Math 7 · 7.SP.B.3

Comparing two populations: the gap between means in MADs

The means of two groups differ by 4 grams. Is that a big difference or a small one?

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Comparing two distributionsMath 7 Mean absolute deviationMath 6

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Comparing two populations: the gap between means in MADs

Why this exists

Two teams do a standing long jump, and one team’s mean jump is 20 cm longer. Is that a lot?

The 20 cm alone cannot say. If every jumper lands within a few centimeters of the team’s mean, a 20 cm gap splits the teams cleanly apart. If each team’s jumps scatter across a meter, the same 20 cm is lost in the scatter. And if two tomato varieties differ by 4 g in mean mass, is that a bigger difference than the 20 cm? Centimeters and grams cannot be compared at all.

One fix answers both. Measure the gap between the means with a ruler made from the data itself: the spread. This lesson uses the mean absolute deviation as the ruler and turns “the means differ by 20 cm” into “the means differ by 2.5 MADs”, a plain number that says how much the two groups overlap. It builds on comparing two distributions, where center and spread were described side by side.

Definition

Two measures from earlier lessons do the work, recalled here only.

The Mean of a data set is the sum of its values divided by how many values there are.
The mean absolute deviation, or \text{MAD}, is the mean of the distances from each value to the mean: how far a typical value sits from the middle.

Take two groups whose MADs are about the same. The gap between the means in MADs is the distance between the two means divided by a MAD:

\text{gap} = \frac{\text{larger mean} - \text{smaller mean}}{\text{MAD}}

The top is the larger mean minus the smaller, so the gap is never negative; say in words which group has the larger mean. When the two MADs are equal, divide by that MAD. When they are close but not equal, divide by either one and say which you used: the two answers are close, so the choice does not change what the number says.

When the MADs are clearly different, one group’s ruler does not fit the other, and the gap in MADs is not a fair measure. Describe the two groups’ centers and spreads instead.

The gap has no units: centimeters divided by centimeters is a plain number. That is what lets a gap in jumps be set beside a gap in tomatoes.

Intuition

Figure 1 shows the long jumps of two teams of ten on one scale. Sigrid’s team has mean 150 cm, Ulrich’s team has mean 170 cm, and on each team a typical jump is 8 cm from the team’s mean: both MADs are 8 cm.

Figure 1Two teams’ standing long jumps. The triangles mark the means. The ruler underneath measures the gap between the means in steps of one MAD.

The ruler under the plots is the whole idea. It starts at one mean and lays down steps of one MAD toward the other. Two full steps reach 166, and half a step more reaches 170: the gap is 2.5 MADs.

The halfway point between the means, 160 cm, is 1.25 MADs from each mean, further out than a typical jump of either team. So most of each team stays on its own side of 160, and the plots overlap only at their edges.

Shrink the gap, or widen the spreads, until the means are less than one MAD apart. Now the halfway point sits well inside both clusters and the two plots lie almost on top of each other, as in Figure 2.

Figure 2Two varieties of tomato. The means differ by 4 g, less than one MAD.

The bands below are guides for describing what the dot plots show, not sharp cut-offs.

Gap in MADs	What the two dot plots look like
less than 1	heavy overlap; the shift is hard to see
from 1 to 2	a visible shift, with a good deal of overlap
2 or more	the groups sit mostly apart; overlap only at the edges

Results

No units, and a change of units leaves it alone. In millimeters instead of centimeters, the distance between the means and the MAD are both ten times larger, and the gap is the same.

Few values lie far out. At most half the values of a data set lie 2 or more MADs from its mean. The MAD really is a typical distance, which is what makes it a fair ruler.

Samples. When the two groups are random samples from two larger populations, a common guide is that a gap of 2 or more MADs between the sample means suggests the populations differ. A gap under 1 shows neither that they differ nor that they are the same.

