Download Teacher Notes for Means and MADS

Teacher Notes
Mean Absolute Deviation Activity
from
Means and MADS
Gary D. Kader
Mathematics Teaching in the Middle School, Vol. 4, No. 6, March 1999
Materials needed: Chart paper, post it notes, projection device, visuals for projecting,
copies of the 8 distributions for students or groups of students
PART I
1. Present students with the following scenario:
Nine people were asked, “How many people are in your family?” One result from
the poll is that the average family size for the nine people was five.
2. Show students Distribution 2. Explain that this line plot shows the distribution of the data collected from
this poll. Explain that a distribution is “...the values of the variable together with the frequency of each
value.” (Navigating through Data Analysis in Grades 6-8, p. 31) In this example the variable is the number
of people in families. Make sure that the students see in this line plot that there are three families with 2
members, one family with three members, etc. Also review how to calculate the mean.
Distribution 2
X
X
X
X
X
X
2
3
4
5
6
7
X
X
8
9
X
10
11
3. Explain that there are many different possible distributions for nine families where the average (mean)
family size would be five.
4. Put students in groups and give them chart paper and 9 sticky notes. Have them create
another distribution that would have a mean average of five. To keep things simple, tell them the smallest
family size will be 2 and the largest family size will be 11 (like in the example).
5. As the groups work, circulate to observe their strategies. (NOTE: Make sure that students
DO NOT write an X on the sticky note!!!) Have them post their chart paper on the wall when they finish.
6. When all groups have finished have each group explain how they arrived at their distribution.
PART II
1. Show students eight different distributions that have a mean of five.
2. Present the following questions for discussion:
1 – If the mean family size is the only information you are given, how does that limit what you know
about the distribution?
Possible responses:
If you only have the mean, it limits you to not knowing how many families are at each of the
numbers or how many people are in each family.
It limits what I know by not telling you what the other family numbers are.
2 – What additional information about the data could be given to identify which one of these eight
distributions is the one that matches the results from the survey we talked about earlier?
Possible responses:
We could be given how far above or below the mean the numbers are.
We could be told how many families are at each of the numbers.
3. Explain to students that these eight line plots demonstrate different variations for the distributions. Finding
a way to measure or summarize this variation is a major goal of statistics. One way to measure variation
is to look at how each number differs from the mean.
4. Ask students the following questions:
1 – Of all eight of these distributions, which one shows data values that differ the least from the mean
value, 5? Explain.
Obviously the first distribution differs the least.
2 – Of all eight of these distributions, which shows data values that differ the most from the mean value,
5? Explain.
Students should see that the last distribution has data values that differ the most from 5.
5. Think. Pair. Share. (You may want to give them a copy of the 8 distributions for this task).
Have students perform the following task individually. After seeing that most students have finished,
tell them to share and explain their results decision to a partner. Then have some students share and justify
their decision to the whole group. Try to get a class consensus on the order.
Task: On the basis of how different the data values appear to be from the mean of 5, how would
you order, from least to most, the other six distributions (distributions 2 – 7)? Make a list
of the order in which you think these distributions should be. Be prepared to explain
your reasoning.
Possible Order: 4, 7, 3, 5, 2, 6, but some variation may be appropriate.
PART III
1. Explain to students that they were very subjective in their ordering of the distributions because they were just
visually inspecting the line plots to make their decisions. In statistics, we quantify this difference from the
mean so that it is no longer a subjective decision.
One way to quantify is to calculate the difference between each data value and the mean. This is called the
deviation.
Show students the equation: Deviation from the mean = value – mean
(NOTE: Some students may actually have done this previously.)
2. Have each group of students get their chart paper back and calculate the deviation for each data value and
write it on the post it note. You might want to demonstrate this with one or two of the data values on the
chart papers before they get their chart paper down. (NOTE: Be sure to do an example where the deviation
is negative.) Have groups put their chart paper back on the wall when finished.
3. Using the distribution on one of the charts, ask students to find the sum of all the deviations in that
distribution. They should get 0.
Ask each group to pick a different distribution from the charts and find the sum of the deviations. They
should all be 0.
Explain that since the deviations always sum to zero, it it not very useful for quantifying the variation.
Instead we add the absolute deviations. The absolute deviation is the distance each value is from the mean.
It is just the absolute value of the deviation. Use Distribution 2 to demonstrate how to find the sum of the
absolute deviations. (You could just write the absolute deviation for each data value beside the X, and then
add.)
Have students add the absolute deviations for all eight distributions and order the distributions from the least
sum to the most. Compare the results to the previous class consensus.
