Regression revisited
FPP 11 and 12 and a little more
Statistical modeling
 Often researchers seek to explain or predict one variable
from others.
 In most contexts, it is impossible to do this perfectly: too
much we don’t know.
 Use mathematical models that describe relationships as best
we can.
 Incorporate chance error into models so that we can
incorporate uncertainty in our explanations/predictions
Linear regression
 Linear regression is probably the most common statistical
model
 Idea is like regression lines from Chapter 10. But, the slope
and intercept from a regression line are estimates of that
true line (just like a sample mean is an estimate of a
population mean).
 Hence, we can make inference (confidence intervals and
hypothesis tests) for the true slope and true intrecept
Linear regression
 Often relationships are described reasonably well by a linear
trend.
 Linear regression allows us to estimate these trends
 Plan of attack
 Pose regression model and investigate assumptions
 Estimate regression parameters from data
 Use hypothesis testing and confidence interval ideas to determine if the
relationship between two variables has occurred by chance alone
 Regression with multiple predictors
Regression terminology
 Typically, we label the outcome variable as Y and the
predictor as X .
 Synonyms for outcome variables:
 response variable, dependent variables
 Synonyms for predictor variables
 explanatory variables, independent variables, covariates
Some notation
 Recall the regression line or least squares line notation from
earlier in the class
y    x
 α denotes the population intercept
 βdenotes the population slope

Sample regression line
 If we collect a sample from some population and use sample
values to calculate a regression line, then there is uncertainty
associated with the sample slope and intercept estimates.
 The following notation is used to denote the sample
regression line
yˆ  a  bx
Motivating example
 A forest service official needs to determine the total
volume of lumber on a piece of forest land
 Any ideas on how she might do this?
Motivating example
 A forest service official needs to determine the total
volume of lumber on a piece of forest land
 Any ideas on how she might do this that doesn’t require cutting
down lots of trees?
 She hopes predicting volume of wood from tree diameter for
individual trees will help determine total volume for the piece of
forest land. She investigates, “Can the volume of wood for a tree
be predicted by its diameter?”
Motivating example
 First she randomly samples 31




trees and measures the
diameter of each tree and then
its volume.
Then she constructs a scatter
plot of the data collected and
checks for a linear pattern
Is relationship linear?
We know how to estimate the
slope and intercept of the line
that “best” fits the data
But
Motivating example
 What would happen if the forest service agent took
another sample of 31 trees?
 Would the slope change?
 Would the intercept change?
 What about a third sample of 31 trees?
 a and b are statistics and are dependent on a sample
 We know how to compute them
 They are also estimates of a population intercept and slope

Mathematics of regression model
 To accommodate the added uncertainty associated with the
regression line we add one more term to the model
y i    x i  i ,
where i comes from N(0, )
 This model specification has three assumptions
 1. the average value of Y for each X falls on line
 2. the deviations don’t depend on X
 3. the deviations from the straight line follow a normal curve
 4. all units are independent
The mechanics of regression
 Questions we aim to answer
 How do we perform statistical inference on the intercept and
slope of the regression line?
 What is a typical deviation from the regression line?
 How do we know the regression line explains the data well?
Estimating intercept and slope
 From early in the semester recall that the intercept and slope
estimates for the line of “best” fit are
16.46
br
 0.97
 5.07
 3.14 
SDx
SDy
a  y  bx  30.14  5.08(13.25)  37.02
yˆ  37.02 5.07x

Root mean square error (RMSE)
 What is the typical deviation from the regression line for a given x?
 The typical deviation is denoted by 

 The root mean square error (RMSE) is a measure of the typical
deviation from the regression
 line for a given x
 For the trees data this is 4.28
 A tree with a diameter of 15 inches can be expected to have a volume
of -37.02 + 5.07(15) = 39.03 cubic inches give or take about 4.28
cubic inches
JMP output
Residuals are used to compute
RMSE
 The deviation of each yi from the line is called a residual that
we will denote by di
di  y i  yˆ i
 y i  (a  bx i )
 An estimate of  that is used in most software packages is
denoted by


n
1
2
s 
di

n  2 i1
Significant tests and CIs
 Going back to the example of trees sampled from the plot of land.
 The sampled trees are one possible random sample from all trees in
the plot of land
 Questions:
 What is a likely range for the population regression slope?
 Does the sample regression slope provide enough evidence to say
with conviction that the population slope doesn’t equal zero?
 Why zero?
CI for slope
 Est. ± multiplier*SE
 Same old friend in a new hat
 We will use the sample slope as an estimate
 The multiplier is found from a t-distribution with (n-2)
degress of freedom
 The SE of the slope (not to be confused with RMSE) is
SEb 
s
n
2
(x

