Estimate mu_1, and mu_2, the average number of pellet groups for deer and rabbits, respectively, per 30-foot-square-plots.

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 μ1,andμ2, 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
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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 μ=2.30,s2=0.65,N=10000,n=500
For rabbits: ν=4.52,s2=0.97,N=10000,n=500
a) Estimate μ1. The sample is quivalent to an estimation of the true mean and so μ1=2.30
Place a bound on ths estimation:
B1=2(1nN)s2n
=2(150010000)0.65500
=0.07
We are reasonably confident that the error of stimation is less than 0.07 pellets of deer per plot.
Likewise, estimate μ2. The sample is quivalent to an estimation of the true mean and so μ2=4.52
Place a bound on ths estimation:
B2=2(1nN)s2n
=2(150010000)0.97500
=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:
(μ1μ2)±2V(μ1)+v(μ2)
=(2.304.52)±20.001235+0.01843
=2.22±0.111
The difference between the means 2.22 and an approriate of errors is ±0.111
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