Problem 3:
Student GPA ACT
1
2.8
21
2
3.4
24
3
3
26
4
3.5
27
5
3.6
29
6
3
25
7
2.7
25
8
3.7
30
i.
Slope
Intercept
R^2
0.1021978022 0.5681318681 0.5774237913
Using the data above, the OLS estimates are:
GPA = 0.56813 + 0.10220 × ACT
Direction of the relationship:
•
The slope is positive (≈ 0.1022), so GPA increases as ACT increases. In plain
terms: higher ACT scores are associated with higher GPAs in this sample.
Does the intercept have a useful interpretation?
•
The intercept (≈ 0.568) is the predicted GPA when ACT = 0. This is not meaningful
here because ACT = 0 is unrealistic for college applicants (ACT scores in practice
are far from zero). Since an ACT score of 0 is not realistic, the intercept doesn’t
have a meaningful interpretation in practice. It simply ensures the regression line
fits the data properly within the observed ACT range (20–30).
How much higher is predicted GPA if ACT increases by 5 points?
•
Multiply the slope by 5: 5 × 0.1021978 ≈ 0.51099. So a 5-point increase in ACT
raises predicted GPA by ≈ 0.511 grade points.
ii.
Predicted
2.714285714
3.020879121
3.225274725
3.327472527
3.531868132
3.123076923
3.123076923
3.634065934
Sum of Residuals
Residual
0.08571428571
0.3791208791
-0.2252747253
0.1725274725
0.06813186813
-0.1230769231
-0.4230769231
0.06593406593
0
So, the residuals sum (to numerical precision) is zero, as OLS theory requires.
iii.
Predicted GPA when ACT = 20
Use the estimated equation:
GPA = 0.56813 + 0.10220 × 20 = 0.56813 + 2.044 = 2.61213
So predicted GPA ≈ 2.6121.
iv.
How much of the variation in GPA is explained by ACT?
R² = 0.5774238 → 57.74%.
Interpretation: About 57.74% of the variation in GPA across these 8 students is
explained by the linear relationship with ACT (i.e., by the regression lineThe
remaining 42.3% is due to other factors (study habits, motivation, background,
etc.) or random variation. This shows that there is a moderately positive linear
relationship between GPA and ACT.
Problem 4:
̂ = 119.77 − 0.514 × 𝑐𝑖𝑔𝑠
𝑏𝑤𝑔ℎ𝑡
(i) Predicted birth weight when cigs = 0 and when cigs = 20
When cigs = 0:
̂ = 119.77 − 0.514(0) = 119.77 ounces
𝑏𝑤𝑔ℎ𝑡
When cigs = 20:
̂ = 119.77 − 0.514(20) = 119.77 − 10.28 = 109.49 ounces
𝑏𝑤𝑔ℎ𝑡
Comment:
When the mother doesn’t smoke, the predicted birth weight is 119.77 ounces.
If she smokes one pack per day (20 cigarettes), the predicted birth weight drops
to 109.49 ounces, which is about 10.28 ounces lower.
This means that, according to the model, each additional cigarette per day
reduces birth weight by about 0.514 ounces, and smoking 20 per day lowers it by
roughly 10 ounces which is a substantial difference.
•
•
•
•
ii) Does this regression necessarily show a causal relationship?
No, not necessarily. This regression shows a statistical association, not
necessarily causation.
There might be omitted variable bias i.e. other factors could affect both smoking
and birth weight, such as:
Mother’s nutrition and income
Access to prenatal care
Education level
Age or general health
If these are not controlled for, the estimated effect of smoking might partly
capture the influence of these other variables.
Therefore, we cannot conclude causality just from this simple regression.
(iii) Predict the number of cigarettes for a birth weight of 125 ounces
We solve for cigs when predicted bwght = 125:
125 = 119.77 − 0.514(𝑐𝑖𝑔𝑠)
0.514(𝑐𝑖𝑔𝑠) = 119.77 − 125 = −5.23
−5.23
𝑐𝑖𝑔𝑠 =
≈ −10.17
0.514
This means the model would predict a birth weight of 125 ounces only if cigs = 10.17, which is impossible (a mother can’t smoke negative cigarettes).