x
)
 i
i1
CI slope
 A 95% confidence interval for
the population slope between
diameter of tree and volume is
 b  multiplier*SEb
 5.07  multiplier*0.249567
 Where does the multiplier
value come from?
 We use the the t-table and find
column with n-2 degrees of
freedom and match with desired
confidence level
 But with 31-2=29 d.f. we are
not able to use t-table. So use
normal
 5.07  1.96*0.249567
 (4.58, 5.56)
CI of slope
 We found that a 95% confidence interval for  is (4.58, 5.56).
 What is the interpretation of this interval?
 95% confident that the population slope that describes the
relationship between a tree’s diameter and its lumber volume is
between 4.58 and 5.56 inches.
 What does the statement “95% Confidence” mean (this is the same
thing as statistical confidence)
 We are confident that the method used will give correct results in
95% of all possible samples.
 That is, 95% of all samples will produce confidence intervals that contain the
true population slope.
Hypothesis test for existence of
linear relationship
 What parameter (αorβ) should we test to determine
whether X is useful in predicting Y?
 We want to test:
 H0: There is NO linear relationship between two numerical
variables (X is NOT useful for predicting Y)
 Ha: There is a linear relationship between two numerical
variables (X is useful for predicting Y)
 Draw the picture
Hypothesis test for existence of
linear relationship
 What parameter (αorβ) should we test to determine
whether X is useful in predicting Y?
 We want to test:
 H0: There is NO linear relationship between two numerical
variables (X is NOT useful for predicting Y)
 Ha: There is a linear relationship between two numerical
variables (X is useful for predicting Y)
 Draw the picture
 The hypothesis can also be stated as
 Ho:  = 0 vs. Ho:   0
Hypothesis test
 The test statistic is
t
est. hyp. 5.07  0

 20.31
SE
.2495
 To find the p-value associated with this test statistic, we find the
area under a t-curve with (n-2) degrees of freedom.
 According to JMP, this p-value equals smaller than 0.0001.
 According to the table it is smaller than 0.0005
 Hence, there is strong evidence against the null. Conclude that
the sample regression slope is not consistent with a population
regression slope being equal to zero. There does appear to be a
relationship between the diameter of a tree and its volume.
JMP output
How well does regression model fit
data?
 Do determine this we need to check the assumptions made
when using the model.
 Recall that the regression assumptions are
 1. The average value of Y for each X falls on a line (i.e. the
relationship between Y and X is linear)
 2. The deviations (RMSE) are the same for all X
 3. For any X, the distribution of Y around its mean is a normal
curve.
 4. All units are independent
Check the regression fit to the data
 When the assumptions are true, values of the residuals should
reflect chance error.
 That is, there should be only random patterns in the
residuals.
 Check this by plotting the residuals versus the predictor
 If there is a non-random pattern in this plot, assumptions
might be violated
Diagnosing residual plots
 When pattern in residuals around the horizontal line at zero
is:
 Curved (e.g. parabolic shape):
 Assumption 1 (slide 25) is violated
 Fan-shaped:
 Assumption 2 (slide 25) is violated
 Filled with many outliers:
 Assumption 2 (slide 25) is violated
Possible patterns in Residual Plots
Residual plot
 Do the residuals look
randomly scattered?
 Or is there some
pattern?
 Is there spread of the
points similar at
different values of
diameter
One number summary of
regression fit
 R2 is the percentage of variation in Y’s explained by the
regression line
 R2 lies between 0 and 1
 Values near 1 indicate regression predicts y’s in data set very
closely
 Values near 0 indicate regression does not predict the y’s in
the data set very closely
Interpretation in tree example
 We get a R2 = 0.93. Hence, the regression line between
diameter and volume explains 93% of the variability volume
Caution about R2
 Don’t rely exclusively on R2 as a measure of the goodness of
fit of the regression.
 It can be large even when assumptions are violated
 Always check the assumptions with residual plots before
accepting any regression model
Predictions from regression
 To predict an outcome for a unit with unobserved Y but
known X, use the fitted regression model
yˆ  a  bx
 Example from the tree data:
 Predict volume from a tree that has a 15 inch

yˆ  37.02  5.07x
 32.02  5.07(15)
 44.03 in3
Recall warnings
 Predicting Y at values of X beyond the range of the X’s in the data is dangerous
(extrapolation)
 Association doesn’t imply causation
 Influential points/outliers
 Fit model with and with out point to see if estimates change
 Often we aren’t interested in the intercept
 Ecological inference
 Regression fits for aggregated data tend to show stronger relationships
 With census data there is no sampling variability (we’ve exhausted the
population)
 There is no standard error
 Sometimes census data are viewed as a random sample from a hypothetical
“super-population”. In this case the census data provide inferences about the
super-population
 When using time as the X variable care must be taken as the independent unit
assumption is often not valid
 Most likely will need to use special models