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Chapter 7: Q. 19 (page 312)

A variable is said to be uniformly distributed or to have a uniform distribution with parameters \(a\) and \(b\) if its distribution has the shape of the horizontal line segment with equation \(y=\frac{1}{(b-a)}\) for \(a<x<b\). The mean and standard deviation of such a variable are \(\frac{(a+b)}{2}\) and \(\frac{(b-a)}{\sqrt{12}}\) respectively. The basic random number generator on a computer or calculator which returns a number between \(0\) and \(1\), simulates a variable having a uniform distribution with parameters \(0\) and \(1\).

a. Sketch the distribution of a uniformly distributed variable with parameters \(0\) and \(1\). Observe from your sketch that such a variable is far from being normally distributed.

b. Use the technology of your choice to simulate \(2000\) samples of two random numbers between \(0\) and \(1\).

c. Find the sample mean of each of the \(2000\) samples obtained in part (b).

d. Determine the mean and standard deviation of the \(2000\) sample means.

e. Theoretically what are the mean and the standard deviation of all possible sample means for samples of size \(2\)? Compare your answers to those you obtained in part (d).

f. Obtained a histogram of the \(2000\) sample means. Is the histogram bell shaped? Would you expect it to be?

g. Repeat parts (b)-(f) for a sample size of \(35\).

Short Answer

Expert verified

Part a.

Part b. The samples will be like

Part c. The mean will be \(\mu =[49.3615, 49.1953, 48.7853,49.1773]\)

Part d. The standard deviation and mean will be

\(\mu_{\bar{x}}=49.6020\)

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Part e. \(\mu_{\bar{x}}=49.34\)

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Part f.

Step by step solution

01

Part a. Step 1. Given information

The value of variable is

\(a=0\)

\(b=1\)

\(n=2\)

02

Part a. Step 2. Calculation

Calculate the mean and standard deviation using given relation

\(mean=\frac{a+b}{2}=0.5\)

\(\sigma=\frac{(b-a)}{\sqrt{12}}=0.2887\)

The compute the value of the function

\(y=\frac{1}{b-a}=1\)

So, we can see that the function value is 1 then sketch a graph using MATLAB.

Program:

Query:

  • First, we have defined given variables.
  • Then calculate the mean and standard deviation.
  • Then calculate the value of the function.
  • Sketch a graph.
03

Part b. Step 1. Calculation

So, according to the question I have used MATLAB to generate the samples.

We use the function \(鈥渘orminv (rand (1000,4), 49.34, 10.14)鈥漒).

After using the above function in MATLAB, we will get the \(1000\) sample for \(4\) dancers.

Program:

Query:

  • First, we have defined the number of samples.
  • Then using function 鈥渘orminv鈥� to find out the samples for dancers.
04

Part c. Step 1. Calculation

So, according to the question I have used MATLAB to generate the samples.

We use the function \(鈥渘orminv (rand (1000,4), 49.34, 10.14)鈥漒).

Then calculate the mean

\(\mu=\frac{\sum_{i-1}^{1000}x_{i}}{n}\)

After solving the equation, we will get

\(\mu =[49.3615, 49.1953, 48.7853,49.1773]\)

Program:

Query:

  • First, we have defined the number of samples.
  • Then using function 鈥渘orminv鈥� to find out the samples for dancers.
  • Then calculate the mean using function 鈥渕ean鈥�.
05

Part d. Step 1. Calculation

So, according to the question I have used MATLAB to generate the samples.

We use the function \(鈥渘orminv (rand (1000,4), 49.34, 10.14)鈥漒).

Then calculate the mean

\(\mu=\frac{\sum_{i-1}^{1000}x_{i}}{n}\)

After solving the equation, we will get

\(\mu =[49.3615, 49.1953, 48.7853,49.1773]\)

Calculate the mean of means

\(\mu_{\bar{x}}=\bar{\mu}\)

and standard deviation

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Program:

Query:

  • First, we have defined the number of samples.
  • Then using function 鈥渘orminv鈥� to find out the samples for dancers.
  • Then calculate the mean using function 鈥渕ean鈥�.
  • Calculate the mean of means and standard deviation.
06

Part e. Step 1. Calculation

The mean of the sampling distribution is equal to the population mean.

\(\mu_{\bar{x}}=\mu=49.34\)

The standard deviation of the sampling distribution is equal to the population standard deviation divide by the square root of the sample size.

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

07

Part f. Step 1. 

So, according to the question I have used MATLAB to generate the samples.

We use the function \(鈥渘orminv (rand (1000,4), 49.34, 10.14)鈥漒).

Then calculate the mean

\(\mu=\frac{\sum_{i-1}^{1000}x_{i}}{n}\)

After solving the equation, we will get

\(\mu =[49.3615, 49.1953, 48.7853,49.1773]\)

Calculate the mean of means

\(\mu_{\bar{x}}=\bar{\mu}\)

\(\mu_{\bar{x}}=49.6020\)

and standard deviation

\(\sigma_{\bar{x}}=\frac{10.14}{\sqrt{4}}=5.07\)

Program:

Query:

  • First, we have defined the number of samples.
  • Then using function 鈥渘orminv鈥� to find out the samples for dancers.
  • Then calculate the mean using function 鈥渕ean鈥�.
  • Calculate the mean of means and standard deviation.
  • Then sketch a histogram of the samples.

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Most popular questions from this chapter

Suppose that a simple random sample is taken without replacement from a finite population of size N.

Part (a): Show mathematically that Equations (7.1) and (7.2) are identical for samples of size 1.

Part (b): Explain in words why part (a) is true.

Part (c): Without doing any computations, determine r for samples of size N without replacement. Explain your reasoning.

Part (d): Use Equation(7.1) to verify your answer in part (c).

Although, in general, you cannot know the sampling distribution of the sample mean exactly, by what distribution can you often approximate it?

7.54 Unbiased and Biased Estimators. A statistic is said to be an unbiased estimator of a parameter if the mean of all its possible values equals the parameter. otherwise, it is said to be a biased estimator. An unbiased estimator yields, on average, the correct value of the parameter, whereas a biased estimator does not.
a. Is the sample mean an unbiased estimator of the population mean? Explain your answer.
b. Is the sample median an unbiased estimator of the population median? (Hint: Refer to Example 7.2 on pages 292-293. Consider samples of size 2.)

7.49 Mobile Homes. According to the U.S. Census Bureau publication Manufactured Housing Statistics, the mean price of new mobile homes is \(65,100. Assume a standard deviation of \)7200. Let x~ denote the mean price of a sample of new mobile homes.
a. For samples of size 50, find the mean and standard deviation of x炉. Interpret your results in words.
b. Repeat part (a) with n=100.

Explain why increasing the sample size tends to result in a smaller sampling error when a sample means is used to estimate a population mean.
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