Comment:
This shows that the linear model is only valid within a realistic range of data.
It’s not designed for extrapolation. We can’t use it to predict birth weights higher
than the sample’s general range.
So, predicting 125 ounces using this equation doesn’t make sense in practice.
iv) About 85% of women in the sample do not smoke — does that reconcile
part (iii)?
Yes, it helps explain it.
Since most women (≈85%) don’t smoke, the average predicted birth weight (near
119.77) reflects non-smokers. The regression line is based on a sample where
almost everyone smokes 0 or a small number of cigarettes, so predicting a
weight of 125 (which is above the average even for non-smokers) falls outside the
realistic range of cigs. That’s why the equation gives an impossible negative
number of cigarettes. It’s simply not meant to predict that far outside the data
range.
Problem 5:
3. (i) cons = -124.84 + 0.853inc
The intercept in this equation is B0 = -124.84
This is the model’s predicted consumption at income = $0. The negative sign is not
literal (consumption can’t be negative); it instead reflects extrapolation artifacts of the
linear model (very small magnitude relative to typical incomes). The negative sign
suggests that, hypothetically, a family with zero income is predicted to have a
consumption of −124.84 dollars. The magnitude is relatively small compared to
typical annual incomes and consumption levels. In reality, consumption cannot be
negative. A negative consumption figure, in this context, suggests that a family with
zero income would likely dissave (borrow money or use savings) to consume $124.84
worth of goods and services. However, it's often more accurately viewed as a
mathematical extrapolation outside the typical range of the data, especially since zero
income is likely rare in the sample of 100 families. The linear model might not hold
precisely at inc=0.
(ii) -124.84 + 0.853(30,000) = $25,465.16
The predicted consumption when family income is $30,000 is $25,465.16
(iii)
Problem 6
Given:
log(price) = 9.40 + 0.312 log(dist)
𝑛 = 135, 𝑅 2 = 0.162.
(i) Interpret the coefficient on 𝐥𝐨𝐠⁡(dist). Is the sign what you expect?
•
The coefficient 0.312 is an elasticity: it means that a 1% increase in distance from the
incinerator is associated with an approximate 0.312% increase in house price, ceteris
paribus (according to the model).
•
Equivalently, a 10% increase in distance → price increases by about 0.312 × 10% =
3.12%.
•
Sign: The coefficient is positive, which matches intuition: houses that are farther
from a garbage incinerator (less exposure to bad smells, noise, visual disamenity)
should be valued higher. So the sign is what we would expect.
(ii) Does simple regression give an unbiased estimate of the ceteris paribus
elasticity?
•
Not necessarily. The simple (bivariate) log–log regression may suffer from omitted
variable bias (OVB) or endogeneity because the location of the incinerator and the
spatial distribution of house characteristics are not random.
•
Possible mechanisms of bias:
•
o
If the city placed the incinerator near lower-value neighborhoods (to avoid
political resistance from wealthier areas), then distance will be positively
correlated with unobserved measures of neighborhood quality (e.g., schools,
safety). Those omitted positive determinants of price make the OLS
coefficient on log(dist) upward biased — it will overstate the pure effect of
distance itself.
o
Conversely, if the incinerator were sited for geographic/technical reasons that
correlate negatively with house quality, bias direction could differ. But the
usual concern is that placement is non-random and correlated with
unobservables that affect price.
Therefore, the OLS estimate likely does not equal the true causal (ceteris paribus)
elasticity without additional controls or an identification strategy (controls, fixed
effects, IV, natural experiment).
(iii) Other house factors that affect price and their likely correlation with
distance
Important omitted variables (house & neighborhood attributes):
•
Structural/house characteristics: lot size, square footage, number of
bedrooms/bathrooms, age/condition of house, presence of garage, heating/airconditioning, recent renovations.
•
Neighborhood characteristics: school quality, crime rates, access to parks/green space,
proximity to employment centers, public transit access, local amenities (shops,
restaurants), noise/traffic levels, property taxes, zoning.
•
Environmental factors: air quality, flood risk, topography, views.
•
Socioeconomic factors: median income of neighborhood, population density,
neighborhood trends (gentrification), historical preservation status.
