1. Describing Bivariate Numerical Data
Chapter 4
Describing Bivariate Numerical Data
Created by Kathy Fritz

2. Forensic scientists must often estimate the age of an unidentified crime victim. Prior to 2010, this was usually done by analyzing teeth and bones, and the resulting estimates were not very reliable. A study described in the paper “Estimating Human Age from T-Cell DNA Rearrangements” (Current Biology [2010]) examined the
This line can be used to estimate the age of a crime victim from a blood test.
relationship between age and a measure based on a blood test.
Age and the blood test measure were recorded for 195 people ranging in age from a few weeks to 80 years. A scatterplot of the data appears to the right.
Do you think there is a relationship? If so, what kind? If not, why not?

3. Correlation
Pearson’s Sample Correlation Coefficient
Properties of r

4. Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
Yes

5. Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
Yes

6. Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
No, looks curved

7. Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
No, looks parabolic

8. Does it look like there is a relationship between the two variables?
If so, is the relationship linear? No

9. Linear relationships can be either positive or negative in direction.
Are these linear relationships positive or negative?
Negative
Positive

10. When the points in a scatterplot tend to cluster tightly around a line, the relationship is described as strong.
Try to order the scatterplots from strongest relationship to the weakest.
These four scatterplots were constructed using data from graphs in Archives of General Psychiatry (June 2010).
A, C, B, D A B C D

11. Pearson’s Sample Correlation Coefficient
Usually referred to as just the correlation coefficient
Denoted by r
Measures the strength and direction of a linear relationship between two numerical variables
The strongest values of the correlation coefficient are r = +1 and r = -1.
The weakest value of the correlation coefficient is r = 0.
An important definition!

12. Properties of r
The sign of r matches the direction of the linear relationship.
r is positive
r is negative

13. Properties of r
The value of r is always greater than or equal to -1 and less than or equal to +1.
Strong correlation
Moderate correlation
Weak correlation

14. Properties of r
3. r = 1 only when all the points in the scatterplot fall on a straight line that slopes upward. Similarly, r = -1 when all the points fall on a downward sloping line.

15. Properties of r
4. r is a measure of the extent to which x and y are linearly related
Find the correlation for these points:
Compute the correlation coefficient?
Sketch the scatterplot.
Does this mean that there is NO relationship between these points? x 2 4 6 8 10 12 14 y 40 20
r = 0
r = 0, but the data set has a definite relationship!

16. Properties of r
The value of r does not depend on the unit of measurement for either variable.
Calculate r for the data set of mares’ weight and the weight of their foals.
Mare Weight (in Kg)
Foal Weight (in Kg)
556
129.0
638
119.0
588
132.0
550
123.5
580
112.0
642
113.5
568
95.0
104.0
616
93.5
549
108.5
504
515
117.5
551
128.0
594
127.5
Mare Weight (in lbs)
Foal Weight (in Kg)
1223.2
129.0
1403.6
119.0
1293.6
132.0
1210.0
123.5
1276.0
112.0
1412.4
113.5
1249.6
95.0
104.0
1355.2
93.5
1207.8
108.5
1108.8
1111.0
117.5
1212.2
127.5
1306.8
r =
r =
Change the mare weights to pounds by multiply Kg by 2.2 and calculate r.

17. Calculating Correlation Coefficient
The correlation coefficient is calculated using the following formula:
where
and

18. Here is the scatterplot: Does the relationship appear linear? Explain.
The web site (The Education Trust) publishes data on U.S colleges and universities. The following six-year graduation rates and student-related expenditures per full-time student for 2007 were reported for the seven primarily undergraduate public universities in California with enrollments between 10,000 and 20,000.
Here is the scatterplot:
Does the relationship appear linear?
Explain.
Expenditures
8810
7780
8112
8149
8477
7342
7984
Graduation rates
66.1
52.4
48.9
48.1
42.0
38.3
31.3

19. College Expenditures Continued:
To compute the correlation coefficient, first find the z-scores. x y zx zy
zxzy
8810
66.1
1.52
1.74
2.64
7780
52.4
-0.66
0.51
-0.34
8112
48.9
0.04
0.19
0.01
8149
48.1
0.12
8477
42.0
0.81
-0.42
7342
38.3
-1.59
-0.76
1.21
7984
31.3
-0.23
-1.38
0.32
To interpret the correlation coefficient, use the definition –
There is a positive, moderate linear relationship between six-year graduation rates and student-related expenditures.
Use the formula to compute the correlation coefficient by using the lists in the graphing calculator.
Show students how to calculate r by using the linear regression command in the calculator.

20. Will the sum of zxzy be positive or negative?
How the Correlation Coefficient Measures the Strength of a Linear Relationship
zx is negative
zy is positive
zxzy is negative
zx is positive
zy is positive
zxzy is positive
zx is negative
zy is negative
zxzy is positive
Will the sum of zxzy be positive or negative?

21. Will the sum of zxzy be positive or negative?
How the Correlation Coefficient Measures the Strength of a Linear Relationship
zx is negative
zy is positive
zxzy is negative
zx is positive
zy is positive
zxzy is positive
zx is negative
zy is negative
zxzy is positive
zx is negative
zy is positive
zxzy is negative
Will the sum of zxzy be positive or negative?

22. Will the sum of zxzy be positive or negative or zero?
How the Correlation Coefficient Measures the Strength of a Linear Relationship
Will the sum of zxzy be positive or negative or zero?

23. These variables are both strongly related to the age of the child
Association does NOT imply causation.
Does a value of r close to 1 or -1 mean that a change in one variable causes a change in the other variable?
Consider the following examples:
The relationship between the number of cavities in a child’s teeth and the size of his or her vocabulary is strong and positive.
Consumption of hot chocolate is negatively correlated with crime rate.
Causality can only be shown by carefully controlling values of all variables that might be related to the ones under study. In other words, with a well-controlled, well-designed experiment.
Should we all drink more hot chocolate to lower the crime rate?
Both are responses to cold weather
Discuss why these variables are related See page 207.
So does this mean I should feed children more candy to increase their vocabulary?
These variables are both strongly related to the age of the child

24. Linear Regression
Least Squares Regression Line

25. Suppose there is a relationship between two numerical variables.
Let x be the amount spent on advertising and y be the amount of sales for the product during a given period.
You might want to predict product sales (y) for a month when the amount spent on advertising is $10,000 (x).
The other variable, denoted by x, is the predictor variable (sometimes called independent or explanatory variable).
The letter y is used to denoted the variable you want to predict, called the response variable (or dependent variable).

26. The equation of a line is:
Where:
b – is the slope of the line
it is the amount by which y increases when x increases by 1 unit
a – is the intercept (also called y-intercept or vertical intercept)
it is the height of the line above x = 0
in some contexts, it is not reasonable to interpret the intercept

27. The Deterministic Model
We often say x determines y.
Notice, the y-value is determined by substituting the
x-value into the equation of the line.
Also notice that the points fall on the line.
But, when we fit a line to data, do all the points fall on the line?

28. How do you find an appropriate line for describing a bivariate data set?
The point (15,44) has a deviation of +4.
To assess the fit of a line, we need a way to combine the n deviations into a single measure of fit.
To assess the fit of a line, we look at how the points deviate vertically from the line.
What is the meaning of this deviation?
y = x
This point is (20,45).
The predicted value for y when x = 20 is:
𝑦 = (20) = 50
The deviation of the point (20,45) from the line is: = -5
What is the meaning of a negative deviation?

29. Least squares regression line
The least squares regression line is the line that minimizes the sum of squared deviations.
The most widely used measure of the fit of a line y = a + bx to bivariate data is the sum of the squared deviations about the line.
The slope of the least squares regression line is:
𝑏= 𝑥− 𝑥 𝑦− 𝑦 𝑥− 𝑥 2
and the y-intercept is:
𝑎= 𝑦 −𝑏 𝑥
The equation of the least square regression line is:
𝑦 =𝑎+𝑏𝑥

30. Let’s investigate the meaning of the least squares regression line
Let’s investigate the meaning of the least squares regression line. Suppose we have a data set that consists of the observations (0,0), (3,10) and 6,2).
Use a calculator to find the least squares regression line
(0,0)
(3,10)
(6,2)
Find the sum of the squares of the deviations from the line
What is the sum of the deviations from the line?
Will the sum always be zero? 6
Find the vertical deviations from the line
Hmmmmm . . .
Why does this seem so familiar? -3
The line that minimizes the sum of squared deviations is the least squares regression line. -3
Sum of the squares = 54

31. Sketch a scatterplot for this data set.
Pomegranate, a fruit native to Persia, has been used in the folk medicines of many cultures to treat various ailments. Researchers are now investigating if pomegranate's antioxidants properties are useful in the treatment of cancer.
In one study, mice were injected with cancer cells and randomly assigned to one of three groups, plain water, water supplemented with .1% pomegranate fruit extract (PFE), and water supplemented with .2% PFE. The average tumor volume for mice in each group was recorded for several points in time.
(x = number of days after injection of cancer cells in mice assigned to plain water and y = average tumor volume (in mm3) x y
Sketch a scatterplot for this data set.
Number of days after injection
Average tumor volume

32. Does the intercept have meaning in this context? Why or why not?
Interpretation of slope:
The average volume of the tumor increases by approximately mm3 for each day increase in the number of days after injection.
Computer software and graphing calculators can calculate the least squares regression line.
Does the intercept have meaning in this context? Why or why not?

33. Pomegranate study continued
Predict the average volume of the tumor for 20 days after injection.
Predict the average volume of the tumor for 5 days after injection.
It is unknown whether the pattern observed in the scatterplot continues outside the range of x-values.
Why?
This is the danger of extrapolation. The least squares line should not be used to make predictions for y using x-values outside the range in the data set.
Can volume be negative?

34. If r = 1, what do you know about the location of the points?
Why is the line used to summarize a linear relationship called the least squares regression line?
An alternate expression for the slope b is:
𝑏=𝑟 𝑠 𝑦 𝑠 𝑥
This terminology comes from the relationship between the least squares line and the correlation coefficient.
The least squares regression line passes through the point of averages 𝑥 , 𝑦 .
Suppose that a point on the line is one standard deviation above the mean of x. The value of this point would be 𝒙 + 𝒔 𝒙 . Substitute this value for x in our equation.
Using the point-slope form of a line and r = 1, we can substitute the alternative slope and the point of averages.
𝑦 − 𝑦 =1 𝑠 𝑦 𝑠 𝑥 𝑥− 𝑥
which is 𝑦 = 𝑦 +1 𝑠 𝑦 𝑠 𝑥 𝑥− 𝑥
Notice that when r = 1, the y-value will be one standard deviation above the mean of y, 𝒚 + 𝒔 𝒚 , for an x-value one standard deviation above the mean of x, ( 𝒙 + 𝒔 𝒙 ).
If r = 1, what do you know about the location of the points?
𝑦 = 𝑦 +1 𝑠 𝑦 𝑠 𝑥 𝑥 + 𝑠 𝑥 − 𝑥
𝑦 = 𝑦 +1 𝑠 𝑦

35. Why is the line used to summarize a linear relationship called the least squares regression line?
Using the least squares line, the predicted y is pulled back in (or regressed) toward 𝑦 .
Let’s investigate what happens when r < 1.
𝑦 = 𝑦 +𝑟 𝑠 𝑦 𝑠 𝑥 𝑥− 𝑥
𝑦 = 𝑦 𝑠 𝑦 𝑠 𝑥 𝑥 + 𝑠 𝑥 − 𝑥
𝑦 = 𝑦 + 0.5𝑠 𝑦
Suppose r = 0.5 and 𝒙= 𝒙 + 𝒔 𝒙 . Substitute these values in our equation.
Notice that when r = 0.5, the y-value will be one-half standard deviation above the mean of y, 𝒚 +𝟎.𝟓 𝒔 𝒚 , for an x-value one standard deviation above the mean of x, ( 𝒙 + 𝒔 𝒙 ).
What would happen if r = 0.4? ? ?

36. If you want to predict x from y, can you use the least squares line of y on x?
The slope of the least squares line for predicting x will be
𝒓 𝒔 𝒙 𝒔 𝒚 not 𝒓 𝒔 𝒚 𝒔 𝒙 . Also, the intercepts of the lines are almost always different.
The regression line of y on x should not be used to predict x, because it is not the line that minimizes the sum of the squared deviations in the x direction.

37. Assessing the Fit of a Line
Residuals
Residual Plots
Outliers and Influential Points
Coefficient of Determination
Standard Deviation about the Line

38. Assessing the fit of a line
Important questions are:
Is the line an appropriate way to summarize the relationship between x and y ?
Are there any unusual aspects of the data set that you need to consider before proceeding to use the least squares regression line to make predictions?
If you decide that it is reasonable to use the line as a basis for prediction, how accurate can you expect predictions to be?
Once the least squares regression line is obtained, the next step is to examine how effectively the line summarizes the relationship between x and y.
This section will look at graphical and numerical methods to answer these questions.

39. Residuals 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙=𝑦− 𝑦
Recall, the vertical deviations of points from the least squares regression line are called deviations.
These deviations are also called residuals.
𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙=𝑦− 𝑦

40. Distance from Debris (x)
In a study, researchers were interested in how the distance a deer mouse will travel for food (y) is related to the distance from the food to the nearest pile of fine woody debris (x). Distances were measured in meters.
Minitab was used to fit the least squares regression line. The regression line is: 𝑦 =− 𝑥
If the point is below the line the residual will be negative.
Distance from Debris (x)
Distance Traveled (y)
6.94
0.00
5.23
6.13
5.21
11.29
7.10
14.35
8.16
12.03
5.50
22.72
9.19
20.11
9.05
26.16
9.36
30.65
14.76
-14.76
9.23
-3.10
9.16
2.13
15.28
-0.93
18.70
-6.67
10.10
12.62
22.04
-1.93
21.58
4.58
22.59
8.06
Distance traveled
Distance to debris
If the point is above the line the residual will be positive.
Calculate the predicted y and the residuals.

41. Residual plots
A residual plot is a scatterplot of the (x, residual) pairs.
Residuals can also be graphed against the predicted y-values
Isolated points or a pattern of points in the residual plot indicate potential problems.
A careful look at the residuals can reveal many potential problems.
A residual plot is a graph of the residuals.

42. Distance from Debris (x)
Deer mice continued
Distance from Debris (x)
Distance Traveled (y)
6.94
0.00
5.23
6.13
5.21
11.29
7.10
14.35
8.16
12.03
5.50
22.72
9.19
20.11
9.05
26.16
9.36
30.65
14.76
-14.76
9.23
-3.10
9.16
2.13
15.28
-0.93
18.70
-6.67
10.10
12.62
22.04
-1.93
21.58
4.58
22.59
8.06
This should remind students of the fact that the sum of the deviations from the mean is always zero.
Plot the residuals against the distance from debris (x)

43. Deer mice continued Are there any isolated points?
Is there a pattern in the points?
The points in the residual plot appear scattered at random.
This indicates that a line is a reasonable way to describe the relationship between the distance from debris and the distance traveled.

44. Deer mice continued
Residual plots can be plotted against either the x-values or the predicted y-values.

45. Residual plots continued
Let’s examine the accompanying data on x = height (in inches) and y = average weight (in pounds) for American females, ages (from The World Almanac and Book of Facts).
The residual plot displays a definite curved pattern. x 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 y
113
115
118
121
124
128
131
134
137
141
145
150
153
159
164
The scatterplot appears rather straight.
Even though r = 0.99, it is not accurate to say that weight increases linearly with height

46. Let’s examine the data set for 12 black bears from the Boreal Forest.
x = age (in years) and y = weight (in kg)
Sketch a scatterplot with the fitted regression line.
If the point affects the placement of the least-squares regression line, then the point is considered an influential point. x
10.5
6.5
28.5
7.5
5.5
11.5
9.5 Y 54 40 62 51 55 56 42 59 50
What would happen to the regression line if this point is removed?
Do you notice anything unusual about this data set?
This observation has an x-value that differs greatly from the others in the data set.

47. An observation is an outlier if it has a large residual.
Black bears continued x
10.5
6.5
28.5
7.5
5.5
11.5
9.5 Y 54 40 62 51 55 56 42 59 50
Notice that this observation falls far away from the regression line in the y direction.
An observation is an outlier if it has a large residual.

48. Coefficient of Determination
The coefficient of determination is the proportion of variation in y that can be attributed to an approximate linear relationship between x & y
Denoted by r2
The value of r2 is often converted to a percentage.
Suppose that you would like to predict the price of houses in a particular city from the size of the house (in square feet). There will be variability in house price, and it is this variability that makes accurate price prediction a challenge.
If you know that differences in house size account for a large proportion of the variability in house price, then knowing the size of a house will help you predict its price.

49. Why do we square the deviations?
Let’s explore the meaning of r2 by revisiting the deer mouse data set.
x = the distance from the food to the nearest pile of fine woody debris
y = distance a deer mouse will travel for food x
6.94
5.23
5.21
7.10
8.16
5.50
9.19
9.05
9.36 y
6.13
11.29
14.35
12.03
22.72
20.11
26.16
30.65
Suppose you didn’t know any x-values. What distance would you expect deer mice to travel?
Why do we square the deviations?
To find the total amount of variation in the distance traveled (y) you need to find the sum of the squares of these deviations from the mean.
Total amount of variation in the distance traveled (y) is
SSTo = m2

50. Why do we square the residuals?
Deer mice continued
x = the distance from the food to the nearest pile of fine woody debris
y = distance a deer mouse will travel for food x
6.94
5.23
5.21
7.10
8.16
5.50
9.19
9.05
9.36 y
6.13
11.29
14.35
12.03
22.72
20.11
26.16
30.65
Distance traveled
Distance to debris
Now let’s find how much variation there is in the distance traveled (y) from the least squares regression line.
Why do we square the residuals?
The amount of variation in the distance traveled (y) from the least squares regression line is
SSResid = m2
To find the amount of variation in the distance traveled (y), find the sum of the squared residuals.

51. Deer mice continued
x = the distance from the food to the nearest pile of fine woody debris
y = distance a deer mouse will travel for food
How does the variation in y change when we used the least squares regression line?
Total amount of variation in the distance traveled (y) is
SSTO = m2.
The amount of variation in y values from the regression line is
SSResid = m2
If the relationship between the two variables is negative, then you would use
− .32 =−0.566
Approximately what percent of the variation in distance traveled (y) can be explained by the linear relationship?
r2 = 32%

52. Standard Deviation about the Least Squares Regression Line
The coefficient of determination (r 2) measures the extent of variability about the least squares regression line relative to overall variability in y. This does not necessarily imply that the deviations from the line are small in an absolute sense.
The standard deviation about the least squares regression line is
𝑠 𝑒 = 𝑆𝑆𝑅𝑒𝑠𝑖𝑑 𝑛−2
The value of se can be interpreted as the typical amount an observation deviates from the least squares regression line.

53. The standard deviation (s):
Partial output from the regression analysis of deer mouse data:
The standard deviation (s):
This is the typical amount by which an observation deviates from the least squares regression line
Predictor
Coef
SE Coef T P
Constant
-7.69
13.33
-0.58
0.582
Distance to debris
3.234
1.782
1.82
0.112
S =
R-sq = 32.0%
R-sq(adj) = 22.3%
Analysis of Variance
Source DF SS MS F
Regression 1
247.68
3.29
Resid Error 7
526.27
75.18
Total 8
773.95
The y-intercept (a):
This value has no meaning in context since it doesn't make sense to have a negative distance.
The slope (b):
The distance traveled to food increases by approximately meters for an increase of 1 meter to the nearest debris pile.
The coefficient of determination (r2):
Only 32% of the observed variability in the distance traveled for food can be explained by the approximate linear relationship between the distance traveled for food and the distance to the nearest debris pile.
SSResid
SSTo

54. Interpreting the Values of se and r2
A small value of se indicates that residuals tend to be small. This value tells you how much accuracy you can expect when using the least squares regression line to make predictions. A large value of r2 indicates that a large proportion of the variability in y can be explained by the approximate linear relationship between x and y. This tells you that knowing the value of x is helpful for predicting y. A useful regression line will have a reasonably small value of se and a reasonably large value of r2.

55. A study (Archives of General Psychiatry[2010]: ) looked at how working memory capacity was related to scores on a test of cognitive functioning and to scores on an IQ test. Two groups were studied – one group consisted of patients diagnosed with schizophrenia and the other group consisted of healthy control subjects.
For the patient group, the typical deviation of the observations from the regression line is about 10.7, which is somewhat large. Approximately 14% (a relatively small amount) of the variation in the cognitive functioning score is explained by the linear relationship.
For the control group, the typical deviation of the observations from the regression line is about 6.1, which is smaller. Approximately 79% (a much larger amount) of the variation in the cognitive functioning score is explained by the regression line.
Thus, the regression line for the control group would produce more accurate predictions than the regression line for the patient group.

56. Putting it All Together
Describing Linear Relationships
Making Predictions

57. Steps in a Linear Regression Analysis
Summarize the data graphically by constructing a scatterplot
Based on the scatterplot, decide if it looks like the relationship between x an y is approximately linear. If so, proceed to the next step.
Find the equation of the least squares regression line.
Construct a residual plot and look for any patterns or unusual features that may indicate that line is not the best way to summarize the relationship between x and y. In none are found, proceed to the next step.
Compute the values of se and r2 and interpret them in context.
Based on what you have learned from the residual plot and the values of se and r2, decide whether the least squares regression line is useful for making predictions. If so, proceed to the last step.
Use the least squares regression line to make predictions.

58. Revisit the crime scene DNA data
Recall the scientists were interested in predicting age of a crime scene victim (y) using the blood test measure (x).
Step 1: Scientist first constructed a scatterplot of the data.
Step 2: Based on the scatterplot, it does appear that there is a reasonably strong negative linear relationship between and the blood test measure.

59. Step 4: A residual plot constructed from these data showed a few observations with large residuals, but these observations were not far removed from the rest of the data in the x direction. The observations were not judged to be influential. Also there were no unusual patterns in the residual plot that would suggest a nonlinear relationship between age and the blood test measure. Step 5: se = 8.9 and r2 = Approximately 83.5% of the variability in age can be explained by the linear relationship. A typical difference between the predicted age and the actual age would be about 9 years.

60. Step 6: Based on the residual plot, the large value of r2, and the relatively small value of se, the scientists proposed using the blood test measure and the least squares regression line as a way to estimate ages of crime victims.
Step 7: To illustrate predicting age, suppose that a blood sample is taken from an unidentified crime victim and that the value of the blood test measure is determined to be The predicted age of the victim would be
𝑦 =−33.65−6.74 −10 =33.75 years

61. Modeling Nonlinear Relationships

62. Choosing a Nonlinear Function to Describe a Relationship
Equation
Looks Like
Quadratic
Square root
Reciprocal
𝑦 =𝑎+ 𝑏 1 𝑥+ 𝑏 2 𝑥 2
𝑦 =𝑎+𝑏 𝑥
𝑦 =𝑎+𝑏 1 𝑥

63. Choosing a Nonlinear Function to Describe a Relationship
The common log (base 10) may also be used.
Function
Equation
Looks Like
Log
Exponential
Power
While statisticians often use these nonlinear regressions, in AP Statistics, we will linearize our data using transformations. Then we can use what we already know about the least squares regression line.
𝑦 =𝑎+𝑏 ln𝑥
𝑦 = 𝑒 𝑎+𝑏𝑥
𝑦 =𝑎 𝑥 𝑏

64. Models that Involve Transforming Only x
The square root, reciprocal, and log models all have the form
𝑦 =𝑎+𝑏(some function of 𝑥)
Where the function of x is square root, reciprocal, or log.
This suggest that if the pattern in the scatterplot of (x, y) pairs looks like one of these curves, an appropriate transformation of the x values should result in transformed data that shows a linear relationship.
Read “x prime” X’
Model
Transformation
Square root
𝑥 ′ = 𝑥
Reciprocal
𝑥 ′ = 1 𝑥
Log
𝑥 ′ = ln 𝑥
Let’s look at an example.

65. First look at a scatterplot of the data.
Is electromagnetic radiation from phone antennae associated with declining bird populations? The accompanying data on x = electromagnetic field strength (Volts per meter) and y = sparrow density (sparrows per hectare)
Field Strength
Sparrow Density
0.11
41.71
0.20
33.60
0.29
24.74
0.40
19.50
0.50
19.42
0.61
18.74
1.01
24.23
1.10
22.04
0.70
16.29
0.80
14.69
0.90
1.20
16.97
1.30
12.83
1.41
13.17
1.50
4.64
1.80
2.11
1.90
0.00
3.01
3.10
3.41
First look at a scatterplot of the data.
The data is curved and looks similar to the graph of the log model.

66. Field Strength vs. Sparrow Density Continued
Ln Field Strength
Sparrow Density
-2.207
41.71
-1.609
33.60
-1.238
24.74
-.0916
19.50
-0.693
19.42
-0.494
18.74
0.001
24.23
0.095
22.04
-0.357
16.29
-0.223
14.69
-0.105
0.182
16.97
0.262
12.83
0.344
13.17
0.405
4.64
0.588
2.11
0.642
0.00
1.102
1.131
1.227
Sparrow Density = 14.8 – ln (Field Strength)
Predictor
Coef
SE Coef T P
Constant
14.805
1.238
11.96
0.000
Ln (field strength)
1.389
-7.59
S =
R-Sq = 76.2%
R-Sq(adj) = 74.9%
Second, we will transform the data by using
𝑥 ′ = ln 𝑥
. . . and graph the scatterplot of y on x’
Notice that the transformed data is now linear. We can find the least squares regression line.

67. Field Strength vs. Sparrow Density Continued
Sparrow Density = 14.8 – ln (Field Strength)
Predictor
Coef
SE Coef T P
Constant
14.805
1.238
11.96
0.000
Ln (field strength)
1.389
-7.59
S =
R-Sq = 76.2%
R-Sq(adj) = 74.9%
The value of R 2 for this model is and se = 5.5.
A residual plot from the least squares regression line fit to the transformed data, shown below, has no apparent patterns or unusual features. It appears that the log model is a reasonable choice for describing the relationship between sparrow density and field strength.

68. Field Strength vs. Sparrow Density Continued
Sparrow Density = 14.8 – ln (Field Strength)
Predictor
Coef
SE Coef T P
Constant
14.805
1.238
11.96
0.000
Ln (field strength)
1.389
-7.59
S =
R-Sq = 76.2%
R-Sq(adj) = 74.9%
This model can now be used to predict sparrow density from field strength. For example, if the field strength is 1.6 Volts per meter, what is the prediction for the sparrow density?
𝑦 =14.805− ln 𝑥
=14.805−10.546( ln 1.6 )
=9.685 sparrows per hectare

69. Models that Involve Transforming y
Let’s consider the remaining nonlinear models, the exponential model and the power model.
Exponential Model
𝑦= 𝑒 𝑎+𝑏𝑥
Power Model
𝑦=𝑎 𝑥 𝑏
ln 𝑦 = ln 𝑒 𝑎+𝑏𝑥
=𝑎+𝑏𝑥
ln 𝑦 = ln 𝑎 𝑥 𝑏
= ln 𝑎 +𝑏 ln 𝑥
Using properties of logarithms, it follows that . . .
Using properties of logarithms, it follows that . . .
Notice that using the transformations below, the exponential and power models are linearized.
Model
Transformation
Exponential
𝑦 ′ = ln 𝑦
Power
𝑥 ′ = ln 𝑥

70. In a study of factors that affect the survival of loon chicks in Wisconsin, a relationship between the pH of lake water and blood mercury level in loon chicks was observed. The researchers thought that it is possible that the pH of the lake could be related to the type of fish that the loons ate. A scatterplot of the data is shown below.
The scatterplot of ln(blood mercury level) on lake pH appears linear.
The linear model is
ln 𝑦 =1.06−0.396𝑥.
The curve appears to be exponential, therefore use
𝑦 ′ = ln 𝑦
to transform the data.
Ln(blood mercury level)= Lake pH
Predictor
Coef
SE Coef T P
Constant
1.0550
0.5535
1.91
0.065
Lake pH
0.0826
-4.79
0.000
S =
R-Sq = 39.6%
R-Sq(adj) = 37.8%

71. Choosing Among Different Possible Nonlinear Models
Often there is more than one reasonable model that could be used to describe a nonlinear relationship between two variables.
How do you choose a model?
1) Consider scientific theory. Does it suggest what model the relationship is?
2) In the absence of scientific theory, choose a model that has small residuals (small se) and accounts for a large proportion of the variability in y (large R 2).

72. Common Mistakes

73. Avoid these Common Mistakes
Correlation does not imply causation. A strong correlation implies only that the two variables tend to vary together in a predictable way, but there are many possible explanations for why this is occurring other than one variable causing change in the other.
The number of fire trucks at a house that is on fire and the amount of damage from the fire have a strong, positive correlation.
So, to avoid a large amount of damage if your house is on fire – don’t allow several fire trucks to come to your house?
Don’t fall into this trap!

74. Avoid these Common Mistakes
A correlation coefficient near 0 does not necessarily imply that there is no relationship between two variables. Although the variables may be unrelated, it is also possible that there is a strong but nonlinear relationship.
Be sure to look at a scatterplot!

75. Avoid these Common Mistakes
The least squares regression line for predicting y from x is NOT the same line as the least squares regression line for predicting x from y.
The ages (x, in months) and heights (y, in inches) of seven children are given. x y
To predict height from age:
ℎ𝑒𝑖𝑔ℎ𝑡 = (𝑎𝑔𝑒)
To predict age from height:
𝑎𝑔𝑒 =− (ℎ𝑒𝑖𝑔ℎ𝑡)

76. Avoid these Common Mistakes
Beware of extrapolation. Using the least squares regression line to make predictions outside the range of x values in the data set often leads to poor predictions.
Predict the height of a child that is 15 years (180 months) old.
ℎ𝑒𝑖𝑔ℎ𝑡 = =81.6 𝑖𝑛𝑐ℎ𝑒𝑠
It is unreasonable that a 15 year-old would be 81.6 inches or 6.8 feet tall

77. Avoid these Common Mistakes
Be careful in interpreting the value of the intercept of the least squares regression line. In many instances interpreting the intercept as the value of y that would be predicted when x = 0 is equivalent to extrapolating way beyond the range of x values in the data set.
The ages (x, in months) and heights (y, in inches) of seven children are given. x y

78. Avoid these Common Mistakes
Remember that the least squares regression line may be the “best” line, but that doesn’t necessarily mean that the line will produce good predictions.
This has a relatively large se – thus we can’t accurately predict IQ from working memory capacity.

79. Avoid these Common Mistakes
It is not enough to look at just r2 or just se when evaluating the regression line. Remember to consider both values. In general, your would like to have both a small value for se and a large value for r2.
This indicates that deviations from the line tend to be small.
This indicates that the linear relationship explains a large proportion of the variability in the y values.

80. Avoid these Common Mistakes
The value of the correlation coefficient, as well as the values for the intercept and slope of the least squares regression line, can be sensitive to influential observations in the data set, particularly if the sample size is small.
Be sure to always start with a plot to check for potential influential observations.