Question

7.58 Class Project Simulation. This exercise can be done individually or, better yet, as a class project.
a. Use a random-number table or random-number generator to obtain a sample (with replacement) of four digits between 0 and 9 .
Do so a total of 50 times and compute the mean of each sample.
b. Theoretically, what are the mean and standard deviation of all possible sample means for samples of size 4?
c. Roughly what would you expect the mean and standard deviation of the 50 sample means you obtained in part (a) to be? Explain your answers.
d. Determine the mean and standard deviation of the 50 sample means you obtained in part (a).
e. Compare your answers in parts (c) and (d). Why are they different?

Short answer

(a) Depending on the first number chosen, the solutions may differ.
(b) The mean and standard deviation of all possible sample means for samples of size $4$are $4.5$and $2.8733$.

(c) The mean and standard deviation for sample mean $50$are $4.5$and $1.4367$.

(d) Depending on the first number chosen, the solutions may differ.

(e) Parts (c) and (d) have the most significant differences due to sampling error.

Step-by-step solution

Part (a) Step 1: Given information

To use a random-number table or random-number generator to obtain a sample (with replacement) of four digits between $0$and $9$.

Part (a) Step 2: Explanation

Let, the given value:
$n=4$
Using the random number table, produce four digits between $0$and $9$. In the random number table, the first four digits in row $15$ are chosen as follows:
$0,7,8,5$
The sum of all values divided by the number of values can be used to compute the mean.
$\overline{\mathrm{x}}=\frac{0+7+8+5}{4}$

$=\frac{20}{4}$

$=5$

Repeat the previous exercise $50$times with different rows and columns, or use a random number generator to get the desired results.

As a result, depending on the first number chosen, the solutions may differ.

Part (b) Step 1: Given information

To find the mean and standard deviation of all possible sample means for samples of size $4$.

Part (b) Step 2: Explanation

Let, the given value:

$n=4$

The population in the provided question is made up of digits $0,1,2,3,4,5,6,7,8,9$. The population mean can be found by multiplying the total of all values by the number of values.
$\mu =\frac{0+1+2+3+4+5+6+7+8+9}{10}$

$=\frac{45}{10}$

$=4.5$

The variance can be computed by dividing the number of observations by the sum of divisions already squared with regard to the mean.
$\sigma =\sqrt{\frac{(0-4.5{)}^2+\dots +(9-4.5{)}^2}{10}}$

$=\sqrt{\frac{82.5}{10}}$

$\approx 2.8733$

As a result, the mean and standard deviation are $4.5$ and $2.8733$.

Part (c) Step 1: Given information

To calculate approximately the mean and standard deviation of the $50$ sample means that obtained in part (a).

Part (c) Step 2: Explanation

The sample mean is the same as the population mean because the mean of the distribution is sampling.
The mean is determined as:
${\mu }_{\overline{x}}=\mu$

$=4.5$

And the standard deviation is determined as:

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

$=\frac{2.8733}{\sqrt{4}}$

$\approx 1.4367$

As a result, the mean and standard deviation for sample size $50$ are $4.5$and $1.4367$.

Part (d) Step 1: Given information

To determine the mean and standard deviation of the $50$ sample means that obtained in part (a).

Part (d) Step 2: Explanation

Determine the standard deviation as follows:

$\overline{x}=\frac{5+4.75+3.75+\dots .+4.25+3.75+4.5}{50}$

$=4.54$

The square root of the sum of divisions already squared with regard to the mean divided by $n-1$ is the standard deviation.
$s=\sqrt{\frac{(5-4.54{)}^2+\dots +(4.5-4.54{)}^2}{50-1}}$

$\approx 1.4596$

Depending on the first number chosen, the solutions may differ.

Part (e) Step 1: Given information

To compare your answers in parts (c) and (d). And determine that how they different.

Part (e) Step 2: Explanation

Since,the parts (c) and (d) have the most significant differences due to sampling error.
To be more specific, even though the numbers are fairly near to the expected values, various samples will have different means and standard deviations.