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

1. ] What can the rancher conclude from these samples about how many sheep graze on each acre of the 40-acre pasture?

2. ] If the sheep were equally "spread out" across all of the 40 acres, how many sheep would you expect to find on average on each acre?

3. ] How do the sample statistics compare to the population mean and standard deviation?

4. ] What margin of error is reasonable for these samples?

Answer 1

From the samples, the rancher cannot conclusively determine how many sheep are grazing on each acre of the 40-acre pasture. The samples only provide information on the number of sheep observed in three different areas, which may not be representative of the entire pasture.
If the sheep were equally spread out across all 40 acres, we would expect to find 250/40 = 6.25 sheep on average on each acre.
The sample statistics (number of sheep observed in the three samples) cannot be used to accurately compare to the population mean and standard deviation, since they are just a small sample of the entire population. However, assuming that the sample means are unbiased and normally distributed, we can use them to make inferences about the population mean and standard deviation.
To determine the margin of error for these samples, we need to know the sample size, the confidence level, and the population standard deviation. Since these values are not given in the problem, we cannot calculate a reasonable margin of error. However, a larger sample size and higher confidence level would generally result in a smaller margin of error.