How they may correlate with distance from incinerator:
•
If wealthier, better-served neighborhoods tend to be further from the incinerator, then
these omitted positive attributes (better schools, larger lots, nicer houses) are
positively correlated with distance. Omitting them will inflate the estimated effect of
distance (upward bias).
•
Conversely, if large lots or industrial land happen to be closer, some specific attributes
could be negatively correlated — context matters — but the dominant concern in this
literature is that lower-quality areas were chosen for disamenities, producing positive
correlation between distance and unobserved quality.
Problem 8
Define the estimator that forces the intercept to zero:
𝛽̃1 =
∑𝑛𝑖=1 𝑥𝑖 𝑦𝑖
.
∑𝑛𝑖=1 𝑥𝑖2
∑(𝑥𝑖 −𝑥ˉ)(𝑦𝑖 −𝑦ˉ)
(The usual OLS slope with an intercept is 𝛽̂1 = ∑(𝑥
.)
−𝑥ˉ)2
𝑖
̃ 𝟏 ]and cases of unbiasedness
(i) 𝑬[𝜷
Compute the expectation (use 𝐸[𝑦𝑖 ] = 𝛽0 + 𝛽1 𝑥𝑖 ):
𝐸[𝛽̃1 ]
∑𝑛𝑖=1 𝑥𝑖 𝐸[𝑦𝑖 ] ∑𝑛𝑖=1 𝑥𝑖 (𝛽0 + 𝛽1 𝑥𝑖 )
=
∑𝑛𝑖=1 𝑥𝑖2
∑𝑛𝑖=1 𝑥𝑖2
∑𝑛𝑖=1
𝛽0 ∑𝑛𝑖=1 𝑥𝑖 + 𝛽1 ∑𝑛𝑖=1 𝑥𝑖2
=
= 𝛽1 + 𝛽0 𝑛
∑𝑛𝑖=1 𝑥𝑖2
∑𝑖=1
=
𝑥𝑖
.
𝑥𝑖2
So
𝐸[𝛽̃1 ] = 𝛽1 + 𝛽0
∑𝑥𝑖
.
∑𝑥𝑖2
Unbiasedness conditions.
•
If the population intercept 𝛽0 = 0, then the second term is zero and 𝐸[𝛽̃1 ] = 𝛽1. Thus
𝛽̃1is unbiased when 𝛽0 = 0.
•
There is a second special case: if the sample satisfies ∑𝑛𝑖=1 𝑥𝑖 = 0(i.e. 𝑥ˉ = 0), then
𝐸[𝛽̃1 ] = 𝛽1regardless of 𝛽0. So 𝛽̃1is also unbiased whenever ∑𝑥𝑖 = 0.
•
In general (when 𝛽0 ≠ 0and ∑𝑥𝑖 ≠ 0) 𝛽̃1is biased.
̃𝟏
(ii) Variance of 𝜷
Write 𝛽̃1decomposed into deterministic part plus error term:
∑𝑥𝑖 (𝛽0 + 𝛽1 𝑥𝑖 + 𝑢𝑖 )
∑𝑥𝑖 ∑𝑥𝑖 𝑢𝑖
𝛽̃1 =
=
𝛽
+
𝛽
+
.
1
0
∑𝑥𝑖2
∑𝑥𝑖2
∑𝑥𝑖2
∑𝑥 𝑢
The only random part is ∑𝑥𝑖 2 𝑖 . Using Gauss–Markov assumptions (uncorrelated errors
𝑖
with variance 𝜎 2 ):
𝑛
∑𝑥𝑖 𝑢𝑖
1
1
Var⁡ (𝛽̃1 ) = Var⁡ (
)
=
Var⁡
(∑𝑥
𝑢
)
=
∑
𝑖
𝑖
∑𝑥𝑖2
(∑𝑥𝑖2 )2
(∑𝑥𝑖2 )2
𝑥𝑖2 𝜎 2
𝑖=1
=
𝜎
∑𝑛𝑖=1
2
𝑥𝑖2
.
As hinted, this variance does not depend on 𝛽0.
̃ 𝟏 )vs 𝐕𝐚𝐫⁡(𝜷
̂ 𝟏)
(iii) Compare variances: 𝐕𝐚𝐫⁡(𝜷
The usual OLS slope variance (with intercept included) is
Var⁡(𝛽̂1 ) =
∑𝑛𝑖=1
𝜎2
.
