Question

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?

Answer 1

Sample 1 MOE = 37.24%

Sample 2 MOE = 29.4%

Sample 3 MOE = None

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.