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finished h2 and added R log

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caes 2017-02-02 23:47:24 -05:00
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3 changed files with 182 additions and 30 deletions
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hw2/.RData Normal file
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hw2/.Rhistory Normal file
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auto = read.table("auto.data",header=T,na.strings="?")
auto
attach(auto)
horsepower
plot(mpg ~ horsepower)
fit = lm(mpg ~ horsepower)
fit
summary(fit)
lines(fit)
abline(fit)
abline(fit,col="red")
plot(mpg ~ horsepower,pch="x")
abline(fit,col="red")
abline(fit,col="red",size="2")
abline(fit,col="red",lt="2")
help(abline)
abline(fit,col="red",lwd="2")
abline(fit,col="red",lwd="4")
summary(lm)
summary(fit)
help(predict)
predict(fit,horsepower=98)
predict(fit)
predict(fit,98)
predict(fit[98])
help(predict)
help(predict.lm)
predict(lm(mpg ~ 98)
predict(lm(mpg ~ horsepower)
predict(lm(mpg ~ 98))
pre
predict(fit,data.frame(p=98(
predict(fit,data.frame(p=98))
predict(fit,data.frame(p=c(98)))
predict(fit,data.frame(p=c(98)),interval="confidence")
predict(fit,data.frame(horesepower=c(98)),interval="confidence")
predict(fit,interval="confidence")
fit
predict(fit,data.frame(c(98))interval="confidence")
predict(fit,data.frame(p=c(98))interval="confidence")
predict(fit,data.frame(p=c(98)),interval="confidence")
predict(fit,data.frame(p=c(98)),interval="confidence")
names(fit)
coef(fit)
confint(fit)
predict(fit,data.frame(horsepower=c(1,98)),interval="confidence")
predict(fit,data.frame(horsepower=c(98)),interval="confidence")
predict(fit,data.frame(horsepower=98),interval="confidence")
predict(fit,data.frame(horsepower=98),interval="prediction")
dev.print(pdf,"mpg_horsepower_regression.pdf")
plot(fit)
par(mfrow=c(2,2))
plot(fit)
dev.print(pdf,"fit_quality.pdf")
save
save()
q()

153
hw2/answers
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@ -17,19 +17,19 @@
which of these predictors have a strong relationship with sales which of these predictors have a strong relationship with sales
of this product. of this product.
TV marketing and radio marketing both have a strong relationship TV marketing and radio marketing both have a strong relationship
to sales, according to their linear regression p-values, but to sales, according to their linear regression p-values, but
newspaper advertising does not appear to be effective, given newspaper advertising does not appear to be effective, given
that the linear model does not account for much of the variation that the linear model does not account for much of the variation
in sales across that domain. We can conclude that cutting back in sales across that domain. We can conclude that cutting back
on newspaper advertising will likely have little effect on the on newspaper advertising will likely have little effect on the
sales of the product, and that increasing TV and radio sales of the product, and that increasing TV and radio
advertising budgets likely will have an effect. Furthermore, we advertising budgets likely will have an effect. Furthermore, we
can see that radio advertising spending has a stronger can see that radio advertising spending has a stronger
relationship with sales, as the best-fit slope is significantly relationship with sales, as the best-fit slope is significantly
more positive than the best fit for TV advertising spending, so more positive than the best fit for TV advertising spending, so
increasing the radio advertising budget will likely be more increasing the radio advertising budget will likely be more
effective. effective.
@ -40,43 +40,70 @@
dollars). Suppose we use least squares to fit the model, and get dollars). Suppose we use least squares to fit the model, and get
β₀ = 50, β₁ = 20, β₂ = 0.07, β₃ = 35, β₄ = 0.01, β₅ = −10. β₀ = 50, β₁ = 20, β₂ = 0.07, β₃ = 35, β₄ = 0.01, β₅ = −10.
(a) Which answer is correct, and why? This is the model: ŷ = 50 + 20 X₁ + 0.07 X₂ + 35 X₃ + 0.01 X₄ +
i. For a fixed value of IQ and GPA, males earn more on -10 X₅
average than females.
ii. For a fixed value of IQ and GPA, females earn more on For fixed IQ and GPA, we can infer that the starting salary
average than males. for a female sharing an IQ and GPA with her male counterpart
will make (35*1 - 10*(GPA*1)) more starting salary units
than her male counterpart. This means that at very low
GPAs(maybe this includes people who didn't attend school?),
males have a lower starting wage, and as GPA grows, males
make a larger starting salary from that point, overtaking
females at GPA=3.5. Therefore,
iii. For a fixed value of IQ and GPA, males earn more on (a) Which answer is correct, and why? → iii. For a fixed value
average than females provided that the GPA is high enough. of IQ and GPA, males earn more on average than females
provided that the GPA is high enough.
iv. For a fixed value of IQ and GPA, females earn more on This one is correct.
average than males provided that the GPA is high enough.
