1. DESCRIPTIVE STATISTICS QUIZ
On a separate piece of paper, create the descriptive statistics for the set below, then create a box-plot.
Set = { 0, 5, 22, 28, 28, 28, 29, 31, 32, 32, 39, 42 }
Descriptive Statistics: Observations, Range, Median, Q3, Q1, IQR, IQL, Outliers, Max, Min, Mean, Mode.

2. Mean absolute deviation
A step-by-step walkthrough

3. Set up
Take a set of numbers, any set of numbers, it doesn’t matter if they are inches or kilometers, ounces or ages, weights or heights, as long as they are all heights, or all ages, or all basketball scores. It also does not matter what order they are arranged in, the just have to be numbers that are in some way or another related to each other.
We will use the numbers: 22, 15, 35, 24, 15, 40, 49, 20, 28, 38.
But in this case, we will designate the first five numbers as
Set A = { 22, 15, 35, 24, 15 },
and the second five numbers as
Set B = { 40, 49, 20, 28, 38 }.
Each number in each set is an observation. Since there are five numbers in each set, there are five observations per set.

4. Mean absolute deviation: step one: finding the mean
The MEAN is a mathematical term synonymous with arithmetic average.
To take a mean, we add up all of the numbers and divide by the number of observation.
In this case, we will use the same sets.
Set A = { 22, 15, 35, 24, 15 }.
Set B = { 40, 49, 20, 28, 38 }.
First, we add the numbers together:
= 111
= 175
Then, we divide by the number of observations, which we have already determined is five.
111
175 =
22.2 = 35 5 5
The MEAN of Set A = 22.2.
The MEAN of Set B = 35.

5. Mean absolute deviation: step two: calculating deviations
The deviation is the difference between the observation and the mean.
To find the deviations, subtract the mean from each observation: observation – mean = deviation.
Let’s re-order the sets to make it easier:
SET A: Mean = 22.2
SET B: Mean = 35
Observations
Mean
Deviation
Observations
Mean
Deviation 22 40 – = – = 15 – = 49 – = 35 – = 20 – =
Try it before I give you the answers. Then match your answers with mine on the next two slides. 24 = 28 – – = 15 = 38 – – =

6. Mean absolute deviation: step two: calculating deviations
The deviation is the difference between the observation and the mean.
To find the deviations, subtract the mean from each observation: observation – mean = deviation.
Let’s re-order the sets to make it easier:
SET A: Mean = 22.2
SET B: Mean = 35
Observations
Mean
Deviation
Observations
Mean
Deviation 22 40 –
22.2 = – 35 = 15 – = 49
22.2 – 35 = 35 – = 20
22.2 – = 35
Try it before I give you the answers. Then match your answers with mine on the next slide. 24 28 –
22.2 = – 35 = 15 = 38 –
22.2 – 35 =

7. Mean absolute deviation: step two: calculating deviations
The deviation is the difference between the observation and the mean.
To find the deviations, subtract the mean from each observation: observation – mean = deviation.
Let’s re-order the sets to make it easier:
SET A: Mean = 22.2
SET B: Mean = 35.0
Observations
Mean
Deviation
Observations
Mean
Deviation 22 40 –
22.2 = =
-0.2 – 35 5 15 –
22.2 = 49
-7.2 – 35 = 14 35 –
22.2 = 20
12.8 – = 35
-15 24 –
22.2 = 28
1.8 – 35 = -7 15 = 38 –
22.2
-7.2 – 35 = 3

8. Mean absolute deviation: step three: determining absolute deviations
The absolute deviation is the distance between the observation and the mean.
Effectively, this means taking the absolute value of each deviation: | deviation | = absolute deviation.
SET A:
SET B:
Absolute Value
Absolute Value
Deviation
Deviation
Deviation
Deviation
-0.2 5 = =
-7.2 = 14 =
12.8 =
-15 =
Try it before I give you the answers. Then match your answers with mine on the next slides.
1.8 = -7 =
-7.2 = 3 =

