Chapter 6, Problem 6DP

Formulas Confirm the two formulas hold true for the central limit theorem for the population containing the elements {1, 5, 10}. First, compute the population mean and standard deviation for the data set. Next, create a list of all 9 of the possible two-element samples that can be created with replacement: {1, 1}, {1, 5}, etc. For each of the 9 compute the sample mean. Now find the mean of the sample means. Does it equal the population mean? Compute the standard deviation of the sample means. Does it equal the population standard deviation, divided by the square root of n?

To determine

To compute:  The population means and standard deviation for the data set.

To Create: List of all 9 of the possible two-element samples and calculate sample mean.

To find: The mean of the sample means and check whether the mean of the sample mean is equal to the population

To find: The standard deviation of sample means and checks whether the population standard deviation, divided by the square root of n.

Answer to Problem 6DP

The population means and standard deviation for the data set is, 5.33 and 3.6818.

List of all 9 of the possible two-element samples:

Sample
(1,1)
(1,5)
(1,10)
(5,1)
(5,5)
(5,10)
(10,1)
(10,5)
(10,10)

The sample mean is,

Mean

$\frac{1+1}{2}=1$

$\frac{1+5}{2}=3$

$\frac{1+10}{2}=5.5$

$\frac{5+1}{2}=3$

$\frac{5+5}{2}=5$

$\frac{5+10}{2}=7.5$

$\frac{10+1}{2}=5.5$

$\frac{10+5}{2}=7.5$

$\frac{10+10}{2}=10$

The mean of the sample means is, 5.33.

The comparison is that, the population mean is same as the mean of the sample means.

The standard deviation of sample means is 2.6034 and it is equal to the population standard deviation, divided by the square root of n.

The comparison is that, the standard deviation of sample means is equal to the population standard deviation.

Explanation of Solution

Given info:

The population containing the elements is $\left\{1,5,10\right\}.$

Calculation:

Use the following formulas to compute the population means and standard deviation for the data set.

$\mu =\frac{{\displaystyle \sum _{i=1}^N{x}_i}}{N}$

$\sigma =\sqrt{\frac{{\displaystyle \sum _{i=1}^N{\left(X_i-\mu \right)}^2}}{N}}$

The population mean for the data set is,

$\begin{array}{c}\mu =\frac{{\displaystyle \sum _{i=1}^N{x}_i}}{N}\\ =\frac{1+5+10}{3}\\ =5.33\end{array}$

The population mean for the data set is,

$\begin{array}{c}\sigma =\sqrt{\frac{{\displaystyle \sum _{i=1}^N{\left(X_i-\mu \right)}^2}}{N}}\\ =\sqrt{\frac{{\left(1-5.33\right)}^2+{\left(5-5.33\right)}^2+{\left(10-5.33\right)}^2}{3}}\\ =\sqrt{\frac{40.6667}{3}}\\ =3.6818\end{array}$

Therefore, the population means and standard deviation for the data set is, 5.33 and 3.6818.

The below table shows the list of all 9 of the possible two-element samples and mean of each sample is,

Sample	Mean
(1,1)	$\frac{1+1}{2}=1$
(1,5)	$\frac{1+5}{2}=3$
(1,10)	$\frac{1+10}{2}=5.5$
(5,1)	$\frac{5+1}{2}=3$
(5,5)	$\frac{5+5}{2}=5$
(5,10)	$\frac{5+10}{2}=7.5$
(10,1)	$\frac{10+1}{2}=5.5$
(10,5)	$\frac{10+5}{2}=7.5$
(10,10)	$\frac{10+10}{2}=10$

The mean of the sample means is,

$\begin{array}{c}{\mu }_{\overline{X}}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{1+3+5.5+3+5+7.5+5.5+7.5+10}{9}\\ =\frac{48}{9}\\ =5.33\end{array}$

Here, the mean of the sample mean is 5.33.

Comparison:

The mean of the sample mean is 5.33 and observe that the population mean is same as the mean of the sample means. That is, ${\mu }_{\overline{X}}=\mu =\mathrm{5.33.}$

The standard deviation of sample means is,

$\begin{array}{c}{\sigma }_{\overline{X}}=\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left({\overline{X}}_i-{\mu }_{\overline{X}}\right)}^2}}\\ =\sqrt{\frac{{\left(1-5.33\right)}^2+{\left(3-5.33\right)}^2+{\left(5.5-5.33\right)}^2+\mathrm{...}+{\left(10-5.33\right)}^2}{9}}\\ =\sqrt{\frac{61.0001}{9}}\\ =2.6034\end{array}$

The population standard deviation, and it is divided by $\sqrt{n}$ is,

$\begin{array}{c}\sigma =\frac{3.6818}{\sqrt{2}}\left(\text{Since }n=2\right)\\ =2.6034\end{array}$

Comparison:

The standard deviation of sample means is equal to the population standard deviation which is ${\sigma }_{\overline{X}}=\sigma =\mathrm{2.6034.}$