Distribution
1
2
3
4
5
6
7
8
Sum of Absolute Deviations
0
26
22
16
22
24
20
36
Possible ordering: 1, 4, 7, 3, 5, 6, 2, 8
or
1, 4, 7, 5, 3, 6, 2, 8
NOTE: Since Distribution 3 and Distribution 5 have the same sum there are two possible orderings.
PART IV
1. Explain that with the previous 8 distributions the number of data values was the same, 9. Therefore adding
the absolute deviations was sufficient to quantify the variation.
Ask: How would it be different if we had distributions that had different numbers of data values?
Students should see that simply adding would not be sufficient because it would be possible for one distribution to have a
greater sum than another simply because there were more data values. Use the following example to illustrate:
Distribution I
X
2
3
X
4
5
6
7
8
9
10
11
Distribution II
2
3
X
X
X
X
X
X
X
X
4
5
6
7
8
9
10
11
In this distribution the mean is
5 and the sum of the absolute
deviations is 4.
In this distribution the
mean is still 5 but the sum
of the absolute deviations
is 5. Even though
visually there appears to
be less difference from
the mean, the sum of the
absolution deviations is
higher because there are
more data values.
Explain to students that since data sets often have different numbers of data values that we find the average
of the absolute deviations instead of just finding the sum. To find the average you simply divide the sum by
the number of data values. For example, in Distribution I, we would divide 4 by 2 to get 2, and for
Distribution II we divide 5 by 8 to get .625. This is called the Mean Absolute Deviation (MAD).
Since the MAD for Distribution II is less than the MAD for Distributions I, we know that Distribution II is
less different from the mean. “A small MAD indicates that the data values are similar to the mean. A large
MAD indicates that the data values are quite different from the mean.” (Kader, MTMS, Vol. 4, No. 6,
March, 1999).
Show the following equation:
Mean Absolute Deviation =
𝑆𝑢𝑚 𝑜𝑓 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛𝑠
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑉𝑎𝑙𝑢𝑒𝑠
2. Assign the following task for students to complete individually:
Task: The following line plots show distributions that all have the same mean. Calculate the mean and
then find the MAD for each distribution. Order the distributions from the least amount of
variation to the most.
PART IV, cont.
Distribution A
X
X
1
2
3
X
4
5
6
7
8
Distribution B
X
X
X
X
4
5
X
X
1
2
3
X
6
7
8
7
8
Distribution C
X
X
X
X
X
X
X
1
2
3
4
X
X
X
5
6
Solution:
The mean for each distribution is 4.
Distribution A: MAD = 4/3  1.33
Distribution B: MAD = 12/7  1.71
Distribution C: MAD = 12/10 = 1.20
From least variability to most the order is C, A, B.
Additional Notes for Teachers:
When describing a data set we look at the center and the spread. To describe the center we use a measure of
central tendency – mean, median, or mode. To describe the spread we use a measure of variability. The range
is a simple measure of variability. Other measures are the mean absolute deviation (MAD) and standard
deviation.
As explained by Kader in his MTMS article, statisticians do not commonly use the MAD in describing the
variation in data. They prefer to use standard deviation. However, understanding standard deviation may be
difficult for middle school students to understand. The MAD is a measure of variability that is accessible to
students at this level and can later be used as a starting point for understanding standard deviation.
Distribution 1
X
X
X
X
X
X
X
X
X
2
3
4
5
6
7
8
9
10
11
Distribution 2
X
X
X
X
X
X
2
3
4
5
X
X
X
3
4
5
6
7
8
X
X
X
X
X
X
X
X
X
2
3
4
5
6
7
X
X
X
X
X
X
2
3
4
5
6
X
X
8
9
10
X
X
9
10
11
8
9
10
11
X
X
X
7
8
9
X
X
X
X
7
8
X
X
X
X
7
8
7
X
11
Distribution 3
X
X
X
2
X
Distribution 4
Distribution 5
6
X
10
11
9
10
11
9
10
11
Distribution 6
X
X
X
X
X
2
3
4
5
6
Distribution 7
X
X
X
X
2
3
X
4
5
6
Distribution 8
X
X
X
X
X
X
2
X
X
X
3
4
5
6
7
8
9
10
11
Distribution I
X
2
3
X
4
5
6
X
X
X
X
X
X
X
X
4
5
6
7
8
9
10
11
7
8
9
10
11
Distribution II
2
3
Distribution A
X
X
1
2
X
3
4
5
6
7
8
Distribution B
X
X
1
2
3
X
X
X
X
4
5
X
6
7
8
7
8
Distribution C
X
X
X
X
X
X
X
1
2
3
4
X
X
X
5
6