Question

**30 POINTS PLS HELP** 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 can the rancher conclude from these samples about how many sheep graze on each acre of the 40-acre pasture? (PLS ANSWER THIS PORTION.)*********

Answer 1

250/40 = 6.25 or 6 sheep per acre

n = 6sample proportion: signified by ρSample 1: 4 → 4/6 = 0.67Sample 2: 1 → 1/6 = 0.17Sample 3: 9 → 9/6 = 1.50

multiply the sample proportion by 1-ρSample 1: 0.67(1-0.67) = 0.67(0.33) = 0.2211Sample 2: 0.17(1-0.17) = 0.17(0.83) = 0.1411Sample 3: 1.50(1-1.5) = 1.5(-0.5) = -0.75

divide the result by n. n = 6Sample 1: 0.2211/6 = 0.03685Sample 2: 0.1411/6 = 0.02352Sample 3: -0.75/6 = -0.125

square root of the quotient to get the standard error.Sample 1: √0.03685 = 0.1919Sample 2: √0.02352 = 0.1534Sample 3: √-0.125 = invalid

value 95% confidence 1.96.

Sample 1: 1.96 * 0.1919 = 0.3761 or 37.61% margin of errorSample 2: 1.96 * 0.1534 = 0.3007 or 30.07% margin of erroR

Explanation:

Answer 2

250/40 = 6.25 or 6 sheep per acre

n = 6sample proportion: signified by ρSample 1: 4 → 4/6 = 0.67Sample 2: 1 → 1/6 = 0.17Sample 3: 9 → 9/6 = 1.50

multiply the sample proportion by 1-ρSample 1: 0.67(1-0.67) = 0.67(0.33) = 0.2211Sample 2: 0.17(1-0.17) = 0.17(0.83) = 0.1411Sample 3: 1.50(1-1.5) = 1.5(-0.5) = -0.75

divide the result by n. n = 6Sample 1: 0.2211/6 = 0.03685Sample 2: 0.1411/6 = 0.02352Sample 3: -0.75/6 = -0.125

square root of the quotient to get the standard error.Sample 1: √0.03685 = 0.1919Sample 2: √0.02352 = 0.1534Sample 3: √-0.125 = invalid

value 95% confidence 1.96.

Sample 1: 1.96 * 0.1919 = 0.3761 or 37.61% margin of errorSample 2: 1.96 * 0.1534 = 0.3007 or 30.07% margin of erroR