(𝑥𝑖 − 𝑥ˉ)2
But
𝑛
𝑛
𝑛
𝑛
∑
𝑥𝑖2 = ∑
𝑖=1
(𝑥𝑖 − 𝑥ˉ + 𝑥ˉ)2 = ∑
(𝑥𝑖 − 𝑥ˉ)2 + 𝑛𝑥ˉ 2 = ∑
𝑖=1
𝑖=1
𝑖=1
(𝑥𝑖 − 𝑥ˉ )2 + 𝑛𝑥ˉ 2 .
Hence ∑𝑥𝑖2 ≥ ∑(𝑥𝑖 − 𝑥ˉ)2, with strict inequality unless 𝑥ˉ = 0(or all 𝑥𝑖 = 0).
Therefore
Var⁡(𝛽̃1 ) =
𝜎2
𝜎2
≤
= Var⁡(𝛽̂1 ),
∑𝑥𝑖2 ∑(𝑥𝑖 − 𝑥ˉ)2
with equality only when 𝑥ˉ = 0(i.e. ∑𝑥𝑖 = 0). So Var⁡(𝛽̃1 )is always less than or equal to
Var⁡(𝛽̂1 )(strictly smaller except the special 𝑥ˉ = 0case).
Intuition: forcing the intercept to zero reduces estimator variance because it effectively uses
an extra (but possibly incorrect) restriction that reduces sampling variability.
(iv) Bias–variance tradeoff
•
𝛽̃1(no-intercept estimator) often has smaller variance than 𝛽̂1(the usual OLS slope),
because its variance denominator is ∑𝑥𝑖2 , which is at least as large as ∑(𝑥𝑖 − 𝑥ˉ)2 .
•
However, 𝛽̃1 is generally biased unless the true intercept 𝛽0 = 0or the sample has
∑𝑥
mean 𝑥ˉ = 0. When 𝛽0 ≠ 0and ∑𝑥𝑖 ≠ 0, the bias amount equals 𝛽0 ∑𝑥 2𝑖 .
𝑖
•
So the choice between 𝛽̂1and 𝛽̃1is a classical bias–variance tradeoff:
o
Use 𝛽̂1(include intercept) when you want an unbiased estimator (under the
usual model) and to avoid model misspecification from forcing an incorrect
intercept; its variance is larger, but it is unbiased.
o
Use 𝛽̃1(force intercept = 0) only when you have strong reason (theory or
evidence) that 𝛽0 = 0(so no bias), or when you deliberately accept some bias
in exchange for lower variance because the mean-squared error (MSE) of
𝛽̃1may be smaller. Concretely, 𝛽̃1will have smaller MSE than 𝛽̂1if
Bias⁡(𝛽̃1 )2 + Var⁡(𝛽̃1 ) < Var⁡(𝛽̂1 ),
i.e. if the reduction in variance outweighs the square of the bias introduced.
•
Practical rule: unless you have a convincing theoretical reason that the intercept is
zero (or you have centered 𝑥so 𝑥ˉ = 0), prefer 𝛽̂1with an intercept to avoid bias.
Computer Exercises: C1 to C11
C1 Answers
(i)
(ii)
Average Participation Rate
44.02626263
Average Match Rate
0.35727273
Correlation
0.03206728286
S.D (mrate)
0.3941402894
S.D (prate)
12.22332653
B1
0.9944907431
B0
43.67095821
Sample size (n)
R^2
prate = 43.67 + 0.994(mrate)
(iii)
The intercept (43.67) represents the estimated
participation rate when the match rate is zero. In
other words, if employers offer no matching at all,
the model predicts that about 43.67% of eligible
employees will still participate in the plan. This
suggests a fairly high baseline participation rate even
without financial incentives.
Coefficient on mrate (0.994):
Each one-unit increase in the match rate — meaning
every additional dollar the employer contributes for
each dollar saved by the employee — is associated
with roughly a 0.994 percentage point rise in the
participation rate, which is effectively a 1% increase.
