Math Core

Are seventh graders taller than sixth graders? Do plants grow faster with fertilizer? To answer questions like these, you compare two groups of data. In this lesson you'll compare two populations using their centers, their spreads, and how much their data overlap.

Center and spread

To describe a data set with a couple of numbers, you need a measure of center (a typical value) and a measure of spread (how much the values vary).

Mean: add the values and divide by how many there are.
Mean absolute deviation (MAD): the average distance of the values from the mean.
Median: the middle value when the data are in order.
Interquartile range (IQR): the distance between the first and third quartiles, the spread of the middle half of the data.

Definition

Mean absolute deviation

To find the MAD, find the mean, find how far each value is from the mean (always a positive distance), and then take the mean of those distances.

Use the mean and MAD when the data are fairly symmetric. Use the median and IQR when the data are lopsided or have outliers, since a few extreme values can pull the mean a long way.

Comparing with dot plots

Two science classes each timed how many seconds $10$ students took to solve a puzzle.

Class A: seconds to solve the puzzle

Class B: seconds to solve the puzzle

The two dot plots have the same shape, but Class B's is shifted to the right. There is a little overlap between $6$ and $7$ seconds, but most Class B students took longer than most Class A students.

Worked example: Finding the means and MADs

Find the mean and MAD for each class above.

Class A: the data are $3, 4, 4, 5, 5, 5, 5, 6, 6, 7$ .

\text{mean} = \dfrac{3 + 4 + 4 + 5 + 5 + 5 + 5 + 6 + 6 + 7}{10} = \dfrac{50}{10} = 5.

The distances from $5$ are $2, 1, 1, 0, 0, 0, 0, 1, 1, 2$ , which add to $8$ . So the MAD is $\dfrac{8}{10} = 0.8$ .

Class B: the data are $6, 7, 7, 8, 8, 8, 8, 9, 9, 10$ .

\text{mean} = \dfrac{80}{10} = 8.

The distances from $8$ are $2, 1, 1, 0, 0, 0, 0, 1, 1, 2$ , which also add to $8$ . So the MAD is $0.8$ .

The means differ by $8 - 5 = 3$ seconds, while each class varies by only $0.8$ seconds on average.

Measuring the difference in MADs

A difference of $3$ seconds is large here because the data are tightly bunched. If each class had values scattered from $0$ to $20$ seconds, the same $3$ -second gap would hardly be noticeable. So you judge the gap between centers compared to the spread.

Difference in means as a multiple of the MAD

When two data sets have similar spreads, divide the difference in means by the MAD:

\dfrac{\text{difference in means}}{\text{MAD}}

This tells you how many MADs apart the centers are. As a rule of thumb, a result of about $2$ or more means the two groups are noticeably different with little overlap. A result below $1$ means there is a lot of overlap.

For the puzzle data, $\dfrac{3}{0.8} = 3.75$ . The means are almost $4$ MADs apart, so Class B really did tend to take longer.

Worked example: Using medians

The times, in minutes, for $9$ bus rides on each of two routes are listed below.

Route 1: $10, 12, 14, 15, 15, 16, 18, 20, 22$
Route 2: $18, 20, 21, 23, 25, 26, 28, 30, 31$

Compare the typical ride times using medians.

Solution. Each list is already in order and has $9$ values, so the median is the $5$ th value. Route 1's median is $15$ minutes and Route 2's median is $25$ minutes. A typical ride on Route 2 takes about $25 - 15 = 10$ minutes longer. The two data sets overlap only between $18$ and $22$ minutes.

Comparing populations with samples

Usually you can't measure every member of two populations. Instead, you take a random sample from each and compare the samples. Because samples vary, you should only claim a difference between the populations when the samples show a clear gap.

Worked example: Sleep in two grades

A researcher took a random sample of $50$ seventh graders and a random sample of $50$ eleventh graders and asked how many hours they slept last night.

sample	mean	MAD
seventh graders	$8.6$ hours	$0.6$ hours
eleventh graders	$7.4$ hours	$0.6$ hours

Is it reasonable to conclude that seventh graders in general sleep more than eleventh graders?

Solution. The difference in means is $8.6 - 7.4 = 1.2$ hours. Compared to the MAD:

\dfrac{1.2}{0.6} = 2.

The sample means are $2$ MADs apart, which is a noticeable difference. Since the samples were random, it's reasonable to infer that seventh graders tend to sleep more than eleventh graders. (That doesn't mean every seventh grader sleeps more than every eleventh grader; the groups still overlap.)

Common mistake

Don't compare centers alone. A difference of $3$ points could be huge or tiny depending on the spread. Always look at the spread too, either with a dot plot or by dividing the difference in means by the MAD.

Tip

A quick picture check: draw both dot plots on the same scale, one above the other. If you can slide one onto the other only by moving it a long way compared to its width, the groups are clearly different.