9. Mean absolute deviation: step three: determining absolute deviations
The absolute deviation is the distance between the observation and the mean.
Effectively, this means taking the absolute value of each deviation: | deviation | = absolute deviation.
SET A:
SET B:
Absolute Value
Absolute Value
Deviation
Deviation
Deviation
Deviation
-0.2
| -0.2 | = 5
| 5 | =
-7.2
| -7.2 | = 14
| 14 | =
12.8
|12.8 | =
|-15 |
-15 =
Try it before I give you the answers. Then match your answers with mine on the next slide.
1.8
| 1.8 | = -7
| -7 | =
-7.2
| -7.2 | = 3
| 3 | =

10. Mean absolute deviation: step three: determining absolute deviations
The absolute deviation is the distance between the observation and the mean.
Effectively, this means taking the absolute value of each deviation: | deviation | = absolute deviation.
SET A:
SET B:
Absolute Value
Absolute Value
Deviation
Deviation
Deviation
Deviation
-0.2
| -0.2 |
0.2 5
| 5 | = = 5
-7.2
| -7.2 | =
7.2 14
| 14 | = 14
12.8
|12.8 | =
12.8
|-15 |
-15 = 15
1.8
| 1.8 | =
1.8 -7
| -7 | = 7
-7.2
| -7.2 | =
7.2 3
| 3 | = 3

11. Mean absolute deviation: step four: find the mean of the absolute deviations
The last step is to find the mean of the absolute deviations. This means addinf the absolute deviations of a set together and dividing by the number of observations.
Absolute Deviation
Absolute Deviation
SET A:
SET B:
0.2 5
7.2 14
12.8 15
1.8 7
Try it before I give you the answers. Then match your answers with mine on the next slides.
7.2 3

12. Mean absolute deviation: step four: find the mean of the absolute deviations
The last step is to find the mean of the absolute deviations. This means adding the absolute deviations of a set together and dividing by the number of observations.
Absolute Deviation
Absolute Deviation
SET A:
SET B:
0.2 5
7.2 14
12.8 15
1.8 7
7.2 3
29.2 ≈5.40
44 ≈6.63
29.2 44
The mean absolute deviation of Set A,
{ 22, 15, 35, 24, 15 }, is approximately 5.4.
The mean absolute deviation of Set B,
{ 40, 49, 20, 28, 38 }, is approximately 6.63.

13. Mean absolute deviation: Putting the the mean absolute deviations to use
Now that we have determined the mean absolute deviation for Set A is approximately 5.4 and the mean absolute deviation for Set B is approximately 6.63, we can decide if set A or Set B is more consistent. Remember that for consistency, we look to see which set is closer to zero. What this really tells us is which observations, taken as a group, has the least deviation from the absolute mean.
SET A:
{ 22, 15, 35, 24, 15 }
SET B:
{ 40, 49, 20, 28, 38 }
𝑀𝐴𝐷≈5.40
𝑀𝐴𝐷≈6.63
As 5.40 is closer to zero than 6.63, Set A is more consistent than Set B.

14. Summary Step 1 Step 2 Step 3 Step 4 Step 5 Find the mean for each set.
To find the mean absolute deviations for Set A = { 22, 15, 35, 24, 15 } and Set B = { 40, 49, 20, 28, 38 }:
Step 1
Find the mean for each set.
Step 2
Determine the deviations for each set.
Step 3
Determine the absolute deviations for each set.
Step 4
Find the mean of the absolute deviations for each set.
Step 5
Once the MAD is determined for each set, the set with the MAD closest to zero is the more consistent set.

15. challenge Try to find the mean absolute deviation of the sets:
Set A = { 22, 15, 35, 24, 15 }
Set B = { 40, 49, 20, 28, 38 }
Set C = { 19, 27, 37, 39, 48 }
Set D = { 17, 17, 17, 17, 17 }
Set E = { 21, 25, 39, 37, 50 }
Set F = { 17, 19, 25, 23, 21 }
until you get it right every time.