Why it is true

Units. Call the distance between the means d and the MAD M. Changing units multiplies every value in both groups by the same positive number c (c = 10 for centimeters to millimeters). Each mean is then c times as large, and so is d. Every distance from a value to its mean is c times as large, so M is too. The factor cancels:

\frac{c \times d}{c \times M} = \frac{d}{M}

Few values far out. Take Sigrid’s team: ten jumps with MAD 8 cm. The MAD is the mean of the ten distances from the mean, so those distances add to 10 \times 8 = 80 cm. A jump 2 MADs or more from the mean is at least 16 cm away, and the total of 80 has room for at most 80 \div 16 = 5 such distances: half the jumps. The same count works for any data set. In fact only two jumps, 134 and 166, are that far out; the bands in the table come from looking at dot plots of ordinary single-humped data, not from a proof.

Samples. Two random samples from one population have different means: sampling variability. But a mean averages out the high and low values in its sample, so sample means wobble much less than single values do. With samples that are not tiny, two sample means from one population usually land within 2 MADs of each other, so a gap of 2 or more makes a real difference the likelier explanation. This is a guide from experience, not a guarantee.

Worked examples

Examples 1 to 3 go with the first day, where the MADs are given. Examples 4 to 7 go with the second: raw data, several groups, spreads that differ, and samples.

Example 1

Two teams on the long jump

Figure 1 shows the standing long jumps of Sigrid’s team and Ulrich’s team. Sigrid’s team has mean 150 cm and Ulrich’s team has mean 170 cm, and both teams have a MAD of 8 cm. Find the gap between the means in MADs, and describe how much the teams overlap.

Solution

The MADs are equal, so one MAD serves as the ruler for both. The distance between the means is 170 - 150 = 20 cm, and in MADs it is

\frac{20}{8} = 2.5

That is in the band “2 or more”, so the teams should sit mostly apart, and the dot plots agree. Only three of Sigrid’s team’s jumps (158, 162 and 166) reach Ulrich’s team’s shortest, 156 cm, and only four of Ulrich’s team’s (156, 160, 162 and 164) are no longer than Sigrid’s team’s longest, 166 cm. The overlap is the strip from 156 to 166 cm, at the edge of each team.

Ulrich’s team jumps further on the whole, by 2.5 MADs. That is not the same as every jumper on Ulrich’s team beating every jumper on Sigrid’s: 166 beats 156.

Example 2

Two varieties of tomato

Figure 2 shows the masses of ten tomatoes of each of two varieties. Variety A has mean 112 g and Variety B has mean 116 g, and both have a MAD of 6 g. Find the gap between the means in MADs, describe the overlap, and compare with Example 1.

Solution

The MADs are equal, so divide the distance between the means by 6:

\frac{116 - 112}{6} = \frac{4}{6} = \frac{2}{3} \approx 0.67

That is under 1: heavy overlap. Variety A runs from 100 to 124 g, and nine of Variety B’s ten tomatoes fall inside that stretch. Variety B is heavier on average, but by less than a typical tomato’s distance from its own variety’s mean.

The raw differences, 20 cm and 4 g, cannot be compared. The gaps in MADs can: 2.5 is far more than 0.67, so the teams are much more clearly separated than the varieties.

Example 3

Headlines in two newspapers

Vivianne counts the words in the front-page headlines of two newspapers over a month. Newspaper P has mean 9 words with MAD 1.5 words. Newspaper Q has mean 12 words with MAD 1.4 words. Find the gap between the means in MADs.

Solution

The MADs, 1.5 and 1.4 words, are about the same, so either can be the ruler. Using Newspaper P’s MAD of 1.5:

\frac{12 - 9}{1.5} = \frac{3}{1.5} = 2

Newspaper Q’s MAD would give \displaystyle \frac{3}{1.4} \approx 2.1, almost the same, which is why either is allowed. Newspaper Q’s headlines are longer on average, by about 2 MADs, so the two papers’ headline lengths overlap little.

