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Simple Linear Regression: Homework Solutions - Prof. Rhonda R. Decook, Assignments of Statistics
Solutions to homework problems on simple linear regression. It covers topics such as the formula for variance, better estimation of the slope when the relationship is more apparent, and the relationship between pcb levels in fish and their age with some non-linear relationship. The document also includes r code for data analysis and visualization.
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22s:
Homework 3: Solutions
Assigned Wednesday, September 17
Due Wednesday, September 24 at classtime
Simple linear regression
1. The formula for the variance:
σ
(
n
+
x¯
∑
n
i=
(x
i
− x¯)
)
=
σ
∑
n
i=
x
i
n
∑
n
i=
(x
i
− x¯)
Looking at this formula, we see that the variability in the estimated intercept is small-
est when ¯x = 0. If ¯x = 0, then we have x-values above and below x=0, and we can get
a pretty good estimate of
ˆ
Y
i
when x
i
= 0, which is in the ‘middle’ of the data. If most
of the x-values are far away from x=0 (as in the pictures below), it is much harder to
do a good job at estimating the
ˆ
Y
i
when x
i
= 0 (i.e. the estimated intercept)... we
don’t have a lot of information near x=0. Depending on which sample you draw, the
place where the fitted line crosses the y-axis can change quite a bit.
0
x
y
slope= 1.
interc= 16.
0
x
y
slope= 2.
interc= 12.
0
x
y
slope= 2.
interc= 13.
0
x
y
slope= 1.
interc= 17.
The above 4 plots show a random sample of size 40 drawn from the same X,Y distri-
bution with a known linear relationship (β 0 =15, β 1 = 2). Though the estimated slopes
are all fairly close, the intercepts have a wider range of estimated values.
2. The slope is better estimated when the relationship is easier to detect, or visually more
apparent. Recall the example we did in class where we took a subset of 20 data points
from a full set of 100 points that showed a strong relationship between x and y. If
we took the middle 20 points, the relationship between x and y was less apparent, the
range of the y-values in that subset was about -5 to 20, but the ˆσ (estimated variability
around the fitted line) was about the same as if we had considered the whole data set.
If we took the outside most 20 data points, the range of the y-values went from -30 to
35 (much larger), and again, ˆσ was about the same. The relationship between x and y
is much more apparent in this second set of subsetted points (containing the extreme
x-values). The fitted line in this second set explains a large % of the variability in the
y-values. The x-values in this second set are much more spread out than in the first
subsetted set that only included the middle data points.
−10 −5 0 5 10
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0
All the data
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Most extreme 20 points
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When there is no variability in X, all the points fall on a vertical line and the slope
is undefined. Mathematically we can see this in the formula because we would be
dividing by 0.
3. This problem is on the relationship between PCB levels in fish and their age.
(a) It looks like there’s some curvature in this relationship. It looks non-linear.
> PCB.data=read.csv("PCB_in_fish.csv")
> attach(PCB.data)
> plot(PCB~Age) ## or you could use plot(Age,PCB)
0
Age
PCB
(b) Some output first:
> lm.out=lm(PCB~Age)
> lm.out$coefficients
(Intercept) Age
-1.451944 1.
> par(mfrow=c(1,2))
> plot(lm.out$fitted.values,lm.out$residuals)
> abline(h=0)
> qqnorm(lm.out$residuals)
> qqline(lm.out$residuals)
-
0
lm.out$fitted.values
lm.out$residuals
-2 -1 0 1 2
-
0
Normal Q-Q Plot
Theoretical Quantiles
Sample Quantiles
If we ignored the curvature in the relationship and went ahead and fit a SLR, we
definitely have some violations of the error assumptions. It looks like variability
increases with the mean (or fitted value), so we’re violating the constant variance
assumption. Also, the qq-plot suggests there is some concern for non-normality,
specifically, there are some unusually large residuals (both positive and negative).
(c) Apply the log
e
transformation...
> plot(log(PCB)~Age,pch=16)
> y.new=log(PCB)
> lm.out=lm(y.new~Age)
> lm.out$coefficients
(Intercept) Age
0.03147247 0.
0
Age
log(PCB)
0.5 1.0 1.5 2.0 2.5 3.
-0.
0.
0.
lm.out$fitted.values
lm.out$residuals
-2 -1 0 1 2
-0.
0.
0.
Normal Q-Q Plot
Theoretical Quantiles
Sample Quantiles
After transformation, the relationship looks fairly linear (first plot), and the con-
stant variance assumption looks pretty good (second plot). There’s a slight de-
parture from normality for the largest residuals, but nothing falls too far from the
line (like it did in the untransformed data).
(d) > summary(lm.out)
Call:
lm(formula = y.new ~ Age)
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.03147 0.20136 0.156 0.
Age 0.25913 0.03080 8.414 6.78e-09 ***
---
Signif. codes: 0 *** 0.001 ** 0.01 * 0.05. 0.1 1
Residual standard error: 0.567 on 26 degrees of freedom
Multiple R-squared: 0.7314,Adjusted R-squared: 0.
F-statistic: 70.8 on 1 and 26 DF, p-value: 6.78e-
Age is a very significant linear predictor for log(PCB). The t-statistic is 8.
with a p-value of 6. 78 × 10
− 9
(e) A 1 year increase in the Age of a fish is associated with a 0.25913 increase in the
mean of the logged-PCB level in fish.
> detach(PCB.data)
4. The Robey data set...
(a) > attach(Robey)
> plot(contraceptors,tfr,pch=16)
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tfr