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Interpreting Multiple Regression: R-squared, Estimates, and Intervals, Lecture notes of Mathematical Statistics

How to interpret the results of a multiple regression analysis, focusing on r-squared values, parameter estimates, and their confidence intervals. Using an example from a study on the proportion of pollen removed by queen and worker bumblebees, it discusses the meaning of r-squared and adjusted r-squared, the interpretation of intercepts and coefficients, and the calculation of confidence intervals for linear combinations of parameters. It also covers the standard error of the mean and the use of the correlation matrix and covariance matrix to find the standard error for a linear combination.

What you will learn

  • How do you interpret the intercept and coefficients in a multiple regression model?
  • What is the meaning of R-squared and adjusted R-squared in multiple regression?
  • How do you calculate confidence intervals for linear combinations of parameters in multiple regression?

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2021/2022

Uploaded on 09/12/2022

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Interpretation in Multiple Regression
Topics:
1.
R-squared and Adjusted R-squared
2.
Interpretation of parameter estimates
3.
Linear combinations of parameter estimates
variance-covariance matrix
standard errors of combinations
standard error for the mean
We will use the final model from last time to illustrate these concepts.
Summaries
of
the
model - least squares estimates with standard errors given below in parentheses:
proportion
2.71
0.89
log
duration
0.57
I
0.38
.14
0.24
= 0.65 with 44 degrees of freedom
R-squared =
0.6068029
R-squared
and
Adjusted
R-squared
:
The
R-squared
value
means
that
61%
of
the
variation
in
the
logit
of
proportion
of
pollen
removed
can
be
explained
by
the
regression
on
log
duration
and
the
group
indicator
variable.
As
R-squared
values
increase
as
we
ass
more
variables
to
the
model,
the
adjusted
R-squared
is
often
used
to
summarize
the
fit
as
it takes into account the the number of variables in the model.
Adjusted R-squared = 1 - Mean Square Error /Total Mean Square
where
Mean
Square
Error
is
2
from
the
regression
model
and
the
Total
mean
square
is
the
sample
variance
of
the
response
(
s
Y
2
2
is
a
good
estimate
if
all
the
regression
coefficients are 0). For this example,
Adjusted R-squared = 1 - 0.65^2/ 1.034 = 0.59.
Intercept
:
the
intercept
in
a
multiple
regression
model
is
the
mean
for
the
response
when
all of the explanatory variables take on the value 0.
In
this
problem,
this
means
that
the
dummy
variable
I
=
0
(code
=
1,
which
was
the
queen
bumblebees)
and
log(duration)
=
0,
or
duration
is
1
second.
For
queenbumblebees,
with
visits
of
1
second,
we
are
95%
confident
that
the
mean
logit(proportion
of
pollen
removed)
is
between
2.71
2.02
0.38
or
between
-
3.49
to
-
1.93.
The
Student
t
quantile 2.02 is based on 44 degrees of freedom; qt(.975, 44).
To
convert
back
to
the
original
units,
we
can
take
the
inverse
of
the
logit
transformation.
I.e.
if
logit(p)
=
log(p/(1-p)),
then
p
=
exp(x)/(1
+
exp(x)).
To
get
the
confidence
interval
pf3
pf4

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Interpretation in Multiple Regression

Topics:

  1. R-squared and Adjusted R-squared
  2. Interpretation of parameter estimates
  3. Linear combinations of parameter estimates

variance-covariance matrix

standard errors of combinations

standard error for the mean

We will use the final model from last time to illustrate these concepts. Summaries of the model - least squares estimates with standard errors given below in parentheses:

logit

proportion   2.71  0.89  log  duration 0.57  I

= 0.65 with 44 degrees of freedom R-squared = 0. R-squared and Adjusted R-squared : The R-squared value means that 61% of the variation in the logit of proportion of pollen removed can be explained by the regression on log duration and the group indicator variable. As R-squared values increase as we ass more variables to the model, the adjusted R-squared is often used to summarize the fit as it takes into account the the number of variables in the model. Adjusted R-squared = 1 - Mean Square Error /Total Mean Square where Mean Square Error is 2 from the regression model and the Total mean square is the sample variance of the response ( sY

