The median-median line - a robust alternative method of fitting a line

The least-squares line is the most common way of fitting a line to data. However, there is an undesirable characteristic of the least-squares method when there are outliers that are far away from the general pattern in the scatterplot. Here we demonstrate this problem and consider an alternative method of fitting a line that is less sensitive to outliers in the data.

Let's return to the dataset that contained two test scores for 27 students in a college class. (The dataset is shown below.) When we plot the Test 1 score against the Test 2 score, we see a positive association and we're interested in using a line to predict a student's Test 2 score from his Test 1 score. We fit a least-squares line that has the equation

Test 2 score = 29.44 + 0.614 Test 1 score.

Test1	Test2	Test1	Test2	Test1	Test2
82	81	95	96	78	78
79	89	98	91	89	75
92	96	87	73	85	72
96	91	80	75	75	76
72	70	76	73	56	69
86	81	72	73	87	91
66	73	94	88	51	63
87	63	64	77	68	64
85	92	65	60	85	85

Let's introduce an outlier in the data. There is one student who received 51 on Test 1 and 63 on Test 2 - his scores are represented by the plotting point on the left side of the figure. Suppose that this student really got a 96 on Test 2 - with this change, the point on the left side is moved in the direction of the arrow. This student is really an outlier since his point is far away from the general group of points in the scatterplot.

Does the introduction of this outlier have any effect on our least-squares line? We recompute the line to be

Test 2 score = 51.22 + 0.357 Test 1 score

and graph this new line on the below scatterplot. Note that the least-squares fit has substantially changed - the single outlier seems to have flattened the least-squares line and the new line no longer seems to be a good fit to the general pattern of points. The problem with the least-squares line is similar to the problem of using a mean as an average of a batch of data in the presence of outliers. A least-squares line, like a mean, can be very sensitive to a few unusual data values that deviate from the main body of data.

A median-median line

There is another way of fitting a good line, the median-median line that is less sensitive to outliers. As the name suggests, this line is based on finding medians, rather than means, in regions of the scatterplot.

To compute this line, we first sort the data by the explanatory variable (here, Test 1) and divide the data into three equal-size groups (or nearly equal-size groups) by the ordered values of this variable. In this example, we have 27 points and we divide the data into the

Left group - the 9 leftmost points
Middle group - the 9 points in the center of the scatterplot
Right group - the 9 rightmost points

The three groups of points are shown in the below table. Also in the below figure, we divide the points in the scatterplot by vertical lines, showing the left, middle, and right groups.

	Left Group		Middle Group		Right Group

	Test1	Test2	Test1	Test2	Test1	Test2
	51	63	76	73	87	63
	56	69	78	78	87	73
	64	77	79	89	87	91
	65	60	80	75	89	75
	66	73	82	81	92	96
	68	64	85	92	94	88
	72	70	85	72	95	96
	72	73	85	85	96	91
	75	76	86	81	98	91

Summary Point	66	70	82	81	92	91

For each group, we find a summary point that is the

(median of explanatory variable, median of response variable).

To illustrate, for the Left Group, the median of the explanatory variable Test 1, is

median {51, 56, 64, 65, 66, 68, 72, 72, 75} = 66

and the median of the response variable Test 2 is

median {63, 69, 77, 60, 73, 64, 70, 73, 76} = 70.

So the Left summary point is (66, 70). In a similar fashion, we find the summary points for the Middle and Right groups. We denote these points by (xL, yL), (xM, yM) , and (xR, yR). They are shown in the table and plotted in the figure.

D7.graph11.png

Now we can compute the median-median line:

1. The slope of the line is found by finding the slope between the Left and Right summary points.

Here the left and right summary points are (66, 70) and (92, 91) and so the slope is

2. Using each point, we can solve for the intercept by the usual formula. The intercept of the median-median line is the average of these three intercepts:

In this example, the intercept is computed to be

Summing up, our median-median line is

Test 2 score = 16.05 + 0.81 x (Test 1 score).

This line is displayed in the below scatterplot.

What was the point of introducing this new way of fitting a line? Remember that the least-squares line could be distorted by the introduction of a single outlying point. Suppose, as above, that we change the Test 2 grade of that one student from 63 to 96. The median-median line for this new dataset is

Test 2 score = 26.3 + 0.69 x (Test 1 score).

We've plotted the original median-median line and the new one in the below figure. Note that the median-median line has changed, but this line is much less affected by the outlier than the least-squares line. To say it a little differently, the median-median is more robust or insensitive to outliers than is the least-squares line. In practice, both lines will give similar fits with no outliers present, but the lines can give very different results when there are outliers in the data.

PRACTICE: the median-median line

The table below gives the average gestation period (in days) and the average longevity (in years) for twelve animal species.

Animal	Gestation	Longevity	Summary points
Mouse	21	3
Kangaroo	36	7
Guinea pig	68	4
Lion	100	15
Beaver	105	5
Sheep	154	12
Baboon	187	20
Cow	284	15
Ass	365	12
Camel	406	12
Giraffe	457	10
Elephant	660	35

A scatterplot of the two variables is shown below. If we are interested in predicting an animal's longevity based on its gestation period, a least-squares fit is given by

LONGEVITY = .0318 GESTATION + 5.0.

This best line is drawn on the graph.

D7.graph14.png

1. The dataset has been arranged by increasing values of the gestation period. Divide the dataset into three groups and find the summary point for each group.

2. Using the three summary points, find the equation of the median-median line.

3. Graph the median-median line on the scatterplot.

4. Are the least-squares and median-median lines similar in this example? If they are different, describe any special features of this dataset that might cause the two fits to be different.

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