Example 4

Arm spans, from the raw data

Thaddeus-Lee measures the arm spans of six 11-year-olds and six 14-year-olds, in centimeters:

Age group	Arm spans (cm)
11-year-olds	141, 143, 145, 147, 149, 151
14-year-olds	152, 156, 157, 158, 161, 164

Find each group’s mean and MAD, then the gap between the means in MADs. Figure 3 shows the data.

Figure 3Arm spans of two age groups. The gap between the means is four MADs, and the plots do not overlap at all.

Solution

The means. Add the values and divide by 6:

\frac{141 + 143 + 145 + 147 + 149 + 151}{6} = \frac{876}{6} = 146

\frac{152 + 156 + 157 + 158 + 161 + 164}{6} = \frac{948}{6} = 158

The MADs. Find each value’s distance from its own group’s mean, then take the mean of those distances. A distance is never negative: 141 is 5 from 146. The 11-year-olds’ distances from 146 are 5, 3, 1, 1, 3, 5, and the 14-year-olds’ distances from 158 are 6, 2, 1, 0, 3, 6.

\frac{5 + 3 + 1 + 1 + 3 + 5}{6} = \frac{18}{6} = 3

\frac{6 + 2 + 1 + 0 + 3 + 6}{6} = \frac{18}{6} = 3

Both MADs are 3 cm.

The gap.

\frac{158 - 146}{3} = \frac{12}{3} = 4

The means are 4 MADs apart, far into the top band. Figure 3 shows no overlap at all: the longest arm span among the 11-year-olds, 151 cm, is shorter than the shortest among the 14-year-olds, 152 cm.

Example 5

Three classes and one puzzle

Three classes time how long their students take to solve the same puzzle. The results, in minutes:

Class	Mean	MAD
A	14	2
B	15	2.5
C	19	2.5

Find the gap in MADs for each pair of classes, and say which classes differ clearly.

Solution

The three MADs are about the same, so one ruler can serve every pair: 2.5 minutes, the MAD of Classes B and C. For A and C, for instance, the distance is 19 - 14 = 5, and \displaystyle \frac{5}{2.5} = 2.

Pair	Distance	Gap in MADs
A and B	1	0.4
A and C	5	2
B and C	4	1.6

Classes A and B are 0.4 MADs apart: their times overlap heavily. Class C was slower than both. It differs clearly from Class A, at 2 MADs, and less sharply from Class B, at 1.6 MADs, where the shift is visible but the times still overlap a good deal.

Example 6

Eggs from two farms

Wystan weighs ten eggs from each of two farms, in grams:

Farm	Egg masses (g)
Farm 1	56, 57, 57, 57, 58, 58, 59, 59, 59, 60
Farm 2	49, 52, 55, 58, 61, 61, 64, 67, 70, 73

Find each farm’s mean and MAD, and decide whether the gap in MADs is a fair summary. Figure 4 shows the data.

Figure 4Eggs from two farms. The spreads are very different, so one MAD cannot serve as the ruler for both.

Solution

Farm 1. The total is 580, so the mean is \displaystyle \frac{580}{10} = 58. The distances from 58 are 2, 1, 1, 1, 0, 0, 1, 1, 1, 2, which add to 10, so the MAD is 1 g.

Farm 2. The total is 610, so the mean is \displaystyle \frac{610}{10} = 61. The distances from 61 are 12, 9, 6, 3, 0, 0, 3, 6, 9, 12, which add to 60, so the MAD is 6 g.

Is it fair? The MADs, 1 g and 6 g, are clearly different, so no single ruler fits both farms. The distance between the means, 61 - 58 = 3 g, would be

\frac{3}{6} = 0.5 \qquad\qquad \frac{3}{1} = 3

MADs, heavy overlap or clear separation, depending only on which farm’s MAD was picked. The ratio is not a fair measure here.

Describe instead. Farm 2’s eggs are 3 g heavier on average, but the striking difference is spread: Farm 1’s eggs all lie between 56 and 60 g, while Farm 2’s vary from 49 to 73 g, with Farm 1’s whole cluster inside that range. This is the method of comparing two distributions.