is a good estimate if all the regression coefficients are 0). For this example, Adjusted R-squared = 1 - 0.65^2/ 1.034 = 0.59. Intercept : the intercept in a multiple regression model is the mean for the response when all of the explanatory variables take on the value 0. In this problem, this means that the dummy variable I = 0 (code = 1, which was the queen bumblebees) and log(duration) = 0, or duration is 1 second. For queenbumblebees, with visits of 1 second, we are 95% confident that the mean logit(proportion of pollen

removed) is between  2.71  2.02  0.38 or between - 3.49 to - 1.93. The Student t

quantile 2.02 is based on 44 degrees of freedom; qt(.975, 44). To convert back to the original units, we can take the inverse of the logit transformation. I.e. if logit(p) = log(p/(1-p)), then p = exp(x)/(1 + exp(x)). To get the confidence interval

for the proportion just apply the inverse transformation. So for queen bumblebees with visits lasting 1 second, we are 95% confident that the mean proportion of pollen removed is between 0.03 and 0.13. [exp(-3.49)/(1 + exp(-3.49)) to exp(-1.93)/(1 + exp(- 1.93))] Note: while this is the interpretation of the intercept, we are extrapolating. Regression Coefficients : Typically the coefficient of a variable is interpreted as the change in the response based on a 1-unit change in the corresponding explanatory variable keeping all other variables held constant. In some problems, keeping all other variables held fixed is impossible (i.e. A quadratic model, or the model with different slopes for queen and worker bees). For this example, we have the estimated coefficient of log(duration) is 0.89. Because we have taken the log transformation of duration, the interpretation of the coefficient is easier to understand by looking at a doubling of duration (review page 208 chapter 8). A doubling of the duration of visit corresponds to a β 1 log(2) change in the mean logit(proportion of pollen removed) or 0.89*log(2) = 0.62. The 95% confidence interval

for β 1 is 0.89  2.02  0.14 or 0.61 to 1.17. The interval under the doubling of

duration is obtained by multiplying this interval by log(2). So a 95% confidence interval for the change in the mean logit(proportion pollen removed) is 0.42 to 0.81. Further simplification is not possible. Dummy variable coefficients: A 1 unit change for a dummy variable implies going from level 0 to level 1, so the the interpretation of the dummy variable coefficient is the amount by which the mean logit(proportion) for worker bees exceeds the mean logit(proportion) for queen bumble bees. i.e the logit of the proportion pollen removed for worker bees is 0.56 higher than the logit for queen bumble bees. A 95% confidence interval for the amount is 0.09 to 1.05. (this is the case for parallel regression lines; if we still had the interaction variable we could not make this statement, since the interaction of the dummy*log(duration) cannot be held constant). In the model derivation, we said that the intercept plus the dummy variable coefficient corresponded to the intercept for the worker bees, which is estimated as - 2.71 + .57 or - 2.14. This can be translated ac to the original scale as we did the intercept for the queen bumble bees. As this has a more interesting meaning, let's find a confidence interval for β 0 + β 2. To do this we need to find the standard error for a linear combination.

Linear Combination of Parameters

To find the variance (and then standard deviation) of the estimator of β 0 + β 2 we need to take into account the individual variances plus how the estimates will vary together from sample to sample (their covariance). The variance of the sum is the sum of the variances plus 2 times the covariance. We can get the covariance from the correlation of the estimates (recall the correlation is the covariance divided by the product of the standard deviations, so the covariance is the correlation times the product of the standard deviations. Since the standard deviations are unknown, we use the estimated covariance matrix calculated using the standard errors. In the Results options for Regression, check

The variance of the mean at this point is found by 

i  0

p

j  0

p

cov!

i ,!

j # Ci C j

which in this case simplifies to

var!

0 #$ 1 % var!

1 #log 2

2

2 $ cov!

1 #$ 1 $ log 2 #& 0.

For more details see section 10.4.3 and exercises 21-23. This is how the se.fit variable is obtained. From the SE(mean) we can get the SE(prediction)

SE prediction Y ' X ( x ))( * Xhat +

SE -. / Y 0 X 1 x 2

2

To get a prediction interval first calculate the prediction interval in the logit scale, then transform the interval using the inverse transformation applied to each endpoint of the interval. Putting this all together we can find the estimates and prediction intervals in the original units. 10 30 50 70 duration of visit (seconds)

Proportion pollen removed code=1 Queens code=2 Workers Estimated Mean for Queens Estimated Mean for Workers 95% Prediction Intervals Queens 95% Prediction Intervals Workers Proportion of Pollen Removed for Queen Bumblebees and Worker Honeybees