Description
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What is a simple random sample? If all sampling units have the same probability of being in the sample, is this always a simple random sample? Why or why not?
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Use R to take a simple random sample of size n = 4 (assume without replacement) from this population: 10, 11, 13, 11, 10, 6, 22, 15, 14, 23. Please include your output in the homework. Using this sample, compute your estimator of the population mean and nd its standard error (what IS a standard error?) and a 95 percent con dence interval.
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We want to know the total number of moose in an area. Suppose we have divided the region into N = 200 quadrats, our guess is that the standard deviation of the moose counts is around s = 3 moose AND we would like a margin of error of less than plus or minus 100 moose. How many sampling units do we have to visit?
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Finally (see problem 4), suppose we had decided to sample n = 12 of the quadrats and got a sample average of 14 moose per quadrat and a sample variance of s2 = 125 square moose per quadrat. Find an estimate of the total number of moose, along with its standard error and then construct a 95 percent con dence interval for the total number of moose.
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I wish to estimate the total number of squirrels in a large region. I’ll do that by dividing the region into N = 1500 transects (long, narrow plots, with one end against a road), each 10 m wide and 1 Km long. I select n = 10 of these to visit (I’ll walk the transects and count animals. I’ll assume that I count every one in the transect and ignore all animals outside the transect. This might work if I alarm them as I go by.) I get the following counts: 12; 20; 8; 42; 23; 18; 6; 8; 13; 17.
(a) Find an estimator of the total number of squirrels in the entire region and the standard error, along with a 95 percent con dence interval.
(b) If I divide the estimated total by the total area of the region (in square Kilometers) I’ll get a density in squirrels per Km2. Find a 95 percent con dence interval for this density (hint – how do you adjust the standard error from part a?).
3
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To use the typical estimator 2 standard errors to nd a 95 percent interval for a proportion, you need the sampling distribution of the sample proportion to be close to normal. You can assume this
is the case if n (est proportion) > 10 and n (1 est proportion) > 10.
Suppose we are looking for the proportion of spruce trees in a low land forest that have a certain genetic trait. We somehow make a list of N = 1320 trees in the area. We take a SRS of size n = 120 trees and nd that 13 have the genetic trait.
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Is this sample size su cient for us to assume the sampling distribution of p.hat is approximately normal? Why or why not?
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Find a 95 percent con dence interval for the true proportion of trees in the region with the trait.
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I wish to conduct a political poll in a small town. The town has a total of 450 residents and I can actually take a SRS of them. I would really like a margin of error of, at most, plus or minus 5 percent. What sample size do I need to take? What sample size would I need to take if I perversely decided to use SRS WITH replacement?
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( This will require the use of some R programming.) Use boot-strapping on the data in problem 2 to get a 95 percent con dence interval for the population mean.
For problem 9, consider the following data, which are the actual number of trees in each of the N = 70 plots in the region:
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1
3
4
5
3
3
0
7
5
0
1
4
1
4
2
3
4
3
4
4
3
3
0
1
2
1
5
5
3
3
11
9
9
15
10
6
6
12
12
12
18
14
13
7
10
13
3
14
11
11
15
11
13
12
12
13
19
10
11
14
13
6
9
7
15
14
9
16
13
12
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Take a SRS (without replacement) of size n = 18 from the population in the table. Find a 95 percent con dence interval for the total over the entire area. Later you will compare this estiamte with the estimate from a strati ed random sample.