Example 7

Samples of two new tomato varieties

A grower has large harvests of two new tomato varieties, C and D. Xiomara picks a random sample of 10 tomatoes from each harvest. The sample from C has mean 131 g and MAD 5 g. The sample from D has mean 119 g and MAD 6 g. Does this suggest the two harvests differ in mean mass?

Solution

The sample means are statistics; the harvests’ means are parameters nobody has measured. The MADs, 5 g and 6 g, are about the same. Using the larger, 6 g:

\frac{131 - 119}{6} = \frac{12}{6} = 2

Another pair of random samples would give different means, but two random samples of 10 from one harvest usually land closer than 2 MADs apart. So the samples suggest that variety C’s harvest really is heavier on average. They do not prove it: an unusual pair of samples can always happen.

Had the sample means been 121 g and 119 g, the gap would be \displaystyle \frac{2}{6}, about 0.33 MADs: the size ordinary sampling variability produces. That would show neither that the harvests differ nor that they are the same.

Boundaries

The spreads must be about the same. If one MAD is clearly larger than the other, as with the eggs in Example 6, the gap in MADs depends on which MAD is chosen and is not a fair measure. Describe the centers and the spreads separately instead.

A MAD of 0 cannot be a ruler. If every value in a group is the same, its MAD is 0, and division by 0 has no meaning.

The mean and MAD must suit the data. One extreme value drags the mean and inflates the MAD, and a strongly lopsided group has no clear middle for the mean to mark. Compare medians and interquartile ranges instead, as in comparing box plots.

The bands are guides. Two pairs with the same gap can overlap by different amounts if their shapes differ, and nothing special happens at exactly 1 or exactly 2.

A gap of 0 does not mean the groups are alike. It compares centers only.

Samples must be random, and not tiny. A sample picked for convenience can differ from its population for reasons that have nothing to do with chance (see convenience samples). Samples of two or three values vary so much that even a large gap between them is weak evidence.

Where this goes next

The gap in MADs is the first of a family of measurements with one shape: a difference divided by a typical amount of variation. The standard deviation later replaces the MAD as the usual ruler, and standardizing applies the same idea to a single value. For samples, the sampling distribution of a difference of means says how far apart two sample means typically land, and the two-sample t-test divides the gap by exactly that.

Exercises

Exercises 1 to 12 go with the first day (Examples 1 to 3), and Exercises 13 to 25 with the second (Examples 4 to 7). When two MADs are about the same but not equal, the answers divide by the larger; the other MAD gives a close number.

Day 1: finding the gap in MADs

Two groups have means 40 and 52, and both have MAD 4. Find the gap between the means in MADs, and describe the overlap.
The headlines in two magazines have mean lengths 7.5 and 8.5 words, and both have MAD 2 words. Find the gap in MADs and describe the overlap.
Eggs from two farms have mean masses 63 g and 70 g, and both have MAD 3.5 g. Find the gap in MADs.
Two classes solve a puzzle. Their mean times are 18 and 20 minutes, with MADs 1.6 and 1.5 minutes. Are the MADs about the same? Find the gap in MADs.
Two varieties of tomato have mean masses 120 g and 121.5 g, and both have MAD 6 g. Find the gap in MADs and say what the dot plots would look like.
A third team joins the long jump of Example 1. Its mean is 162 cm and its MAD is 8 cm. Find the gap in MADs between the new team and Sigrid’s team, and between the new team and Ulrich’s.

Day 1: working backwards and comparing

Two groups with the same MAD, 6 g, have means 2.5 MADs apart. How many grams apart are the means?
Two groups’ means are 24 cm apart, which is 1.5 MADs. What is the MAD?
Pair P has means 30 and 32 with MAD 2.5. Pair Q has means 30 and 39 with MAD 3. Find each gap in MADs, and say which pair’s dot plots overlap more.

