Mean Absolute Deviation Mean Absolute Deviation O Definition: Mean - PowerPoint PPT Presentation

Mean Absolute Deviation

Mean Absolute Deviation O Definition: Mean Absolute Deviation (MAD) is the average of absolute differences between each value in a set of values, and the average of all values of that set. O The MAD measures the average (absolute) distance of the sample values from the mean of the data values. O Let x = data values O Let 𝒚 = mean.

Mean Absolute Deviation O For example, the average (or mean) of the set of values 1, 2, 3, 4, and 5 is (15 ÷ 5) or 3. The difference between this average (3) and the values in the set is 2, 1, 0, -1, and -2; the absolute difference being 2, 1, 0, 1, and 2. The average of these numbers (6 ÷ 5) is 1.2 which is the mean absolute deviation.

Find the Mean Absolute Deviation Test scores for 6 students were : 85, 92, 88, 80, 91 and 20. 1. Find the me mean an: (85+92+88+80+91+20)/6=76 2. Find the deviation iation from m th the me mean an: 85-76=9 92-76=16 88-76=12 80-76=4 91-76=15 20-76=-56

Find the mean absolute deviation Test scores for 6 students were : 85, 92, 88, 80, 91 and 20. 3. Find the absolute value of each deviation from the mean:          85 76 9 92 76 16 88 76 12         80 76 4 91 76 15 20 76 56

Find the mean absolute deviation Test scores for 6 students were : 85, 92, 88, 80, 91 and 20. 4. Find the sum of the absolute values: 9 + 16 + 12 + 4 + 15 + 56 = 112 5. Divide the sum by the number of data items: 112/6 = 18.7 The me mean an ab absolu olute e deviation viation is 18.7 18.7.

Steps for Finding MAD 1. Find the mean. 2. Subtract the sample mean from each data value. 3. Take the absolute values of the results. 4. Add the absolute values together. 5. Divide by the sample size.

MAD – Practice Problem What is the MAD for the following sample values? 3 8 6 12 0 -4 10

Use a Table Data Values, 𝒚 x – 𝒚 𝒚 − 𝒚 x Total

MAD – Practice Problem Solution MAD = 32/7 = 4.57