Question

The rare booted eagle of western Europe was the focus of a study by S. Suarez et al. to identify optimal nesting habitat for this raptor. According to their paper "Nesting Habitat Selection by Booted Eagles and implications for Management", the distances of such nests to the nearest marshland are normally distributed with mean \(4.66\) km and standard deviation \(0.75\) km.

a. Simulate \(3000\) samples of four distances each.

b. Determine the sample mean and sample standard deviation of each of the \(3000\) samples.

c. For each of the \(3000\) samples, determine the observed value of the standardized version of \(\bar{x}\).

d. Obtain a histogram of the \(3000\) observations in part (c).

e. Theoretically, what is the distribution of the standardized version of \(\bar{x}\)?

f. Compare your results from parts (d) and (e).

g. For each of the \(3000\) samples, determine the observed value of the studentized version of \(\bar{x}\).

h. Obtain a histogram of the \(3000\) observations in part (g).

i. Theoretically, what is the distribution of the studentized version of \(\bar{x}\) ?

j. Compare your results from parts (h) and (i).

k. Compare your histogram from parts (d) and (h). How and why do they differ?

Short answer

Part a.

Then the five batting average will be

Part b. \(\bar{x}=4.66\)

\(\sigma=0.37\)

Part c.

Part d.

Step-by-step solution

Part a. Step 1. Given information

The number of sample \((n)\), mean \((\bar{x})\) and standard deviation \((\sigma)\) is given.

\(\bar{x}=4.66\)

\(\sigma=0.75\)

\(n=3000\times 4=12000\)

Part a. Step 2. Calculation

Generate \(2000\) samples of males using function “norminv” with \(0.031\) sample mean and \(0.27\) standard deviation in MATLAB

\(r=norminv(rand(3000, 4), 4.66, 0.75)\)

After that we will get random 3000 samples.

Program:

Query:

• First, we have defined the number of samples.

• Then create \(2000\) random samples using function “norminv” with sample mean \(0.27\) and standard deviation \(0.031\).

Part b. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=4.66\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.37\)

The solution is \(\bar{x}=4.66\)

\(\sigma=0.37\)

Part c. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=4.66\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.37\)

Calculate the standardized value for the all samples

\(z=\frac{sample mean - 4.66}{0.37}\)

Program:

Query:

• First, we have defined the number of samples.
• Then generate the random samples.
• Then calculate the mean and standard deviation.

• Calculate the \(z-\)score which is standardized form of all samples.

Part d. Step 1. Calculation

Calculate the mean using relation

\(\bar{x}=\frac{\sum_{i-1}^{n}x_{i}}{n}\)

After simplifying, we will get

\(\bar{x}=4.66\)

Calculate the standard deviation using relation

\(\sigma =\sqrt{\frac{\sum (x_{i}-\mu)^{2}}{n}}\)

After simplifying we will get

\(\sigma=0.37\)

Calculate the standardized value for the all samples

\(z=\frac{sample mean - 4.66}{0.37}\)

Program:

Query:

• First, we have defined the number of samples.
• Then generate the random samples.
• Then calculate the mean and standard deviation.
• Calculate the z-score which is standardized form of all samples.
• Then generate the histogram.