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 |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 ?
Solution :
About 1.1 times the MAD.
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