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cricket20 [7]
3 years ago
7

What is the value of k

Mathematics
1 answer:
Maslowich3 years ago
8 0

Answer:

k=10

Step-by-step explanation:

The sum of the two equations given is 115. Set both equal to 115 and solve.

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Number of whole numbers between 45 and 69 is _________ .<br> (a) 31 (b) 23 (c) 25 (d) 28
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The number of whole number is 23 through subtracting the numbers. hope it helps ☺️

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The U.S. Department of Agriculture records data on farm acreage and number of farms by county for every county in the country. T
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Solution :

From the table,

Number of countries = $N_i$

Mean = $\overline Y_i$

Standard deviation = $\sigma_i$

Margin of error of B = 50,000 acres

So we determine the $\text{sample size n}$ and then allocate the four regions of $n_1, n_2, n_3$ and $n_4$.

Under the $\text{proportional allocation}$, the $\text{equation}$ for the value of $n_1$ which yields $V(\overline Y_{st}) = D = \frac{B^2}{H}$  is given by :

$n=\frac{\sum_{K=1}^L N_K \sigma^2_K}{ND+\frac{1}{N}\sum_{K=1}^L N_K \sigma^2_K}$

Let us complete the numerator

$\sum_{K=1}^L N_K \sigma^2_K = N_1\sigma_1^2 + N_2\sigma_2^2 + N_3\sigma_3^2 + N_4\sigma_4^2$

                $= 1052 \times (271)^2 +  210 \times (79)^2 + 1376 \times (244)^2 + 418 \times (837)^2 $

                = 453329920

The bound on the error of estimation is B = 50,000 acres

Hence we get

$D=\frac{B^2}{H}$

  $=\frac{50000^2}{4}$

  = 625,000,000

$n=\frac{\sum_{K=1}^L N_K \sigma^2_K}{ND+\frac{1}{N}\sum_{K=1}^L N_K \sigma^2_K}$

  $=\frac{453329920}{3056 \times 625 + \frac{1}{3056} \times 453329920}$

 = 220.24044

≈ 221 (approximately)

Therefore, the sample size under the proportional allocation from each stratum are given by :

$n_1=n\left[\frac{N_1}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{1052}{3056}\right]$

   ≈ 76

$n_2=n\left[\frac{N_2}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{210}{3056}\right]$

   ≈ 15

$n_3=n\left[\frac{N_3}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{1376}{3056}\right]$

   ≈ 100

$n_4=n\left[\frac{N_4}{\sum_{K=1}^\mu N_K}\right]$

   $=221\left[\frac{418}{3056}\right]$

   ≈ 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.

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