Mean Absolute Deviation Use the same population of {9 cigarettes, 10 cigarettes, 20 cigarettes} from Exercise 45. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not centre about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population?

Three values of the number of cigarettes smoked per day are used to make a population. They are used to extract nine samples selected with replacement.

Mean Absolute Deviation(MAD) is calculated by dividing the absolute value of the sum of the squared differences of the observations from the mean by n. Here, n is the total number of values.

Mathematically,

$MAD=\frac{{\sum }_{i=1}^n\left|x_i-\overline{x}\right|}{n}$

Nine samples of size 2 each are selected as shown below:

Use the following steps to compute the mean of MADs for all samples:

• First, obtain the mean for each sample.
• Using this mean, compute the mean absolute deviation.
• Repeat the process for all samples.
• Compute the aggregate mean of all MADs.

Sample 1:

$$\begin{gathered} {\overline{x}}_1=\frac{9+10}{2} \\ =9.5 \end{gathered}$$

The mean of sample 1 is 9.5.

$$\begin{gathered} MA{D}_1=\frac{\left|9-9.5\right|+\left|10-9.5\right|}{2} \\ =0.5 \end{gathered}$$

The mean absolute deviation of sample 1 is 0.5.

Sample 2:

$$\begin{gathered} {\overline{x}}_2=\frac{9+20}{2} \\ =14.5 \end{gathered}$$

The mean of sample 2 is 14.5.

$$\begin{gathered} MA{D}_2=\frac{\left|9-14.5\right|+\left|20-14.5\right|}{2} \\ =5.5 \end{gathered}$$

The mean absolute deviation of sample 2 is 5.5.

Sample 3:

$$\begin{gathered} {\overline{x}}_3=\frac{10+20}{2} \\ =15.0 \end{gathered}$$

The mean of sample 3 is 15.0.

$$\begin{gathered} MA{D}_3=\frac{\left|10-15.0\right|+\left|20-15.0\right|}{2} \\ =5.0 \end{gathered}$$

The mean absolute deviation of sample 3 is 5.0.

Sample 4:

$$\begin{gathered} {\overline{x}}_4=\frac{9+9}{2} \\ =9.0 \end{gathered}$$

The mean of sample 4 is 9.0.

$$\begin{gathered} MA{D}_4=\frac{\left|9-9.0\right|+\left|9-9.0\right|}{2} \\ =0.0 \end{gathered}$$

The mean absolute deviation of sample 4 is 0.0.

Sample 5:

$$\begin{gathered} {\overline{x}}_5=\frac{10+10}{2} \\ =10.0 \end{gathered}$$

The mean of sample 5 is 10.0.

$$\begin{gathered} MA{D}_5=\frac{\left|10-10.0\right|+\left|10-10.0\right|}{2} \\ =0.0 \end{gathered}$$

The mean absolute deviation of sample 5 is 0.0.

Sample 6:

$$\begin{gathered} {\overline{x}}_6=\frac{20+20}{2} \\ =20.0 \end{gathered}$$

The mean of sample 6 is 20.0.

$$\begin{gathered} MA{D}_6=\frac{\left|20-20.0\right|+\left|20-20.0\right|}{2} \\ =0.0 \end{gathered}$$

The mean absolute deviation of sample 6 is 0.0.

Sample 7:

$$\begin{gathered} {\overline{x}}_7=\frac{10+9}{2} \\ =9.5 \end{gathered}$$

The mean of sample 7 is 9.5.

$$\begin{gathered} MA{D}_7=\frac{\left|10-9.5\right|+\left|9-9.5\right|}{2} \\ =0.5 \end{gathered}$$

The mean absolute deviation of sample 7 is 0.5.

Sample 8:

$$\begin{gathered} {\overline{x}}_8=\frac{20+9}{2} \\ =14.5 \end{gathered}$$

The mean of sample 8 is 14.5.

$$\begin{gathered} MA{D}_8=\frac{\left|20-14.5\right|+\left|9-14.5\right|}{2} \\ =5.5 \end{gathered}$$

The mean absolute deviation of sample 8 is 5.5.

Sample 9:

$$\begin{gathered} {\overline{x}}_9=\frac{20+10}{2} \\ =15.0 \end{gathered}$$

The mean of sample 9 is 15.0.

$$\begin{gathered} MA{D}_9=\frac{\left|20-15.0\right|+\left|10-15.0\right|}{2} \\ =5.0 \end{gathered}$$

The mean absolute deviation of sample 9 is 5.0.

The mean of all nine sample mean absolute deviations is obtained as follows:

$$\begin{gathered} \overline{MAD}=\frac{{\sum }_{i=1}^{9}MA{D}_i}{9} \\ =\frac{0.5+5.5+....+5.0}{9} \\ =2.4. \end{gathered}$$

The mean absolute deviation of the samples is 2.4.

The population mean is obtained as shown below:

$$\begin{gathered} \mu =\frac{9+10+20}{3} \\ =13.0. \end{gathered}$$

The population mean is 13.0.

The population mean absolute deviation is obtained as shown below:

$$\begin{gathered} MA{D}_{pop}=\frac{{\sum }_{i=1}^3\left|x_i-13\right|}{3} \\ =\frac{\left|9-13.0\right|+\left|10-13.0\right|+\left|20-13.0\right|}{3} \\ =4.7 \end{gathered}$$

The population mean absolute deviation is 4.7.

An estimate from the set of samples is an unbiased measure if the mean of the sample measures is the same as the population measure value.

The population mean absolute deviation is 4.7, and the mean absolute deviation of the samples is 2.4.

Since the two values are not equal, it can be concluded that the sample mean absolute deviation is not an unbiased estimator of the population mean absolute deviation.