a. Estimate

b. Estimate the difference in the mean size of pellet groups per plot for the two animals, with an appropriate margin of error.

Deer

Sample mean = 2.30

Sample variance = 0.65

Rabbits

Sample mean = 4.52

Sample variance = 0.97

Nann
2021-08-14
Answered

Resource managers of forest game lands were concerned about the size of the deer and rabbit populations during the winter months in a particular forest. As an estimate of population size, they proposed using the average number of pellet groups for rabbits and deer per 30-foot-square plots. From an aerial photograph, the forest was divided into N= 10,000 30-foot-square grids. A simple random sample of 2 = 500 plots was taken, and the number of pellet groups was observed for rabbits and for deer. The results of this study are summarized in the accompanying table.

a. Estimate$\mu}_{1},{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}{\mu}_{2$ , the average number of pellet groups for deer and rabbits, respectively, per 30-foot-square-plots. Place bounds on the errors of estimation.

b. Estimate the difference in the mean size of pellet groups per plot for the two animals, with an appropriate margin of error.

Deer

Sample mean = 2.30

Sample variance = 0.65

Rabbits

Sample mean = 4.52

Sample variance = 0.97

a. Estimate

b. Estimate the difference in the mean size of pellet groups per plot for the two animals, with an appropriate margin of error.

Deer

Sample mean = 2.30

Sample variance = 0.65

Rabbits

Sample mean = 4.52

Sample variance = 0.97

You can still ask an expert for help

gwibdaithq

Answered 2021-08-15
Author has **84** answers

A simple random sample of n=500 plots was taken and the number of pellet groups observed for rabbits and for deer. The results of this study are summarized in the below:

For deer$\mu =2.30,{s}^{2}=0.65,N=10000,n=500$

For rabbits:$\nu =4.52,{s}^{2}=0.97,N=10000,n=500$

a) Estimate$\mu}_{1$ . The sample is quivalent to an estimation of the true mean and so ${\mu}_{1}=2.30$

Place a bound on ths estimation:

$B}_{1}=2\sqrt{(1-\frac{n}{N})\frac{{s}^{2}}{n}$

$=2\sqrt{(1-\frac{500}{10000})\frac{0.65}{500}}$

$=0.07$

We are reasonably confident that the error of stimation is less than 0.07 pellets of deer per plot.

Likewise, estimate$\mu}_{2$ . The sample is quivalent to an estimation of the true mean and so ${\mu}_{2}=4.52$

Place a bound on ths estimation:

$B}_{2}=2\sqrt{(1-\frac{n}{N})\frac{{s}^{2}}{n}$

$=2\sqrt{(1-\frac{500}{10000})\frac{0.97}{500}}$

$=0.086$

We are reasonably confident that the error of stimation is less than 0.086 pellets of deer per plot.

b) Estimate diference i the mean size of pellet groups per plot and a margin error:

$({\mu}_{1}-{\mu}_{2})\pm 2\sqrt{V\left({\mu}_{1}\right)+v\left({\mu}_{2}\right)}$

$=(2.30-4.52)\pm 2\sqrt{0.001235+0.01843}$

$=-2.22\pm 0.111$

The difference between the means 2.22 and an approriate of errors is$\pm 0.111$

For deer

For rabbits:

a) Estimate

Place a bound on ths estimation:

We are reasonably confident that the error of stimation is less than 0.07 pellets of deer per plot.

Likewise, estimate

Place a bound on ths estimation:

We are reasonably confident that the error of stimation is less than 0.086 pellets of deer per plot.

b) Estimate diference i the mean size of pellet groups per plot and a margin error:

The difference between the means 2.22 and an approriate of errors is

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-Shape of the sampling distribution is always the same shape as the population distribution, no matter what the sample size is.

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-Sampling distributions get closer to normality as the sample size increases.

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C. Sampling distributions of means get closer to normality as the sample size increases.

D. Sampling distribution of the mean is always right skewed since means cannot be smaller than 0.

A. Shape of the sampling distribution of means is always the same shape as the population distribution, no matter what the sample size is.

B. Sampling distributions of means are always nearly normal.

C. Sampling distributions of means get closer to normality as the sample size increases.

D. Sampling distribution of the mean is always right skewed since means cannot be smaller than 0.

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Which of the following statements about the sampling distribution of the sample mean is incorrect?

(a) The standard deviation of the sampling distribution will decrease as the sample size increases.

(b) The standard deviation of the sampling distribution is a measure of the variability of the sample mean among repeated samples.

(c) The sample mean is an unbiased estimator of the population mean.

(d) The sampling distribution shows how the sample mean will vary in repeated samples.

(e) The sampling distribution shows how the sample was distributed around the sample mean.

(a) The standard deviation of the sampling distribution will decrease as the sample size increases.

(b) The standard deviation of the sampling distribution is a measure of the variability of the sample mean among repeated samples.

(c) The sample mean is an unbiased estimator of the population mean.

(d) The sampling distribution shows how the sample mean will vary in repeated samples.

(e) The sampling distribution shows how the sample was distributed around the sample mean.

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(b) the standard deviation of the sampling distribution is a measure of the variability of the sample mean among repeated samples,

(c) the sample mean is an unbiased estimator of the true population mean,

(d) the sampling distribution shows how the sample mean will vary in repeated samples,

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(b) the standard deviation of the sampling distribution is a measure of the variability of the sample mean among repeated samples,

(c) the sample mean is an unbiased estimator of the true population mean,

(d) the sampling distribution shows how the sample mean will vary in repeated samples,

(e) the sampling distributions shows how the sample was distributed around the sample mean.