Item
|
Value
|
Value – Mean = Difference
|
1
|
4
|
4 – 8 = -4
|
2
|
8
|
8 – 8 = 0
|
3
|
12
|
12 – 8 = 4
|
4
|
10
|
10 – 8 = 2
|
5
|
6
|
6 – 8 = -2
|
The problem with this, is that the sum of the difference column, will always equal zero. This is because the mean is the fulcrum of all values, so there should be an equal difference to the left (negative) as to the right (positive). We need to get rid of the negatives. To accomplish this we could multiply by -1, but that would turn the positive values to a negative. A better approach might be to simply ignore the negative sign (known as taking the absolute value), as illustrated in Table 15. This approach will actually lead to the Mean Absolute Deviation (MD or MAD) (Kenney, 1939).
Item
|
Value
|
Value – Mean = Difference
|
Absolute deviation
|
1
|
4
|
4 – 8 = -4
|
|-4| = 4
|
2
|
8
|
8 – 8 = 0
|
|0| = 0
|
3
|
12
|
12 – 8 = 4
|
|4| = 4
|
4
|
10
|
10 – 8 = 2
|
|2| = 2
|
5
|
6
|
6 – 8 = -2
|
|-2| = 2
|
The total absolute deviation is equal to 4 + 0 + 4 + 2 + 2 = 12, so the mean absolute deviation is 12 / 5 = 2.4. On average a value was 2.5 above or below the mean of 8. There is another method to get rid of those negative differences, which will be discussed in the next section.
References
Kenney, J. F. (1939). Mathematics of Statistics; Part one. London: Chapman & Hall. Retrieved from http://archive.org/details/MathematicsOfStatisticsPartI
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