COMPARING DIFFERENCES IN CENTERS TO VARIABILITY

About "Comparing differences in centers to variability"

Comparing differences in centers to variability :

Recall that to find the mean absolute deviation (MAD) of a data set, first find the mean of the data. Next, take the absolute value of the difference between the mean and each data point. Finally, find the mean of those absolute values.

Comparing differences in centers to variability - Example

The tables show the number of minutes per day students in a class spend exercising and playing video games. What is the difference of the means as a multiple of the mean absolute deviations ?

Solution :

Step 1 :

Calculate the mean number of minutes per day exercising.

0 + 7 + 7 + 18 + 20 + 38 + 33 + 24 + 22 + 18 + 11 + 6 = 204

Divide the sum by the number of students.

204 ÷ 12 = 17

Step 2 :

Calculate the mean absolute deviation for the number of minutes exercising.

|0-17| = 17

|7-17| = 10

|18-17| = 1

|20-17| = 3

|38-17| = 21

|33-17| = 16

|24-17| = 7

|22-17| = 5

|18-17| = 1

|11-17| = 6

|6-17| = 11

Find the mean of the absolute values.

17 + 10 + 10 + 1 + 3 + 21+ 16 + 7 + 5 + 1 + 6 + 11 = 108

Divide the sum by the number of students.

108 ÷ 12 = 9

Step 3 :

Calculate the mean number of minutes per day playing video games. Round to the nearest tenth.

13+18+19+30+32+46+50+34+36+30+23+19 = 350

Divide the sum by the number of students.

350 ÷ 12 ≈ 29.2

Step 4 :

Calculate the mean absolute deviation for the numbers of minutes playing video games.

|13-29.2| = 16.2

|18-29.2| = 11.2

|19-29.2| = 10.2

|30-29.2| = 0.8

|32-29.2| = 2.8

|46-29.2| = 16.8

|50-29.2| = 20.8

|34-29.2| = 4.8

|36-29.2| = 6.8

|30-29.2| = 0.8

|23-29.2| = 6.2

|19-29.2| = 10.2

Find the mean of the absolute values. Round to the nearest tenth.

16.2 + 11.2 + 10.2 + 0.8 + 2.8 + 16.8 + 20.8 + 4.8 + 6.8 + 0.8 + 6.2 + 10.2 = 107.6

Divide the sum by the number of students.

107.6 ÷ 12 ≈ 9

Step 5 :

Find the difference in the means.

Subtract the lesser mean from the greater mean.

29.2 - 17 = 12.2

Step 6 :

Write the difference of the means as a multiple of the mean absolute deviations, which are similar but not identical.

Divide the difference of the means by the MAD.

12.2 ÷ 9 ≈ 1.36

The means of the two data sets differ by about 1.4 times the variability of the two data sets.

Comparing differences in centers to variability - Practice question

Question :

The high jumps in inches of the students on two intramural track and field teams are shown below. What is the difference of the means as a multiple of the mean absolute deviations ?