference between mean acreage for the North Central region and the South, with a margin of error.
d. Estimate the difference between mean acreage for the North East and the West, with a margin of error.
e. Estimate the mean acreage per county across the United States, with a margin of error.

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.

3. Find mZS.
Q= 5x + 2)
P= 10x - 3)
(7x - 11)=ºR
(13x - 31)º=S
(8x - 19)=T

PLS HELP!! need the answer asap