Day 1: reasoning

Two pairs of groups both have means 20 cm apart. In the first pair both MADs are 4 cm; in the second both are 25 cm. Explain why the same 20 cm says different things about the two pairs.
Explain why the gap in MADs has no units, and why that matters when comparing a gap in centimeters with a gap in grams.
Ulrich says: “Our means are 2.5 MADs apart, so everyone on my team jumped further than everyone on Sigrid’s team.” Use Figure 1 to say whether he is right.

Day 2: from raw data

Two classes time a puzzle, in minutes. Class X: 10, 12, 13, 15, 20. Class Y: 15, 19, 20, 22, 24. Find each mean and MAD, then the gap in MADs.
Arm spans in centimeters. Group P: 150, 152, 155, 157, 161. Group Q: 154, 157, 159, 160, 165. Find each mean and MAD, then the gap in MADs.
Egg masses in grams. Farm R: 58, 61, 63, 66. Farm S: 51, 54, 56, 59. Find each mean and MAD, then the gap in MADs.
Headline lengths in words. Newspaper M: 6, 7, 7, 8, 9, 11. Newspaper N: 7, 9, 10, 10, 11, 13. Find each mean and MAD, then the gap in MADs.

Day 2: several groups, and when the method fails

Three teams do the standing long jump. Their means are 148, 156 and 168 cm, with MADs 8, 8 and 10 cm. Find the gap in MADs for each pair, and say which pair is most clearly separated.
Two classes time a puzzle. Class 1 has mean 30 minutes and MAD 1 minute. Class 2 has mean 34 minutes and MAD 8 minutes. Is a gap in MADs a fair summary? If not, describe the two classes instead.
A farm’s eggs have mean 52 g and MAD 3 g. A second farm’s eggs have the same MAD and a larger mean. What is the smallest mean the second farm could have for the gap to be 2 MADs or more?

Day 2: samples

Random samples of 15 headlines from each of two newspapers have mean lengths 9.2 and 12.4 words, with MADs 1.6 and 1.4 words. Find the gap in MADs. Does it suggest the newspapers’ headlines differ in mean length?
Random samples of 10 tomatoes from each of two harvests have means 120 g and 123 g, with MADs 6 g and 5 g. Find the gap in MADs. Does this show the two harvests have the same mean mass?

Day 2: reasoning

Use the eggs of Example 6 to explain why the gap in MADs needs the two spreads to be about the same.
Two different pairs of random samples, taken from the same two populations, give gaps of 2.4 MADs and 1.6 MADs. Explain how both can happen.
Every jump in Example 1 is converted from centimeters to millimeters. Find the new means, the new MAD and the new gap in MADs, and explain why the gap did not change.
Can two groups have a gap of 0 MADs and still look very different on a dot plot? Explain, with an example.