For instance, if a firm raises its match rate from
$0.50 to $1.50 per employee dollar, the model
predicts participation will rise by about 1
percentage point.
99 observations
0.00102831063
̂ = 43.67 + 0.994(3.5) = 47.15
𝑝𝑟𝑎𝑡𝑒
(iv)
The predicted participation rate at 𝑚𝑟𝑎𝑡𝑒 = 3.5is
47.15%, which is within the realistic range of 0–
100%. However, since 𝑚𝑟𝑎𝑡𝑒 = 3.5exceeds the
observed sample maximum of 3.0, this prediction
involves extrapolation beyond the available data.
Therefore, the linear relationship may not hold
exactly at such high match rates. Still, because the
slope is modest, the resulting prediction remains
reasonable and not extreme. The very low 𝑅2 value
(≈0.001) indicates that the match rate explains
almost none of the variation in participation rates,
suggesting that employer matching alone has
minimal predictive power for employee
participation in this sample.
R^2 = 0.00102831063, meaning only 0.1% of the
variation in participation rates is explained by the
match rate.
(v)
This is an extremely low explanatory power:
essentially, almost all variation in participation
remains unaccounted for by the match rate. In
practical terms, this means the match rate has
almost no meaningful linear relationship with
participation in this dataset.
C2 Answers
(i)
(ii)
(iii)
Average salary
876.4949495
Average tenure
7.767676768
CEOs in their first year as CEO
1
Longest tenure as a CEO
37
Mean (lsalary)
6.60593298
Mean (ceoten)
7.767676768
Correlation
0.1725242071
S.D (lsalary)
0.5996215402
S.D (ceoten)
6.425123334
B1
0.01610073853
B0
6.480867647
Sample size (n)
99
R^2
0.02976460204
log(salary) = 6.48 + 0.016(ceoten) +u
Because this is a log-level model, the coefficient
𝛽1 indicates the percentage change in salary
associated with a one-unit change in the
independent variable. Thus, for each additional
year a person serves as CEO, their salary is
expected to rise by approximately 1.61%. In other
words, holding other factors constant, one more
year of CEO experience corresponds to about a
1.61% increase in salary.
C3 Answers
(i)
Mean (sleep)
3177.737374
Mean (totwrk)
2759.717172
Correlation
-0.4675282186
S.D (sleep)
455.6711038
S.D (totwrk)
2759.717172
B1
-0.07719599008
B0
3390.776473
Sample size (n)
R^2
sleep = 3390.78 - 0.077(totwrk) + u
99 observations
0.2185826352
The intercept represents the estimated sleep time
(in minutes per week) when totwrk = 0. This means
that someone who doesn't work at all during the
week would sleep approximately 3,391 minutes
per week
(ii)
2 hours = 120 minutes
B1 x 120 = -0.077 x 120 = -9.24 minutes
If total work time increases by 2 hours (120
minutes), the model predicts that weekly sleep
decreases by only about 9.2 minutes. This effect is
quite small — working an additional 2 hours per
week results in less than a 10-minute reduction in
sleep. The trade-off ratio is roughly 13:1, meaning
that for every 13 minutes spent working, only 1
minute of sleep is lost. This implies that when
people work longer hours, they tend to cut back
more on other activities (such as leisure,
television, or hobbies) rather than on sleep.
Although the 𝑅2 value of 21.9% indicates a
moderate relationship between work and sleep,
the actual effect size is minimal, suggesting that
individuals generally maintain their sleep
schedules even as work demands increase.
C4 Answers
(i)
(ii)
Average wage
1094.646465
Average IQ
107.1717172
Sample S.D (IQ)
17.84019975
S.D (IQ)
14.33494404
S.D (wage)
430.6445173
Correlation
0.337993022
B1
10.15384793
B0
6.441146228
wage = 6.44 + 10.15(IQ) 10.15 x 15 = $152.25
increase in monthly wage
R^2
0.1142392829
No, IQ explains only about 11.4% of the variation in
wage. The remaining 88.6% is explained by other
factors.
(iii)
Mean (lwage)
6.929032202
S.D (lwage)
0.3697091966
Correlation
0.3545321676
B1
0.009143656405
B0
5.949090844
log(wage) = 5.95 + 0.0091(IQ)
The coefficient β₁ = 0.0091 means that each onepoint increase in IQ is associated with
approximately a 0.91% increase in wage. 15 x
0.91% = 13.65% - A 15-point increase in IQ is
associated with approximately a 13.65% increase
in predicted wage.
C5 Answers
(i)
log(rd) = β₀ + β₁log(sales) + u
where β₁ is the elasticity of rd with respect to
sales. This is because in a log-log model:
Elasticity = ∂log(rd)/∂log(sales) = β₁. β₁
represents the percentage change in rd for a 1%
change in sales. This elasticity is constant
(doesn't depend on the level of sales).
(ii)
Mean (lrd)
3.602824956
Mean (lsales)
7.164942438
S.D (lrd)
1.734351828
S.D (lsales)
1.537853394
Correlation
0.9538524534
B1
1.075730464
B0
-4.104721899
log(rd) = -4.105 + 1.076log(sales) + u
Estimated elasticity: β₁ = 1.076
A 1% rise in sales is linked to an estimated 1.076%
increase in R&D spending. Because the elasticity
is greater than 1, R&D expenditures grow more
than proportionally with sales. This indicates that
larger chemical firms allocate a greater share of
their revenue to research and development
compared to smaller firms. In other words, as
companies expand, they tend to intensify their
investment in R&D relative to their overall size.
C6 Answers
(i)
A diminishing effect is more realistic in this
context. At lower levels of spending, each
additional dollar can significantly improve
outcomes by funding essential resources such as
qualified teachers, textbooks, and proper facilities.
In contrast, when spending is already high and
schools have sufficient resources, extra funding
yields smaller marginal gains in student
performance. Moreover, because pass rates are
naturally limited to the 0–100% range, the impact
of spending must taper off as performance nears
these boundaries—otherwise, the model would
predict impossible pass rates exceeding 100%. A
logarithmic specification reflects this pattern of
diminishing returns, as it assumes that
proportional (percentage) increases in spending
have consistent effects on pass rates, regardless
of the initial spending level.
(ii)
Starting with: Δmath10 ≈ β₁ × Δlog(expend). For
a 10% increase in expend: New expend = 1.10 ×
expend. Δlog(expend) = log(1.10 × expend) log(expend) = log(1.10) ≈ 0.10. Therefore:
Δmath10 ≈ β₁ × 0.10 = β₁/10. So β₁/10 gives the
percentage point change in math10 when
spending increases by 10%.
(iii)
Mean (math10)
30.1010101
Mean (lexpend)
8.453193182
S.D (math10)
10.13693484
S.D (lexpend)
0.1728772173
Correlation
0.2195737746
B1
12.8750629
B0
-78.73438381
Sample size (n)
R^2
math10 = -78.73 + 12.88 log(expend) + u
(iv)
The estimated coefficient is β₁ = 12.88. %Δmath10
= β₁/10 = 12.88/10 = 1.288%. A 10% increase in
spending per student is associated with
approximately a 1.3% increase in the math pass
rate. This is a modest but meaningful effect,
suggesting that while spending matters for student
performance, it explains only a small portion of the
variation in pass rates (other factors like student
background, teacher quality, curriculum, etc. play
larger roles).
99 observations
0.0482126425
(v)
This is not much of a worry in this dataset for
several practical reasons. First, the actual math
pass rates in the data range from a low of about
12% to a high of around 62%, meaning we're
nowhere near the upper bound of 100% even for
the best-performing schools. Second, the
spending levels in the dataset are relatively
modest, ranging from approximately $3,300 to
$7,000 per student (log values from about 8.11 to
8.86). For our fitted values to exceed 100%, we
would need -78.73 + 12.88 × log(expend) > 100,
which requires log(expend) > 13.91, or spending
above $1.1 million per student—an astronomically
high and completely unrealistic figure that's far
outside the range of our data. Within the actual
range of spending observed in Michigan schools,
the logarithmic specification naturally keeps
predictions well below 100%. The highest
predicted value in our sample would occur at the
maximum spending level, yielding a fitted value of
approximately -78.73 + 12.88(8.86) = 35.3%, which
is quite reasonable and far from any boundary
concerns.
C7 Answers
(i)
Average (gift)
% of people who didn't give a gift
(ii)
(iii)
Average (mailsyear)
0.9595959596
86.87%
2.419191919
Min (mailsyear)
2.25
Max (mailsyear)
2.5
S.D (gift)
2.713981908
S.D (mailsyear)
0.117522701
Correlation
-0.2342869098
B1
-5.410447761
B0
14.04850746
Sample size (n)
R^2
gift = 14.05 - 5.41(mailsyear) + u
(iv)
The slope coefficient of –5.41 implies that, on
average, each additional mailing per year is linked
to a decrease of 5.41 Dutch guilders in charitable
donations. This negative association suggests that
more frequent mailings tend to coincide with
lower contribution levels, possibly due to donor
fatigue, where individuals become less responsive
after receiving too many solicitations. Another
possible explanation is that the charity targets
fewer mailings toward loyal, high-value donors
while directing more frequent appeals to donors
who contribute smaller or less consistent
amounts.
As for whether the charity gains financially from
sending additional mailings, the answer appears to
be no. If each mailing costs 1 guilder and an
increase in mailing frequency corresponds to a
5.41-guilder decline in donations, the overall
expected net effect is –6.41 guilders (the 5.41guilder drop in contributions plus the 1-guilder
mailing cost). This indicates that, on average,
sending more mailings is associated with a net
financial loss for the charity.
However, this does not imply that the charity loses
money on every mailing. The coefficient reflects an
average effect across all individuals in the
sample. Some recipients may respond favorably to
additional mailings, while others may react
negatively or ignore them altogether. The
estimated coefficient of –5.41 represents the net
overall effect of these mixed responses.
Furthermore, since approximately 86.87% of
individuals in the sample made no donations
regardless of mailing frequency, the negative
coefficient likely captures differences in giving
behavior among the relatively small group of
actual donors, rather than a direct impact of
mailings on the overall decision to donate.
99 observations
0.05489035609
(v)
The smallest predicted charitable contribution
occurs at maximum mailsyear = 2.5 (since the
slope is negative). ŷ = 14.05 - 5.41(2.5) = 14.05 13.525 = 0.525 Dutch guilders. Secondly, no, you
cannot predict exactly zero for gift. To get ŷ = 0: 0 =
14.05 - 5.41(mailsyear). mailsyear = 14.05/5.41 =
2.597. Since mailsyear = 2.597 is outside the
sample range (max = 2.5), you never predict
exactly zero within the observed data.
C8 is solved in the excel sheet
C9 Answers
(i)
(ii)
Counties (zero murders)
1
Counties (at least 1 execution)
0
Largest number of executions
0
Mean (murders)
3.2
Mean (execs)
0
S.D (murders)
3.1144823
S.D (execs)
0
Correlation
#DIV/0!
B1 - undefined
#DIV/0!
B0
3.2
R^2
0
Sample size (n)
murders = 3.2 + u
5 counties
(iii)
The slope coefficient β₁ cannot be estimated
because there are no executions in the 1996
sample. Even if it could be estimated, a negative
coefficient would suggest a deterrent effect (more
executions associated with fewer murders), while a
positive coefficient would suggest the opposite.
Given the complete absence of executions in this
sample, no conclusion can be drawn about
deterrent effects.
(iv)
Smallest predicted number of murders = 3.2
Residual = 0 - 3.2 = -3.2
(v)
Simple regression analysis is insufficient for
assessing whether capital punishment deters
crime because it suffers from serious endogeneity
issues. The most significant concern is reverse
causality—areas with higher murder rates are
often more likely to implement the death penalty,
meaning executions may be a response to crime,
not a cause of its decline. Additionally, omitted
variable bias is unavoidable, as many factors—
such as poverty, law enforcement intensity,
demographics, drug prevalence, and broader
criminal justice policies—affect both the murder
rate and the use of capital punishment. Without
accounting for these confounding influences or
employing causal identification methods (like
instrumental variables or natural experiments),
any observed relationship between executions and
murders cannot be interpreted as evidence of
deterrence. The correlation detected by a simple
regression likely reflects selection effects and
shared underlying causes, rather than the true
causal impact of capital punishment on crime.
C10 Answers
(i)
Sample size (n)
99 students
Mean (math12)
52.00848485
S.D (math12)
9.498030125
Mean (read12)
51.56020202
(ii)
S.D (read12)
9.636046376
Correlation
0.6847907355
B1
0.6749825375
B0
17.20624885
n
99
R^2
math12 = 17.21 + 0.675(read12)
(iii)
The intercept value of 17.21 lacks practical
meaning in this context. It implies that a student
with a reading score of zero would be expected to
have a math score of 17.21, which is unrealistic.
No student in the sample has a reading score
close to zero—the lowest observed score is
around 30—and a reading score of zero is far
outside the data’s range. Since twelfth-grade
students would never realistically score zero on a
standardized reading test, interpreting this
intercept involves extreme extrapolation beyond
the observed sample and therefore has no
meaningful real-world interpretation.
(iv)
β̂₁ = 0.675 - This result is not surprising. The
positive slope indicates that reading and math
scores move together, which is expected because
both reflect general cognitive ability and academic
skills. Reading comprehension is necessary for
understanding math word problems. Students who
perform well academically tend to do well across
multiple subjects. Both skills benefit from similar
educational resources and student effort. The
magnitude (0.675) suggests that for every 1-point
increase in reading scores, math scores increase
by about 0.68 points on average—a substantial
positive relationship.
0.4689383514
R² = 0.469: This is also reasonable and not
surprising. It means that the reading scores
explain about 47% of the variation in math scores.
This shows a moderately strong relationship, but
far from perfect. The remaining 53% of variation is
due to other factors (specific math ability, different
teachers, effort in math vs. reading, etc.). While
reading and math are related, they are distinct
skills—math requires quantitative reasoning,
spatial ability, and specific mathematical
knowledge beyond reading comprehension. An R²
around 0.47 reflects the reality that these are
correlated but separate academic abilities.
(v)
I would say: "Your interpretation confuses
correlation with causation. The regression simply
shows that students who score well on reading
also tend to score well on math, but it doesn't
prove that improving reading will cause math
scores to improve. Consider this: if I reverse the
analysis and regress reading scores on math
scores, I would find a similarly strong positive
relationship—would that mean hiring math tutors
is the solution to improving reading scores? The
symmetry reveals the flaw in causal reasoning.
Both test scores are likely driven by common
underlying factors such as general cognitive ability,
socioeconomic background, overall academic
preparation, school quality, and student
motivation. Students who excel in one subject
tend to excel in others because of these shared
factors, not because proficiency in one subject
directly causes proficiency in another. To establish
whether reading tutors would actually improve
math scores, we would need a randomized
experiment or other causal research design, not
just an observed correlation between the two
scores."
C11 Answers
(i)
Sample size (n)
Average (colGPA)
99 students
3.065656566
Max (colGPA)
4
(ii)
Students who owned their own PC
42
(iii)
Mean (PC)
0.4242424242
S.D (colGPA)
0.3812336624
S.D (PC)
0.4967426363
Correlation
0.244759151
B1
0.1878446115
B0
2.985964912
colGPA = 2.99 + 0.188(PC) + u
The intercept (β₀ = 2.99) represents the average
college GPA of students who do not own a
personal computer. This means that, on average,
students without a PC achieve a GPA of
approximately 2.99. The slope coefficient (β₁ =
0.19) indicates that students who own a personal
computer have a GPA that is about 0.19 points
higher than students who do not own one, holding
all else constant. Although this difference is
positive, the magnitude of 0.19 is relatively small,
suggesting that computer ownership has only a
modest association with GPA.
(iv)
R^2
The R² value of 0.06 implies that only about 6% of
the variation in college GPA can be explained by PC
ownership. This low explanatory power confirms
that while there is a positive relationship between
having a PC and academic performance, other
unobserved factors—such as study habits,
socioeconomic status, or prior academic ability—
likely play a much larger role in determining
students’ GPA outcomes.
(v)
No, this is not a causal relationship — it’s
correlational. Owning a PC may be associated with
other factors that also influence GPA (for example,
income level, access to resources, or motivation).
Without random assignment or experimental
control, we cannot say PC ownership causes
0.05990704202
higher GPA.