Created by Kathy Fritz 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 TCell DNA Re ID: 915887
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Slide1
Chapter 4
Describing Bivariate Numerical Data
Created by Kathy Fritz
Slide2Forensic 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
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?
This line can be used to estimate the age of a crime victim from a blood test.
Slide3Correlation
Pearson’s Sample Correlation Coefficient
Properties of
r
Slide4Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
Yes
Slide5Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
Yes
Slide6Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
No, looks curved
Slide7Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
Yes
No, looks parabolic
Slide8Does it look like there is a relationship between the two variables?
If so, is the relationship linear?
No
Slide9Linear relationships can be either
positive or negative in direction.
Are these linear relationships positive or negative?
Positive
Negative
Slide10When 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
B
C
D
A, C, B, D
Slide11Pearson’s Sample Correlation Coefficient
Usually referred to as just the
correlation coefficientDenoted by rMeasures 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!
Slide12Properties of
r
The sign of r matches
the
direction
of the linear relationship.
r is positive
r is negative
Slide13Properties of
r
The value of
r
is always greater than or equal to
-1
and less than or equal to +1.
Weak correlation
Strong correlation
Moderate correlation
Slide14Properties 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.
Slide15Properties 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.
x
2
4
6
8
10
12
14
y
40
20
8
4
8
20
40
r
= 0
r
= 0, but the data set has a
definite
relationship!
Does this mean that there is
NO
relationship between these points?
Slide16Properties of
r
The value of
r
does not depend on the
unit of measurement
for either variable.
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
642
104.0
556
104.0
616
93.5
549
108.5
504
95.0
515
117.5
551
128.0
594
127.5
Calculate
r
for the data set of mares’ weight and the weight of their foals.
r
= -0.00359
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
1412.4
104.0
1223.2
104.0
1355.2
93.5
1207.8
108.5
1108.8
95.0
1111.0
117.5
1212.2
127.5
1306.8
127.5
Change the mare weights to pounds by multiply Kg by 2.2 and calculate
r
.
r
= -0.00359
Slide17Calculating Correlation Coefficient
The correlation coefficient is calculated using the following formula:
where
and
Slide18The web site
www.collegeresults.org (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
Slide19College Expenditures Continued:
To compute the correlation coefficient, first find the
z
-scores.
x
y
z
x
zy
z
x
z
y
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
0.12
0.01
8477
42.0
0.81
-0.42
-0.34
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.
Slide20How the Correlation Coefficient Measures the Strength of a Linear Relationship
z
x
is positive
z
y
is positivezxz
y is positive
z
x
is negative
z
y
is
negative
z
x
z
y
is positive
z
x
is negative
z
y
is positive
z
x
z
y
is
negative
Will the
sum of
z
x
z
y
be positive or negative?
Slide21How the Correlation Coefficient Measures the Strength of a Linear Relationship
z
x
is positive
z
y
is positivezxzy is positive
z
x
is negative
z
y
is
negative
z
x
z
y
is positive
z
x
is negative
z
y
is positive
z
x
z
y
is
negative
Will the
sum of
z
x
z
y
be positive or negative?
z
x
is negative
z
y
is positive
z
x
z
y
is
negative
Slide22How the Correlation Coefficient Measures the Strength of a Linear Relationship
Will the
sum of
z
x
z
y be positive or negative or zero?
Slide23Does 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.
These variables are both strongly related to the age of the childBoth are responses to cold weather
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
.
So does this mean I should feed children more candy to increase their vocabulary?
Should we all drink more hot chocolate to lower the crime rate?
Association does
NOT
imply causation.
Slide24Linear Regression
Least Squares Regression Line
Slide25Suppose 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 letter y is used to denoted the variable you want to predict, called the response variable (or dependent variable).
The other variable, denoted by x, is the
predictor variable (sometimes called independent or explanatory variable).
Slide26Where:
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 = 0in some contexts, it is not reasonable to interpret the interceptThe equation of a line is:
Slide27The Deterministic Model
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.We often say x determines
y.
But, when we fit
a line to data, do all the points fall on the line?
Slide28How do you find an appropriate line for describing a bivariate data set?
y
= 10 + 2
x
To assess the fit of a line, we look at how the points
deviate
vertically from the line.This point
is (20,45).The predicted value for y
when x = 20 is:
= 10 + 2(20) = 50
The
deviation
of the point (20,45) from the line is: 45 - 50 = -5
What is the meaning of a negative deviation?
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
.
What is the meaning of
this deviation
?
Slide29Least squares regression line
The
least squares regression line is the line that minimizes the sum of squared deviations.
The
slope
of the least squares regression line is:
and the
y
-intercept
is:
The
equation
of the least square regression line is:
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.
Slide30(0,0)
(3,10)
(6,2)
Sum of the squares = 54
Use a calculator to find the
least squares regression line
Find the vertical deviations from the line
-3
6
-3
What is the sum of the deviations from the line?
