A rancher’s herd of 250 sheep grazes over a 40-acre pasture. He would like to find out how many sheep are grazing on each acre of the pasture at any given time, so he has some images of the pasture taken by the state department of agriculture’s aerial photography division. Here are three samples of the images. Sample 1 : 4
Sample 2 : 1
Sample 3 : 9

What margin of error is reasonable for these samples?

Sample 1 MOE = 37.24%

Sample 2 MOE = 29.4%

Sample 3 MOE = None

Step-by-step explanation:

We are told that 250 sheep grazes over a 40-acre pasture.

Thus, number of sheeps per acre = 250/40 = 6.25 ≈ 6

Thus, n = 6

For the samples given, we can find sample proportions as;

Sample 1: p_o = 4/6 = 0.67

Sample 2: p_o = 1/6 = 0.17

Sample 3: p_o = 9/6 = 1.5

Formula for test statistic here is;

z = √(p_o(1 - p_o)/n)

For sample 1;

z = √(0.67(1 - 0.67)/6)

z ≈ 0.19

For sample 2;

z = √(0.17(1 - 0.17)/6)

z ≈ 0.15

For sample 3;

z = √(1.5(1 - 1.5)/6)

z = √−0.125

We are not given confidence level but we will use 95%.

Value of z_α at 95% Confidence level is 1.96. Thus, we can find the margin of error for the samples;

For sample 1;

Margin of error at z = 0.19 is;

1.96 × 0.19 = 0.3724 = 37.24%

For sample 2;

Margin of error at z = 0.15 is;

1.96 × 0.15 = 0.294 = 29.4%

For sample 3;

Since test statistic has a negative square root, then it has no margin of error.

x-3y=4

step-by-step e

multiply length by height. multiply the length by the height, and then double the product. work out 10 x 10 results = 100 square inches, then work out 100 x 2 = 200. the answer is 200 square inches.

step-by-step explanation: