How Far Apart? Overlap and Multiples of Variability | Grade 7 Math | LevelKite

Ten bean seedlings grew on a windowsill and ten grew under a lamp. After two weeks, the lamp plants look taller. But how different are the two groups, really?

You will use three ideas you already know:

A data distribution is the way a group's values are spread along a scale: which values occur and how often.
A center is one number that stands for a typical value in the group. Two centers are the mean (the sum of the values divided by the number of values, which is never $0$ for a group that has data) and the median (the middle of the ordered values; with an even count, the mean of the two middle values).
A measure of variability is one number that tells how spread out the values are.

In this lesson you will first read how much two distributions overlap, which means how much they share the same part of the scale. Then you will describe the gap between two centers by comparing it with the variability.

Seedling heights after two weeks

Group	Number of plants	Heights in order (cm)	Lowest (cm)	Highest (cm)
Window plants	10	3, 4, 4, 5, 5, 5, 7, 8, 9, 10	3	10
Lamp plants	10	6, 7, 8, 9, 11, 11, 11, 12, 12, 13	6	13

Line plot

Window plants: each mark is one plant. There is one mark at

3

cm, two at

4

cm, three at

5

cm, and one each at

7

,

8

,

9

and

10

cm.

A line plot shows one mark for each data value, stacked above that value on a number line.

The plot above shows the window plants. Its marks match the table: one mark at $3$ , two at $4$ , three at $5$ , none at $6$ , and one each at $7$ , $8$ , $9$ and $10$ . The plot below shows the lamp plants: one mark each at $6$ , $7$ , $8$ and $9$ , none at $10$ , three at $11$ , two at $12$ and one at $13$ .

Both plots use the same scale. It starts at $0$ and has one tick for each centimeter up to $14$ . So a tick value stands for the same height on both plots. That lets you compare the groups by reading tick values and counting marks. Do not judge by how the two pictures happen to line up on the page.

Line plot

Lamp plants: each mark is one plant. There is one mark each at

6

,

7

,

8

and

9

cm, three at

11

cm, two at

12

cm and one at

13

cm.

Now compare the two plots. What does it mean for two distributions to overlap?

Call the stretch of the scale from a group's lowest value to its highest value the group's span. Two distributions on the same scale overlap on the values that lie inside both spans. The overlap starts at the greater of the two lowest values and ends at the lesser of the two highest values. If that start is greater than that end, the distributions do not overlap at all.

To assess the overlap, count marks on each plot:

Marks inside the overlap show values that fall within both groups' spans.
Marks outside the overlap show values that are clearly separated from the other group: they are lower than all of its values or higher than all of its values.

The more marks sit inside the overlap, the more alike the two groups look. The more marks sit outside it, the more clearly separated they look.

Worked example

Assess the overlap of the two line plots

Use the two line plots. The window plants have marks from $3$ cm to $10$ cm, and the lamp plants have marks from $6$ cm to $13$ cm. How much do the two distributions of heights overlap, and what does that tell you about the plants?

Read each group's span, from its lowest mark to its highest mark, off the tick values. The window marks run from 3 to 10 and the lamp marks run from 6 to 13. Each span is 7 cm wide, so the two groups are spread out by about the same amount.
$\text{Window: } 10 - 3 = 7 \text{ cm} \qquad \text{Lamp: } 13 - 6 = 7 \text{ cm}$
Find the overlap, the stretch inside both spans. It starts at 6, the greater of the two lowest values, because any height below 6 is below every lamp height. It ends at 10, the lesser of the two highest values, because any height above 10 is above every window height. Since 6 is less than 10, the spans do overlap.
$\text{start} = 6 \qquad \text{end} = 10 \qquad 6 \lt 10$
Count the marks inside the overlap, tick by tick from 6 to 10, to see how much of each group lies within both spans. The window plot has no mark at 6 and one mark each at 7, 8, 9 and 10. The lamp plot has one mark each at 6, 7, 8 and 9 and no mark at 10.
$\text{Window: } 0+1+1+1+1 = 4 \qquad \text{Lamp: } 1+1+1+1+0 = 4$
Count the marks outside the overlap to see how much of each group is separated. Window marks below 6: one at 3, two at 4 and three at 5. Lamp marks above 10: three at 11, two at 12 and one at 13. Then check that 4 inside plus 6 outside gives all 10 plants in each group.
$\text{Window: } 1+2+3 = 6 \qquad \text{Lamp: } 3+2+1 = 6 \qquad 4+6 = 10$
Say what the counts mean for the plants. The 6 window plants outside the overlap are shorter than every lamp plant, and the 6 lamp plants outside it are taller than every window plant. Inside the overlap the groups mix: the 10 cm window plant is taller than the lamp plants at 6, 7, 8 and 9 cm.

Answer

The distributions overlap partly, from $6$ cm to $10$ cm. That stretch holds $4$ of the $10$ window plants and $4$ of the $10$ lamp plants, and the other $12$ plants are clearly separated. Lamp plants tend to be taller, but not every lamp plant is taller than every window plant.

The line plots show that lamp plants tend to be taller. A center puts a number on that. You will soon check that the two mean heights are $4$ cm apart. Is $4$ cm a big gap?

It depends on how much the heights inside each group vary. If the heights in each group are spread over many centimeters, a $4$ cm gap between the centers is small next to that spread. If the heights in each group stay close together, the same $4$ cm gap is large next to that spread.

So use the variability as a measuring stick. Ask: how many times does the variability fit into the gap between the centers? Then you can express the gap as a multiple of the variability: the gap equals some number times the variability, such as $2$ times or $1.5$ times. Here that number does not have to be a whole number.

Key concept

Express the gap as a multiple of the variability

Let $d$ be the difference between the two centers: the greater center minus the lesser center, so $d$ is never negative. Let $v$ be the measure of variability, in the same unit as $d$ . It can be a whole number, a fraction or a decimal, but it must not be $0$ . To express the gap as a multiple of the variability, find the number $k$ with $d = k \times v$ . Because $v$ is not $0$ , dividing both sides by $v$ keeps them equal and gives $k$ .

Use this when both groups have about the same variability, so one $v$ fits both. Since $d$ and $v$ share a unit, $k$ has no unit: you say the means are $k$ MADs apart. For the same kind of center and measure, a greater $k$ means the centers are farther apart compared with each group's spread, so the plots tend to overlap less.

k = \frac{d}{v} \quad (v \neq 0)

Worked example

The gap between the means, in MADs

Window heights (cm): 3, 4, 4, 5, 5, 5, 7, 8, 9, 10. Lamp heights (cm): 6, 7, 8, 9, 11, 11, 11, 12, 12, 13. Express the difference between the two mean heights as a multiple of the mean absolute deviation (MAD).

Find each mean: add the 10 heights, then divide by 10, the number of plants. Window: 3 + 4 + 4 + 5 + 5 + 5 + 7 + 8 + 9 + 10 = 60. Lamp: 6 + 7 + 8 + 9 + 11 + 11 + 11 + 12 + 12 + 13 = 100.
$\text{Window: } \frac{60}{10} = 6 \text{ cm} \qquad \text{Lamp: } \frac{100}{10} = 10 \text{ cm}$
Find each MAD: take the distance of every height from its group's mean, add the distances, then divide by 10. Window distances from 6 are 3, 2, 2, 1, 1, 1, 1, 2, 3, 4, which add to 20. Lamp distances from 10 are 4, 3, 2, 1, 1, 1, 1, 2, 2, 3, which also add to 20.
$\text{Window: } \frac{20}{10} = 2 \text{ cm} \qquad \text{Lamp: } \frac{20}{10} = 2 \text{ cm}$
Check the conditions. Both groups have a MAD of 2 cm, so one value of v fits both groups. That value is not 0, so you may divide by it.
$v = 2 \text{ cm}, \quad v \neq 0$
Find the gap d between the centers. Subtract the lesser mean from the greater mean so the difference is positive. Subtracting in the other order would give negative 4, which has the same size but the opposite sign.
$d = 10 - 6 = 4 \text{ cm}$
Find k, the number of MADs that fit into the gap. It must make k times 2 equal 4. Dividing both sides by 2 keeps them equal, and it is allowed because 2 is not 0. Centimeters divided by centimeters leaves a number with no unit.
$k \times 2 = 4 \quad \Rightarrow \quad k = \frac{4}{2} = 2$
Check by multiplying: 2 MADs of 2 cm each should rebuild the 4 cm gap, and they do.
$2 \times 2 \text{ cm} = 4 \text{ cm}$

Answer

The mean height of the lamp plants is $4$ cm greater than the mean height of the window plants. That difference is $2$ times the MAD of $2$ cm, so the means are $2$ MADs apart.

Worked example

The gap between the medians, in IQRs

Use the same heights. Window (cm): 3, 4, 4, 5, 5, 5, 7, 8, 9, 10. Lamp (cm): 6, 7, 8, 9, 11, 11, 11, 12, 12, 13. Express the difference between the two median heights as a multiple of the interquartile range (IQR).

Find the window median and quartiles. The 10 heights are in order, and 10 is an even count, so the median is the mean of the 5th and 6th values, 5 and 5. Splitting at the median gives the lower half 3, 4, 4, 5, 5 and the upper half 5, 7, 8, 9, 10. Q1 is the middle value of the lower half and Q3 is the middle value of the upper half.
$\text{median} = \frac{5+5}{2} = 5, \quad Q_1 = 4, \quad Q_3 = 8, \quad \text{IQR} = 8 - 4 = 4 \text{ cm}$
Do the same for the lamp heights. The 5th and 6th values are 11 and 11. The lower half is 6, 7, 8, 9, 11 and the upper half is 11, 11, 12, 12, 13.
$\text{median} = \frac{11+11}{2} = 11, \quad Q_1 = 8, \quad Q_3 = 12, \quad \text{IQR} = 12 - 8 = 4 \text{ cm}$
Check the conditions and find the gap d. Both IQRs are 4 cm, so one value of v fits both groups, and 4 is not 0, so you may divide by it. Subtract the lesser median from the greater median so the difference is positive.
$v = 4 \text{ cm}, \quad d = 11 - 5 = 6 \text{ cm}$
Divide the gap by the IQR to find k, the number of IQRs that fit into the gap. The quotient is not a whole number, and that is fine.
$k = \frac{d}{v} = \frac{6}{4} = 1.5$
Check by multiplying: 1.5 IQRs of 4 cm each should rebuild the 6 cm gap, and they do.
$1.5 \times 4 \text{ cm} = 6 \text{ cm}$

Answer

The median height of the lamp plants is $6$ cm greater than the median height of the window plants. That difference is $1.5$ times the IQR of $4$ cm, so the medians are $1.5$ IQRs apart.

Takeaway. Two tools work together when you compare two groups measured on the same scale.

Overlap: find the stretch of the scale inside both spans, then count the marks inside it and outside it.
Multiple: subtract the lesser center from the greater center, then divide by a measure of variability that is not $0$ and is about the same for both groups.

For the seedlings, the plots overlap from $6$ cm to $10$ cm with $8$ of the $20$ plants inside, the means are $2$ MADs apart, and the medians are $1.5$ IQRs apart. Both multiples are correct. They use different centers and different measures, so always name the ones you used.

Next, you will use overlap and these multiples to draw informal comparative inferences about two populations, and you will analyze errors in comparisons of two data distributions.