Will
the sum
always be zero?
The line that
minimizes
the sum of
squared deviations is
the
least squares regression line
.
Find the sum of the squares of the deviations from the 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).
Hmmmmm
. . .
Why does this seem so familiar?
Slide31Pomegranate, 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 11 15 19 23 27
y 150 270 450 580 740Sketch a scatterplot for this data set.
Number of days after injection
Average tumor volume
Slide32Interpretation of slope:
The average volume of the tumor increases by approximately 37.25 mm
3
for each day increase in the number of days after injection.
Does the intercept have meaning in this context? Why or why not?
Computer software and graphing calculators can calculate the least squares regression line.
Slide33Pomegranate 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.
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.
It is unknown whether the pattern observed in the scatterplot continues outside the range of
x
-values.
Why?
Can volume be negative?
Slide34Why is the line used to summarize a linear relationship called the least squares
regression line?
An alternate expression for the slope b is:
The least squares regression line passes through
t
he point of averages
This terminology comes from the relationship between the least squares line and the correlation coefficient.
Using the point-slope form of a line and
r
= 1, we can substitute the alternative slope and the point of averages.
which is
If r = 1, what
do you know about the location of the points?
Suppose that a point
on the line is one standard deviation above the mean of
x
. The value of this point would be
. Substitute this value for
x
in our equation.
Notice that when
r
= 1, the
y
-value will be one standard deviation
above the mean of
y
,
, for an
x
-value one standard deviation above the mean of
x
,
.
Slide35Why is the line used to summarize a linear relationship called the least squares
regression line?
Let’s investigate what happens when
r
< 1
.
Suppose
r = 0.5
and
. 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
,
, for an
x
-value
one
standard deviation above the mean of
x
,
.
Using the least squares line, the predicted y is pulled back in (or
regressed
) toward
.
What would
happen if r = 0.4? . . . 0.3? . . . 0.2?
Slide36The
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.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
not
.
Also, the
intercepts
of the lines are almost always different.
Slide37Assessing the Fit of a Line
Residuals
Residual Plots
Outliers and Influential Points
Coefficient of Determination
Standard Deviation about the Line
Slide38Assessing 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.
Slide39Residuals
Recall, the vertical deviations of points from the least squares regression line are called
deviations.
These deviations are also called
residuals
.
Slide40In 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.
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
Calculate the predicted
y
and the residuals.
Distance traveled
Distance to debris
If the point is
below
the line the residual will be
negative
.
If the point is
above
the line the residual will be
positive
.
Minitab was
used to fit the least squares regression line. The regression line is:
Slide41Residual plots
A residual plot is a
scatterplot of the (x, residual) pairs.
Residuals can also be graphed against the
predicted
y
-valuesIsolated 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.
Slide42Deer 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
Plot the
residuals
against the
distance from debris (x)
Slide43Are there any
isolated points
?
Is there
a pattern
in the points?
Deer mice continued
The points in the residual plot appear scattered at random.
This indicates that
a line
is a reasonable way to describe the r
elationship
between the distance
from debris and the distance traveled.
Slide44Residual plots can be plotted against either the
x
-values
or the
predicted
y
-values
.
Deer mice continued
Slide45Residual plots continued
Let’s examine the accompanying data on
x
= height (in inches) and
y
= average weight
(in pounds) for American females, ages 30-39 (from The World Almanac and Book of Facts).
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.
The residual plot displays a
definite curved pattern
.
Even though
r
= 0.99, it is
not accurate
to say that weight increases
linearly
with height
Slide46Let’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.
x
10.5
6.5
28.5
10.5
6.5
7.5
6.5
5.5
7.5
11.5
9.5
5.5
Y
54
40
62
51
55
56
62
42
40
59
51
50
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.
What would happen to the regression line if this point is removed?
If
the point affects the placement of the least-squares regression line, then the point is considered an
influential point
.
Slide47Black bears continued
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.
x
10.5
6.5
28.5
10.5
6.5
7.5
6.5
5.5
7.5
11.5
9.5
5.5
Y
54
40
62
51
55
56
62
42
40
59
51
50
Slide48Coefficient 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 r2The 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.
Slide49Suppose you
didn’t
know any
x
-values. What distance would you
expect deer mice to travel? 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
0
6.13
11.29
14.35
12.03
22.72
20.11
26.16
30.65
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 = 773.95 m
2
Why do we square the deviations?
Slide50Now let’s find how much
variation
there is in the
distance traveled (
y
)
from the least squares regression line.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
0
6.13
11.29
14.35
12.03
22.72
20.11
26.16
30.65
The amount of
variation in the distance traveled (
y
)
from the
least squares regression line
is
SSResid
= 526.27 m
2
To find the amount of variation in the distance traveled (
y
), find the
sum of the squared residuals
.
Distance traveled
Distance to debris
Why do we square the residuals?
Slide51The amount of variation in
y
values from the regression line is
SS
Resid
= 526.27 m
2
Total amount of variation
in the distance traveled (
y
) is
SS
TO
= 773.95 m
2
.
Approximately what
percent of the variation in distance traveled (
y
)
can be explained by the
linear relationship?
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
r
2
= 32%
How does the variation in
y
change when we used the least squares regression line?
If the relationship between
the two variables is negative, then you would use
Slide52The
standard deviation about the least squares regression line is
The value of
s
e
can be interpreted as the
typical amount an observation deviates from the least squares regression line. 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.
Slide53Partial output from the regression analysis of deer mouse data:
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 = 8.67071
R-sq = 32.0%
R-
sq
(
adj
) = 22.3%
Analysis of Variance
Source
DF
SS
MS
F
P
Regression
1
247.68
247.68
3.29
0.112
Resid
Error
7
526.27
75.18
Total
8
773.95
The coefficient of determination (
r
2
):
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.
The standard deviation (s):
This is the
typical
amount by which an observation deviates from the least squares regression line
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 3.234 meters for an increase of 1 meter to the nearest debris pile.
SSResid
SSTo
Slide54A 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 r
2
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.Interpreting the Values of s
e and r2
Slide55A study (Archives of General Psychiatry[2010]: 570-577) 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.
Slide56Putting it All Together
Describing Linear Relationships
Making Predictions
Slide57Steps in a Linear Regression Analysis
Summarize the data graphically by constructing a
scatterplotBased 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 s
e 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.
Slide58Revisit 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.
Slide59Step 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:
s
e
= 8.9 and
r2 = 0.835Approximately 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.
Slide60Step 6: Based on the residual plot, the large value of
r
2
, and the relatively small value of
s
e
, 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 -10. The predicted age of the victim would be
Slide61Modeling Nonlinear Relationships
Slide62Choosing a Nonlinear Function to Describe a Relationship
Function
Equation
Looks Like
Quadratic
Square root
Reciprocal
Slide63Choosing a Nonlinear Function to Describe a Relationship
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.
The common log (base 10) may also be used.
Slide64Models that Involve Transforming Only
x
The square root, reciprocal, and log models all have the form
Where the function of x is square root, reciprocal, or log.
Model
Transformation
Square root
Reciprocal
Log
Model
Transformation
Square root
Reciprocal
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”
Let’s look at an example.
Slide65Is 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
16.29
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
0.00
3.10
14.69
3.41
0.00
First look at a scatterplot of the data.
The data is curved and looks similar to the
graph of the log model.
Slide66Field 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
16.29
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
0.00
1.131
14.69
1.227
0.00
Second, we will transform the data by using
. . .
. . . and graph the scatterplot
of
y
on
x
’
Notice that the transformed data is now linear. We can find the least squares regression line.
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)
-10.546
1.389
-7.59
0.000
S = 5.50641
R-
Sq
= 76.2%
R-
Sq
(
adj
) = 74.9%
Slide67Field 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)
-10.546
1.389
-7.59
0.000
S = 5.50641
R-
Sq
= 76.2%
R-
Sq
(
adj
) = 74.9%
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.
The value of
R
2
for this model
is 0.762
and
s
e
= 5.5
.
Slide68Field 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)
-10.546
1.389
-7.59
0.000
S = 5.50641
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?
Slide69Models that Involve Transforming
y
Let’s consider the remaining nonlinear models, the exponential model and the power model.
Model
Transformation
Exponential
Power
Model
Transformation
Exponential
Power
Exponential Model
Using
properties of logarithms, it follows that . . .
Power Model
Using
properties of logarithms, it follows that . . .
Notice that using the transformations below, the exponential and power models are linearized.
Slide70In 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 curve appears to be exponential,
t
herefore use
t
o transform the data.
The
scatterplot of
ln
(blood mercury level) on lake pH appears linear.
The
linear model is
.
Ln(blood mercury level)= 1.06-0.396 Lake pH
Predictor
Coef
SE
Coef
T
P
Constant
1.0550
0.5535
1.91
0.065
Lake pH
-0.3956
0.0826
-4.79
0.000
S = 0.6056
R-
Sq
= 39.6%
R-
Sq
(
adj
) = 37.8%
Slide71Choosing 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).
Slide72Common Mistakes
Slide73Avoid 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.
Don’t fall into this trap!
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?
Slide74Avoid 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!
Slide75Avoid 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 16 24 42 60 75 102 120 y 24 30 35 40 48 56 60
To predict height from age:
To predict age from height:
Slide76Avoid 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.
It is unreasonable that a 15 year-old would be 81.6 inches or 6.8 feet tall
Slide77Avoid 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 16 24 42 60 75 102 120 y 24 30 35 40 48 56 60
Slide78Avoid 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 s
e
– thus we can’t accurately predict
IQ from working memory capacity.
Slide79Avoid these Common Mistakes
It is not enough to look at just
r2 or just s
e
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.
Slide80Avoid 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.