Outliers, Linear Regression and Correlation | Statistics

Learning objectives

In some data sets, there are values (observed data points) called outliers. Outliers are observed data points that are far from the least-squares line. They have large errors, where the error or residual is not very close to the best-fit line.

Outliers need to be examined closely. Sometimes, they should not be included in the analysis of the data, like if it is possible that an outlier is a result of incorrect data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier.

Besides outliers, a sample may contain one or a few points that are called influential points. Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and determine whether the slope of the regression line is changed significantly.

You also want to examine how the correlation coefficient, r, has changed. Sometimes, it is difficult to discern a significant change in slope, so you need to look at how the strength of the linear relationship has changed. Computers and many calculators can be used to identify outliers and influential points. Regression analysis can determine if an outlier is, indeed, an influential point. The new regression will show how omitting the outlier will affect the correlation among the variables, as well as the fit of the line. A graph showing both regression lines helps determine how removing an outlier affects the fit of the model.

Identifying Outliers

We could guess at outliers by looking at a graph of the scatter plot and best-fit line. However, we would like some guideline regarding how far away a point needs to be to be considered an outlier. As a rough rule of thumb, we can flag as an outlier any point that is located farther than two standard deviations above or below the best-fit line. The standard deviation used is the standard deviation of the residuals or errors.

We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points outside this extra pair of lines are flagged as potential outliers. Or, we can do this numerically by calculating each residual and comparing it with twice the standard deviation. With regard to the TI-83, 83+, or 84+ calculators, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods.

In the third exam/final exam example, you can determine whether there is an outlier. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE (sum of the squared errors) should be smaller and the correlation coefficient ought to be closer to 1 or –1.

Graphical Identification of Outliers

With the TI-83, 83+, or 84+ graphing calculators, it is easy to identify the outliers 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 were equal to 2s or more, 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. Let’s 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 these 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),
where ŷ = –173.5 + 4.83x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit.

Graph the scatter plot with the best-fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the Y= equation editor. Press ZOOM-9 to get a good view. You will see that the only point that is not between Y2 and Y3 is the point (65, 175). On the calculator screen, it is barely outside these lines, but it is considered an outlier because it is more than two standard deviations away from the best-fit line. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam.

Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell whether 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 clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers.

View original figure (external; image not reproduced in this text edition)

Original figure description: The scatter plot of exam scores with a line of best fit.Two yellow dashed lines run parallel to the line of best fit. The dashed lines run above and below the best fit line at equal distances. One data point falls outside the boundary created by the dashed lines—it is an outlier.

Numerical Identification of Outliers

In element-4662, the first two columns include the third exam and final exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83x. The residuals, or errors, that were mentioned in Section 3 of this chapter have been calculated in the fourth column of the table: Observed y value predicted y value = y ŷ.

s is the standard deviation of all the y ŷ = ε values, where n is the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as

s = \sqrt{\frac{S S E}{n - 2}} .

Rather than calculate the value of s ourselves, we can find s using a computer or calculator. For this example, the calculator function 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	ŷ	y – ŷ
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 with the residuals in column four 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 point (65, 175) as an outlier. Recall that recalculation of the least-squares regression line and summary statistics, following deletion of an outlier, may be used to determine whether an outlier is also an influential point. This process also allows you to compare the strength of the correlation of the variables and possible changes in the slope both before and after the omission of any outliers.

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

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

$ŷ = - 355.19 + 7.39 x$ and $r = 0.9121$ .

The slope is now 7.39, compared to the previous slope of 4.83. This seems significant, but we need to look at the change in r-values as well. The new line shows $r = 0.9121$ , which indicates a stronger correlation than the original line, with $r = 0.6631,$ since $r = 0.9121$ is closer to 1. This means the new line is a better fit to the data values. The line can better predict the final exam score given the third exam score. It also means the outlier of (65, 175) was an influential point, since there is a sizeable difference in r-values. We must now decide whether to delete the outlier. If the outlier was recorded erroneously, it should certainly be deleted. Because it produces such a profound effect on the correlation, the new line of best fit allows for better prediction and an overall stronger model.

You may use Excel to graph the two least-squares regression lines and compare the slopes and fit of the lines to the data, as shown in figure_12_06_01.

View original figure (external; image not reproduced in this text edition)

Original figure description: This shows two scatter plots, each with a line of best fit. Each is labeled final exam score on the y-axis and third exam score on the x-axis. The part (a) scatter plot includes the complete data set and has points plotted at (65, 175), (66, 126), (67, 133), (67, 153), (69, 151), (69, 159), (70, 163), (71, 159), (71, 163), (71, 185), and (75, 198). The part (a) scatter plot is labeled y = 4.8274x 173.51 and R2 = 0.43969. The part (b) scatter plot is missing the student 1 data point and has points plotted at (66, 126), (67, 133), (67, 153), (69, 151), (69, 159), (70, 163), (71, 159), (71, 163), (71, 185), and (75, 198). The part (b) scatter plot is labeled y = 7.3878x 355.19 and R2 = 0.8319.

You can see that the second graph shows less deviation from the line of best fit. It is clear that omission of the influential point produced a line of best fit that more closely models the data.

Numerical Identification of Outliers: Calculating s and Finding Outliers Manually

If you do not have the function LinRegTTest on your calculator, then you must calculate the outlier in the first example by doing the following.

First, square each |y – ŷ|.

The squares are

35²
17²
16²
6²
19²
9²
3²
1²
10²
9²
1².

Then, add (sum) all the |y – ŷ| squared terms using the formula

$\underset{i = 1}{Σ} 11 {(| y_{i} - ŷ_{i} |)}^{2} = \underset{i = 1}{Σ} 11 ε_{i}^{2}$ (Recall that y_i – ŷ_i = ε_i).

= 35² + 17² + 16² + 6² + 19² + 9² + 3² + 1² + 10² + 9² + 1²

= 2,440 = SSE.

The result, SSE, is the sum of squared errors.

Next, calculate s, the standard deviation of all the y – ŷ = ε-values where n = the total number of data points.

The calculation is $s = \sqrt{\frac{SSE}{n - 2}}$ .

For the third exam/final exam example, $s = \sqrt{\frac{2440}{11 - 2}} = 16.47.$

Next, multiply s by 2:
(2)(16.47) = 32.94
32.94 is two standard deviations away from the mean of the y – ŷ values.

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 is at least 2s, then we would consider the data point to be too far from the line of best fit. We call that point a potential outlier.

For the example, if any of the |y – ŷ| values are at least 32.94, the corresponding (x, y) data point is a potential outlier.

For the third exam/final exam example, all the |y – ŷ| values are less than 31.29 except for the first one, which is 35:

35 > 31.29. That is, |y – ŷ| ≥ (2)(s).

The point that corresponds to |y – ŷ| = 35 is (65, 175). Therefore, the data point (65, 175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.)

The next step is to compute a new best-fit line using the 10 remaining points. The new line of best fit and the correlation coefficient are

ŷ = –355.19 + 7.39x and r = .9121.

Using this new line of best fit (based on the remaining 10 data points in the third exam/final exam example), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line?

Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam.

The original line predicted that ŷ = –173.51 + 4.83(73) = 179.08, so the prediction using the new line with the outlier eliminated differs from the original prediction.

The consumer price index (CPI) measures the average change over time in prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the nation’s economy to government, businesses, and labor forces, the CPI helps them make economic decisions. The president, U.S. Congress, and the Federal Reserve Board use CPI trends to form monetary and fiscal policies. In the following table, x is the year and y is the CPI.

x	y	x	y
1915	10.1	1969	36.7
1926	17.7	1975	49.3
1935	13.7	1979	72.6
1940	14.7	1980	82.4
1947	24.1	1986	109.6
1952	26.5	1991	130.7
1964	31.0	1999	166.6

Draw a scatter plot of the data.
Calculate the least-squares line. Write the equation in the form ŷ = a + bx.
Draw the line on a scatter plot.
Find the correlation coefficient. Is it significant?
What is the average CPI for the year 1990?

See eip-idp35936320.
Using our calculator, ŷ = 3204 + 1.662x is the equation of the line of best fit.
See eip-idp35936320.
r = 0.8694. The number of data points is n = 14. Use the 95 Percent Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12: In this case, df = 12. The corresponding critical values from the table are ±0.532. Since 0.8694 > 0.532, r is significant. We can use the predicted regression line we found above to make the prediction for x = 1990.
ŷ = 3204 + 1.662(1990) = 103.4 CPI.

View original figure (external; image not reproduced in this text edition)

Original figure description: Scatter plot and line of best fit of the consumer price index data, on the y-axis, and year data, on the x-axis.

If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website (ftp://ftp.bls.gov/pub/special.requests/cpi/cpiai.txt). Our data are taken from the column Annual Avg. (third column from the right). For example, you could add more current years of data. Try adding the more recent years: 2004, CPI = 188.9; 2008, CPI = 215.3; and 2011, CPI = 224.9. See how this affects the model. (Check: ŷ = –4436 + 2.295x; r = 0.9018. Is r significant? Is the fit better with the addition of the new points?)

Try It

The following table shows economic development measured in per capita income (PCINC).

Year	PCINC	Year	PCINC
1870	340	1920	1,050
1880	499	1930	1,170
1890	592	1940	1,364
1900	757	1950	1,836
1910	927	1960	2,132

What are the independent and dependent variables?
Draw a scatter plot.
Use regression to find the line of best fit and the correlation coefficient.
Interpret the significance of the correlation coefficient.
Is there a linear relationship between the variables?
Find the coefficient of determination and interpret it.
What is the slope of the regression equation? What does it mean?
Use the line of best fit to estimate PCINC for 1900 and for 2000.
Determine whether there are any outliers.

a. The independent variable (x) is the year and the dependent variable (y) is the per capita income.

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph is shown. The y-axis is labeled PCINC and the x-axis is labeled year. The y-axis has tick marks at 0, 500, 1000, 1500, 2000, and 2500. The x-axis has tick marks at 1860, 1880, 1900, 1920, 1940, 1960, and 1980.

c. ŷ = 18.61x – 34574; r = 0.9732

d. At df = 8, the critical value is 0.632. The r-value is significant because it is greater than the critical value.

e. There does appear to be a linear relationship between the variables.

f. The coefficient of determination is 0.947, which means that 94.7% of the variation in PCINC is explained by the variation in the years.

g. and h. The slope of the regression equation is 18.61, and it means that per capita income increases by $18.61 for each passing year. ŷ = 785 when the year is 1900, and ŷ = 2,646 when the year is 2000.

i. There do not appear to be any outliers.

95 Percent Critical Values of the Sample Correlation Coefficient Table

Degrees of Freedom: n – 2	Critical Values: + and –
1	0.997
2	0.950
3	0.878
4	0.811
5	0.754
6	0.707
7	0.666
8	0.632
9	0.602
10	0.576
11	0.555
12	0.532
13	0.514
14	0.497
15	0.482
16	0.468
17	0.456
18	0.444
19	0.433
20	0.423
21	0.413
22	0.404
23	0.396
24	0.388
25	0.381
26	0.374
27	0.367
28	0.361
29	0.355
30	0.349
40	0.304
50	0.273
60	0.250
70	0.232
80	0.217
90	0.205
100	0.195

References

Committee on Ways and Means, U.S. House of Representatives. (n.d.). Washington, DC: U.S. Department of Health and Human Services.

Microsoft Bookshelf. (n.d.).

Physicians Desk Reference Staff. (1990). Physicians desk reference. Ohio: Medical Economics Company.

U.S. Bureau of Labor Statistics. (n.d.). Retrieved from https://www.bls.gov/

Chapter Review

To determine whether a point is an outlier, do one of the following:

Input the following equations into the TI 83, 83+, 84, or 84+ calculator:

$\begin{matrix} y_{1} = a + b x \\ y_{2} = a + b x + 2 s \\ y_{3} = a + b x - 2 s \end{matrix}$
where s is the standard deviation of the residuals.
If any point is above y₂ or below y₃, then the point is considered to be an outlier.
Use the residuals and compare their absolute values to 2s, where s is the standard deviation of the residuals. If the absolute value of any residual is greater than or equal to 2s, then the corresponding point is an outlier.
Note: The calculator function LinRegTTest (STATS, TESTS, LinRegTTest) calculates s.

Marcus states that all outliers are influential points. Is he correct? Explain.

No, he is not correct. An outlier is only an influential point if it significantly impacts the slope of the least-squares regression line and the correlation coefficient, r. If omission of this data point from the calculation of the regression line does not show much impact on the slope or r-value, then the outlier is not considered an influential point. For different reasons, it still may be determined that the data point must be omitted from the data set.

Use the following information to answer the next four exercises. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69.

View original figure (external; image not reproduced in this text edition)

Original figure description: A graph is shown. The y-axis has tick marks at 0, 20, 40, 60, 80, and 100. The x axis has tick marks at 0, 2, 4, 6, 8, 10, 12, 14. The line goes up steadily from 68 and ends around 100.

Do there appear to be any outliers?

Yes. There appears to be an outlier at (6, 58).

A point is removed and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why?

What effect did the potential outlier have on the line of best fit?

The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate as a predictor for the data.

Are you more or less confident in the predictive ability of the new line of best fit?

The sum of squared errors (SSE) for a data set of 18 numbers is 49. What is the standard deviation?

s = 1.75

The standard deviation for the SSE for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier?

Homework

Given the information in Table 12.30, which represents the relationship between final exam math grades and final exam history grades, decide whether point (56, 95) is an influential point. Explain how you arrived at your decision. Show all work.

x (final exam math grades)	y (final exam history grades)
54	60
56	68
77	82
74	78
63	69
51	55
88	97
72	77
69	78
56	95

Using LinRegTTest, the output for the original least-squares regression line is $y = 26.14 + 0.7539 x$ and $r = 0.6657 .$

The output for the new least-squares regression line, after omitting the outlier of (56, 95), is $ŷ = 6.36 + 1.0045 x$ and $r = 0.9757 .$

The slope of the new line is quite a bit different from the slope of the original least-squares regression line, but the larger change is shown in the r-values, such that the new line has an r-value that has increased to a value that is almost equal to one.

Thus, it may be stated that the outlier (56, 95) is also an influential point.

In Table 12.31, the height (sidewalk to roof) of notable tall buildings in America is compared with the number of stories of the building (beginning at street level).

Height (in feet)	Stories
1,050	57
428	28
362	26
529	40
790	60
401	22
380	38
1,454	110
1,127	100
700	46

Using stories as the independent variable and height as the dependent variable, make a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
Find the estimated heights for a building that has 32 stories and for a building that has 94 stories.
Based on the data in id1aa0827143, is there a linear relationship between the number of stories in tall buildings and the height of the buildings?
Are there any outliers in the data? If so, which point(s)?
What is the estimated height of a building with six stories? Does the least-squares line give an accurate estimate of height? Explain why or why not.
Based on the least-squares line, adding an extra story is predicted to add about how many feet to a building?
What is the slope of the least-squares (best-fit) line? Interpret the slope.

Ornithologists (scientists who study birds) tag sparrow hawks in 13 different colonies to study their population. They gather data for the percentage of new sparrow hawks in each colony and the percentage of those that have returned from migration.

Percent return: 74, 66, 81, 52, 73, 62, 52, 45, 62, 46, 60, 46, 38
Percent new: 5, 6, 8, 11, 12, 15, 16, 17, 18, 18, 19, 20, 20

Enter the data into a calculator and make a scatter plot.
Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from Part A.
Explain what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70 percent of the adults from the previous year have returned. What is the prediction?

a. and b. Check student solution.

c. The slope of the regression line is 0.3031 with a y-intercept of 31.93. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 32 percent new sparrow hawks, which doesn’t make sense, because if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 30.3 percent.

d. If we examine r₂, we see that only 57.52 percent of the variation in the percentage of new birds is explained by the model and the correlation coefficient, r = .7584 only indicates a somewhat strong correlation between returning and new percentages.

e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66 percent, our observed new percentage, 6 percent, is almost 6 percent less than the predicted new value of 11.98 percent. If we remove this data pair, we see only an adjusted slope of 0.2789 and an adjusted intercept of 30.9816. In other words, although these data generate the largest residual, it is not an outlier, nor is the data pair an influential point.

f. If there are 70 percent returning birds, we would expect to see y = 0.2789(70) + 30.9816 = 0.114 or 11.4 percent new birds in the colony.

The following table shows data on average per capita coffee consumption and death rate from heart disease in a random sample of 10 countries.

Yearly Coffee Consumption (liters)	2.5	3.9	2.9	2.4	2.9	0.8	9.1	2.7	0.8	0.7
No. of Deaths from Heart Disease	221	167	131	191	220	297	71	172	211	300

Enter the data into a calculator and make a scatter plot.
Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from Part A.
Explain what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.
Do the data provide convincing evidence that there is a linear relationship between the amount of coffee consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.

The following table consists of one student athlete’s time (in minutes) to swim 2,000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days.


Swim Time	Heart Rate
34.12	144
35.72	152
34.72	124
34.05	140
34.13	152
35.73	146
36.17	128
35.57	136
35.37	144
35.57	148

Enter the data into a calculator and make a scatter plot.
Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from Part A.
Explain what the slope and y-intercept of the regression line tell us.
How well does the regression line fit the data? Explain your response.
Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain.

Check student solution.
Check student solution.
We have a slope of –1.4946 with a y-intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate decreases by 1.5 beats per minute. If the student is not swimming at all, the y-intercept indicates that his heart rate will be 193.88 beats per minute. Although the slope has meaning (the longer it takes to swim 2000 m, the less effort the heart puts out), the y-intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low.
Because only 1.5 percent of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship.
Point (34.72, 124) generates the largest residual: –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes 2.953, with the y-intercept changing to 247.1616. Although the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y-intercept becomes more meaningful.

A researcher is investigating whether population impacts homicide rate. He uses demographic data from Detroit, Michigan, to compare homicide rates and the population.


Population Size	Homicide Rate per 100,000 People
558,724	8.6
538,584	8.9
519,171	8.52
500,457	8.89
482,418	13.07
465,029	14.57
448,267	21.36
432,109	28.03
416,533	31.49
401,518	37.39
387,046	46.26
373,095	47.24
359,647	52.33

Use a calculator to construct a scatter plot of the data. What is the independent variable? Why?
Use the calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot.
Discuss what the following mean in context:
1. The slope of the regression equation
2. The y-intercept of the regression equation
3. The correlation coefficient, r
4. The coefficient of determination, r²
Do the data provide convincing evidence that there is a linear relationship between population size and homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question.

School	Mid-Career Salary (in thousands of U.S. dollars)	Yearly Tuition (in U.S. dollars)
Princeton	137	28,540
Harvey Mudd	135	40,133
CalTech	127	39,900
U.S. Naval Academy	122	0
West Point	120	0
MIT	118	42,050
Lehigh University	118	43,220
NYU-Poly	117	39,565
Babson College	117	40,400
Stanford	114	54,506

Use the data in the Table 12.35 to determine the linear regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer.

If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = –0.0009x + 160, with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set.

Bring It Together

The average number of people in a family who attended college for various years is given in id1034sdfa0215.

Year	No. of Family Members Attending College
1969	4.0
1973	3.6
1975	3.2
1979	3.0
1983	3.0
1988	3.0
1991	2.9

Using year as the independent variable and number of family members attending college as the dependent variable, draw a scatter plot of the data.
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Does the y-intercept, a, have any meaning here?
Find the correlation coefficient. Is it significant?
Pick two years between 1969 and 1991 and find the estimated number of family members attending college.
Based on the data in id1034sdfa0215, is there a linear relationship between the year and the average number of family members attending college?
Using the least-squares line, estimate the number of family members attending college for 1960 and 1995. Does the least-squares line give an accurate estimate for those years? Explain why or why not.
Are there any outliers in the data?
What is the estimated average number of family members attending college for 1986? Does the least-squares line give an accurate estimate for that year? Explain why or why not.
What is the slope of the least-squares (best-fit) line? Interpret the slope.

c. No. The y-intercept would occur at year 0, which doesn’t exist.

The percent of female wage and salary workers who are paid hourly rates is given in id8522228278 for the years 1979 to 1992.

Year	Percent of Workers Paid Hourly Rates
1979	61.2
1980	60.7
1981	61.3
1982	61.3
1983	61.8
1984	61.7
1985	61.8
1986	62.0
1987	62.7
1990	62.8
1992	62.9

Using year as the independent variable and percent of workers paid hourly rates as the dependent variable, draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Does the y-intercept, a, have any meaning here?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
Find the estimated percentages for 1991 and 1988.
Based on the data, is there a linear relationship between the year and the percentage of female wage and salary earners who are paid hourly rates?
Are there any outliers in the data?
What is the estimated percentage for the year 2050? Does the least-squares line give an accurate estimate for that year? Explain why or why not.
What is the slope of the least-squares (best-fit) line? Interpret the slope.

Check student's solution.
Yes.
No, the y-intercept would occur at year 0, which doesnt exist.
ŷ = −266.8863 + 0.1656x.
0.9448, yes.
62.8233, 62.3265.
Yes.
No, (1987, 62.7).
72.5937, no.
Slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656.

Use the following information to answer the next two exercises. The cost of a leading liquid laundry detergent in different sizes is given in id10105430.

Size (ounces)	Cost ($)	Cost per Ounce
16	3.99
32	4.99
64	5.99
200	10.99

Using size as the independent variable and cost as the dependent variable, draw a scatter plot.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
If the laundry detergent were sold in a 40 oz. size, what is the estimated cost?
If the laundry detergent were sold in a 90 oz. size, what is the estimated cost?
Does it appear that a line is the best way to fit the data? Why or why not?
Are there any outliers in the given data?
Is the least-squares line valid for predicting what a 300 oz. size of the laundry detergent would cost? Why or why not?
What is the slope of the least-squares (best-fit) line? Interpret the slope.

Complete id10105430 for the cost per ounce of the different sizes of laundry detergent.
Using size as the independent variable and cost per ounce as the dependent variable, draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
If the laundry detergent were sold in a 40 oz. size, what is the estimated cost per ounce?
If the laundry detergent were sold in a 90 oz. size, what is the estimated cost per ounce?
Does it appear that a line is the best way to fit the data? Why or why not?
Are there any outliers in the the data?
Is the least-squares line valid for predicting what a 300 oz. size of the laundry detergent would cost per ounce? Why or why not?
What is the slope of the least-squares (best-fit) line? Interpret the slope.

Size (ounces) Cost ($) Cost per ounce

16 3.99 24.94

32 4.99 15.59

64 5.99 9.36

200 10.99 5.50
Check student solution.
There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size.
ŷ = 20.2368 – 0.0819x
r = –.8086
40-oz: 16.96 cents/oz
90-oz: 12.87 cents/oz
The relationship is not linear; the least-squares line is not appropriate.
There are no outliers.
No. You would be extrapolating. The 300-oz size is outside the range of x.
X = –0.08194. For each additional ounce in size, the cost per ounce decreases by 0.082 cents.

Size (ounces)	Cost ($)	Cost per ounce
16	3.99	24.94
32	4.99	15.59
64	5.99	9.36
200	10.99	5.50

According to a flyer published by Prudential Insurance Company, the costs of approximate probate fees and taxes for selected net taxable estates are as follows:

Net Taxable Estate ($)	Approximate Probate Fees and Taxes ($)
600,000	30,000
750,000	92,500
1,000,000	203,000
1,500,000	438,000
2,000,000	688,000
2,500,000	1,037,000
3,000,000	1,350,000

Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
Find the estimated total cost for a net taxable estate of $1,000,000. Find the cost for $2,500,000.
Does it appear that a line is the best way to fit the data? Why or why not?
Are there any outliers in the data?
Based on these results, what would be the probate fees and taxes for an estate that does not have any assets?
What is the slope of the least-squares (best-fit) line? Interpret the slope.

The following are advertised sale prices of color televisions at Anderson’s:

Size (inches)	Sale Price ($)
9	147
20	197
27	297
31	447
35	1,177
40	2,177
60	2,497

Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
Find the estimated sale price for a 32-inch television. Find the cost for a 50-inch television.
Does it appear that a line is the best way to fit the data? Why or why not?
Are there any outliers in the data?
What is the slope of the least-squares (best-fit) line? Interpret the slope.

Size is x, the independent variable, and price is y, the dependent variable.
Check student solution.
The relationship does not appear to be linear.
ŷ = –745.252 + 54.75569x.
r = .8944 and yes, it is significant.
32-inch: $1006.93, 50-inch: $1992.53.
No, the relationship does not appear to be linear. However, r is significant.
No, the 60-inch TV.
For each additional inch, the price increases by $54.76.

id797832aawwee6 shows the average heights for American boys in 1990.

Age (years)	Height (centimeters)
Birth	50.8
2	83.8
3	91.4
5	106.6
7	119.3
10	137.1
14	157.5

Decide which variable should be the independent variable and which should be the dependent variable.
Draw a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. Is it significant?
Find the estimated average height for a 1-year-old. Find the estimated average height for an 11-year-old.
Does it appear that a line is the best way to fit the data? Why or why not?
Are there any outliers in the data?
Use the least-squares line to estimate the average height for a 62-year-old man. Do you think that your answer is reasonable? Why or why not?
What is the slope of the least-squares (best-fit) line? Interpret the slope.

State	No. of Letters in Name	Year Entered the Union	Rank for Entering the Union	Area (square miles)
Alabama	7	1819	22	52,423
Colorado	8	1876	38	104,100
Hawaii	6	1959	50	10,932
Iowa	4	1846	29	56,276
Maryland	8	1788	7	12,407
Missouri	8	1821	24	69,709
New Jersey	9	1787	3	8,722
Ohio	4	1803	17	44,828
South Carolina	13	1788	8	32,008
Utah	4	1896	45	84,904
Wisconsin	9	1848	30	65,499

We are interested in whether there is a relationship between the ranking of a state and the area of the state.

What are the independent and dependent variables?
What do you think the scatter plot will look like? Make a scatter plot of the data.
Does it appear from inspection that there is a relationship between the variables? Why or why not?
Calculate the least-squares line. Put the equation in the form ŷ = a + bx.
Find the correlation coefficient. What does it imply about the significance of the relationship?
Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas?
Use the two points in Part F to plot the least-squares line on your graph from Part B.
Does it appear that a line is the best way to fit the data? Why or why not?
Are there any outliers?
Use the least-squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not?
Delete Hawaii and substitute Alaska for it. Alaska is a state with an area of 656,424 square miles.
Calculate the new least-squares line.
Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case?
Do you think that, in general, newer states are larger than the original states?

Rank is the independent variable and area is the dependent variable.
Check student solution.
There appears to be a linear relationship, with one outlier.
ŷ (area) = 24177.06 + 1010.478x
r = .50047. r is not significant, so there is no relationship between the variables.
Alabama: 46,407.576 square miles, Colorado: 62,575.224 square miles.
The Alabama estimate is closer than the Colorado estimate.
If the outlier is removed, there is a linear relationship.
There is one outlier (Hawaii).
rank 51: 75,711.4 square miles, no.


Alabama	7	1819	22	52,423
Colorado	8	1876	38	104,100
Hawaii	6	1959	50	10,932
Iowa	4	1846	29	56,276
Maryland	8	1788	7	12,407
Missouri	8	1821	24	69,709
New Jersey	9	1787	3	8,722
Ohio	4	1803	17	44,828
South Carolina	13	1788	8	32,008
Utah	4	1896	45	84,904
Wisconsin	9	1848	30	65,499

ŷ = –87065.3 + 7828.532x.
Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier.
Yes, with the exception of Hawaii.

Glossary

outlier: an observation that does not fit the rest of the data

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