Answers

3 MADs; the groups sit mostly apart. \displaystyle \frac{52 - 40}{4} = \frac{12}{4} = 3.
0.5 MADs; heavy overlap. \displaystyle \frac{8.5 - 7.5}{2} = \frac{1}{2}, less than 1.
2 MADs. \displaystyle \frac{70 - 63}{3.5} = \frac{7}{3.5} = 2, so the eggs overlap little.
Yes; 1.25 MADs. The distance is 2, and \displaystyle \frac{2}{1.6} = 1.25 (about 1.33 with the other MAD): a visible shift with a good deal of overlap.
0.25 MADs. \displaystyle \frac{121.5 - 120}{6} = \frac{1.5}{6} = 0.25. The plots would lie almost on top of each other.
1.5 MADs from Sigrid’s team and 1 MAD from Ulrich’s. \displaystyle \frac{162 - 150}{8} = 1.5 and \displaystyle \frac{170 - 162}{8} = 1.
15 g. The distance is the gap times the MAD: 2.5 \times 6 = 15.
16 cm. The MAD is \displaystyle \frac{24}{1.5} = 16.
P: 0.8 MADs; Q: 3 MADs; pair P overlaps more. \displaystyle \frac{32 - 30}{2.5} = 0.8 and \displaystyle \frac{39 - 30}{3} = 3.
With MAD 4 cm the gap is \displaystyle \frac{20}{4} = 5 MADs: the groups are cleanly separated. With MAD 25 cm it is \displaystyle \frac{20}{25} = 0.8 MADs: values stray further than the gap, so the groups overlap heavily.
The top and bottom of the fraction are in the same unit, so the unit cancels. A gap of 20 cm and one of 4 g cannot be compared, but 2.5 MADs and 0.67 MADs can, as in Example 2.
No. The gap describes the teams as groups. In Figure 1, Sigrid’s team’s 166 cm beats Ulrich’s team’s 156, 160, 162 and 164 cm.
X: mean 14, MAD 2.8. Y: mean 20, MAD 2.4. Gap about 2.14 MADs. Distances from 14: 4, 2, 1, 1, 6, total 14; from 20: 5, 1, 0, 2, 4, total 12. Then \displaystyle \frac{6}{2.8} \approx 2.14: Class Y was clearly slower.
P: mean 155, MAD 3.2. Q: mean 159, MAD 2.8. Gap 1.25 MADs. Distances 5, 3, 0, 2, 6, total 16, and 5, 2, 0, 1, 6, total 14. Then \displaystyle \frac{4}{3.2} = 1.25: a visible shift, much overlap.
R: mean 62, MAD 2.5. S: mean 55, MAD 2.5. Gap 2.8 MADs. In both farms the distances from the mean are 4, 1, 1, 4, total 10. Then \displaystyle \frac{7}{2.5} = 2.8: Farm R’s eggs are clearly heavier.
M: mean 8, MAD \displaystyle \frac{4}{3}. N: mean 10, MAD \displaystyle \frac{4}{3}. Gap 1.5 MADs. Distances 2, 1, 1, 0, 1, 3 and 3, 1, 0, 0, 1, 3 each total 8, so each MAD is \displaystyle \frac{8}{6}, and \displaystyle 2 \div \frac{4}{3} = 1.5.
First and second: 1 MAD. First and third: 2 MADs. Second and third: 1.2 MADs. The first and third are most clearly separated. \displaystyle \frac{8}{8} = 1, \displaystyle \frac{20}{10} = 2 and \displaystyle \frac{12}{10} = 1.2, using the larger MAD where the two differ.
No. The MADs, 1 and 8 minutes, are clearly different: the 4 minutes between the means would be \displaystyle \frac{4}{8} = 0.5 or \displaystyle \frac{4}{1} = 4 MADs depending on the choice. Class 2 took 4 minutes longer on average, but the bigger difference is spread: Class 1’s times were typically 1 minute from the mean, Class 2’s typically 8 minutes.
58 g. The distance must be at least 2 \times 3 = 6 g, so the mean must be at least 52 + 6 = 58.
2 MADs; yes. \displaystyle \frac{12.4 - 9.2}{1.6} = \frac{3.2}{1.6} = 2. Random samples this far apart suggest the second newspaper’s headlines are longer on average.
0.5 MADs; no. \displaystyle \frac{123 - 120}{6} = 0.5. A gap this small is what sampling variability produces, so the samples cannot tell whether the harvests differ, and they do not show the means are equal either.
The distance between the means is 3 g: 0.5 MADs with Farm 2’s MAD, heavy overlap, but 3 MADs with Farm 1’s, clear separation. The number would depend on an arbitrary choice.
Each random sample catches different members of its population, so the sample means and MADs, and with them the gap, change from one pair of samples to the next: sampling variability.
Means 1500 mm and 1700 mm, MAD 80 mm, gap 2.5 MADs. Every value is multiplied by 10, so the means, their distance and the MAD are too: \displaystyle \frac{200}{80} = 2.5.
Yes. The data 9, 10, 11 and 0, 10, 20 both have mean 10, so the gap is 0 whichever MAD is used, but the first is tightly bunched and the second widely spread.