12.8 Outliers (modified r. bloom)  (Page 3/2)

Graphical identification of outliers

With the TI-83,83+,84+ graphing calculators, it is easy to identify the outlier graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance was equal to $2s$ or farther, then we would consider the data point to be "too far" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3:

As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s=16.412

Line Y2=-173.5+4.83x-2(16.4) and line Y3=-173.5+4.83X+2(16.4)

Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x=65, y=175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than 2 standard deviations away from the best fit line.

Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph more clear. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers.

Numerical identification of outliers

In the table below, the first two columns are the third exam and final exam data. The third column shows the predicted $\hat{y}$ values calculated from the line of best fit: $\hat{y}$ =-173.5+4.83x. The residuals, or errors, have been calculated in the fourth column of the table: $\text{observed y value}-\text{predicted y value}=y-\hat{y}$ .

$s$ is the standard deviation of all the $y-\hat{y}=\epsilon$ values where $n$ = the total number of data points. If each residual is calculated and squared, and the results are added up, we get the SSE. The standard deviation of the residuals is calculated from the SSE as:

$s=\sqrt{\frac{\text{SSE}}{n-2}}$

Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For this example, our calculator LinRegTTest found s=16.4 as the standard deviation of the residuals:

• 35
• -17
• 16
• -6
• -19
• 9
• 3
• -1
• -10
• -9
• -1

.

$x$	$y$	$\hat{y}$	$y-\hat{y}$
65	175	140	$175-140=35$
67	133	150	$133-150=-17$
71	185	169	$185-169=16$
71	163	169	$163-169=-6$
66	126	145	$126-145=-19$
75	198	189	$198-189=9$
67	153	150	$153-150=3$
70	163	164	$163-164=-1$
71	159	169	$159-169=-10$
69	151	160	$151-160=-9$
69	159	160	$159-160=-1$

We are looking for all data points for which the residual is greater than 2s=2(16.4)=32.8 or less than -32.8. Compare these values to the residuals in column 4 of the table. The only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35.

How does the outlier affect the best fit line?

Numerically and graphically, we have identified the point (65,175) as an outlier. We should re-examine the data for this point to see if there are any problems with the data. If there is an error we should fix the error if possible, or delete the data. If the data is correct, we would leave it in the data set. For this problem, we will suppose that we examined the data and found that this outlier data was an error. Therefore we will continue on to delete the outlier, so that we can explore how it affects the results, as a learning experience.

Compute a new best-fit line and correlation coefficient using the 10 remaining points:

On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, the new line of best fit and the correlation coefficient are:

$\hat{y}=-355.19+\text{7.39x}$ and $r=0.9121$

The new line with $r=0.9121$ is a stronger correlation than the original ( $r$ =0.6631) because $r=0.9121$ is closer to 1. This means that the new line is a better fit to the 10 remaining data values. The line can better predict the final exam score given the third exam score.

• Scatter plot and line of best fit.
• $\hat{y}=-3204+\text{1.662x}$ is the equation of the line of best fit.
• $r=0.8694$
• The number of data points is $n=14$ . Use the 95% Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12. $n-2=12$ . The corresponding critical value is 0.532. Since $0.8694>0.532$ , $r$ is significant.
• $\hat{y}=-3204+1.662\left(1990\right)=103.4$ CPI
• Using the calculator LinRegTTest, we find that s = 25.4 ; graphing the lines Y2=-3204+1.662X-2(25.4) and Y3=-3204+1.662X+2(25.4) shows that no data values are outside those lines, identifying no outliers. (Note that the year 1999 was very close to the upper line, but still inside it.)