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 as 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 of 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 part (c) and (d). Why are they different?

\(n=4\)

Now, generate four digits between \(0\) and \(9\) using random number table.

Here the initial \(4\) digits in row \(15\) in the random number table are selected as follows: \(0,7,8,5\)

Mean can be calculated as the summation of all values divided by the quantity of values

\(\bar{x}=\frac{0+7+8+5}{4}=\frac{20}{4}=5\)

After this, the above exercise can be repeated 50 times using different rows and columns or a random number generator can also be used to obtain the required results.

\(n=4\)

POPULATION PARAMETERS

In the given question the population consists of digits \(0,1,2,3,4,5,6,7,8,9\)

The population mean can be calculated as the summation of all values divided by the quantity of values.

\(\mu=\frac{0+1+2+3+4+5+6+7+8+9}{10}=\frac{45}{10}=4.5\)

The variance could be calculated as the sum of divisions already squared with respect to the mean divided by number of observations

\(\sigma =\sqrt{\frac{(0-4.5)^{2}+...+(9-4.5)^{2}}{10}}=\sqrt{\frac{82.5}{10}}\approx 2.8733\)

The mean of the distribution is sampling of the sample mean is same as the population mean

\(\mu_{x}=\mu=4.5\)

\(\sigma _{x}=\frac{\sigma}{\sqrt{4}}=\frac{2.8733}{\sqrt{4}}\approx1.4367\)

\(\bar{x}=\frac{5+4.75+3.75+...+4.25+3.75+4.5}{50}=4.54\)

The standard deviation is square root of the sum of divisions already squared with respect to the mean divided by \(n-1\)

\(s=\sqrt{\frac{(5-4.54)^{2}+...+(4.5-4.54)^{2}}{50-1}}\approx1.4596\)

The main difference between the results is parts (c) and (d) is because of sampling error. To be more precise, different samples would have different means and different standard deviations even though the values would be very close to the expected values.