(b) Predict the salary of a female with IQ of 110 and a GPA of (b) Predict the salary of a female with IQ of 110 and a GPA of
4.0. 4.0.
(c) True or false: Since the coefficient for the GPA/IQ ŷ = 50 + 20*4.0 + 0.07*110 + 35*1 + 0.01*(4.0*110) - 10*(4.0*1)
interaction term is very small, there is very little evidence of
an interaction effect. Justify your answer.
→ ŷ = 137.1 salary units
(c) True or false: Since the coefficient for the GPA/IQ
interaction term is very small, there is very little evidence of
an interaction effect. Justify your answer.
False. There is still a noticeable effect because the
coefficient for IQ's effect alone is only 7 times greater
than the coefficient of the interaction term. So, this term
holds significant weight compared to the overall
response of the model to IQ.
4. I collect a set of data (n = 100 observations) containing a 4. I collect a set of data (n = 100 observations) containing a
single predictor and a quantitative response. I then fit a linear single predictor and a quantitative response. I then fit a linear
regression model to the data, as well as a separate cubic regression model to the data, as well as a separate cubic
regression, i.e. Y = β₀ + β₁ X + β₂ X² + β₃ X³ + . regression, i.e. Y = β₀ + β₁ X + β₂ X² + β₃ X³ + ε.
(a) Suppose that the true relationship between X and Y is (a) Suppose that the true relationship between X and Y is
linear, i.e. Y = β₀ + β₁ X + . Consider the training residual linear, i.e. Y = β₀ + β₁ X + ε. Consider the training residual
sum of squares (RSS) for the linear regression, and also the sum of squares (RSS) for the linear regression, and also the
training RSS for the cubic regression. Would we expect one to be training RSS for the cubic regression. Would we expect one to be
lower than the other, would we expect them to be the same, or is lower than the other, would we expect them to be the same, or is
there not enough information to tell? Justify your answer. there not enough information to tell? Justify your answer.
For the training data, the cubic regression might return a
better RSS than the linear regression, but this would only
be because the cubic is fitting points that are varied
according to the ε random error. It also may not, depending
on how that random error expressed itself in this case.
(b) Answer (a) using test rather than training RSS. (b) Answer (a) using test rather than training RSS.
For the test error, the RSS will almost certainly be greater
for the cubic model than the linear model, because the
random error ε will likely express itself in a way that is
inconsistent with the noise that the cubic model adopted
during its training. The linear model will be more likely to
have a lower RSS the more test data is used against the
models.
(c) Suppose that the true relationship between X and Y is not (c) Suppose that the true relationship between X and Y is not
linear, but we don’t know how far it is from linear. Consider linear, but we don’t know how far it is from linear. Consider
the training RSS for the linear regression, and also the the training RSS for the linear regression, and also the
@ -84,3 +111,71 @@
lower than the other, would we expect them to be the same, or is lower than the other, would we expect them to be the same, or is
there not enough information to tell? Justify your answer. (d) there not enough information to tell? Justify your answer. (d)
Answer (c) using test rather than training RSS. Answer (c) using test rather than training RSS.
The cubic model will pick up more information because of its
additional degrees of freedom. If the true relationship is
more complex than linear, then the cubic model will likely
have a lower RSS over the linear model. If the model is less
complex than linear (E.G. perhaps it is just a constant
scalar relationship) then the linear model will still be
more likely to have a smaller RSS, because the cubic will
again pick up information from the ε noise that is not
inherent in the real relationship.
8. This question involves the use of simple linear regression on the Auto
data set.
(a) Use the lm() function to perform a simple linear regression
with mpg as the response and horsepower as the predictor. Use
the summary() function to print the results. Comment on the
output. For example:
There is definitely a correlation between horsepower and
mpg. The RSE is ~4.9, which is not insignificant and does
indicate that the response may not be truly linear, but it
is small enough relative to the mpg magnitude that it's
clear a relationship exists. The R² statistics corroborates
this by indicating (it has a small value at ~0.6) that a
large proportion of the mpg variability is explained by the
model. mpg has a negative correlation with horsepower,
indicated by the negative coefficient on the horsepower
factor.
For example, for a vehicle with 98 horsepower, one can
expect with 95% confidence that the mpg will be within 23.97
and 24.96, if the vehicles follow our model. However, after
incorporating the irreducible error, the prediction turns
out to be much less precise, with a 95% prediction interval
spanning 14.8 to 34.1. Some of this variability may also be
reduced by using a quadratic model, from visual inspection
of the plot.
(b) Plot the response and the predictor. Use the abline() function
to display the least squares regression line.
Attached.
(c) Use the plot() function to produce diagnostic plots of the least
squares regression fit. Comment on any problems you see with
the fit.
Attached. From these four plots it's clear there is a lot of
variability that remains unexplained by the linear model.
The standardized residuals plotted against the fitted values
shows clearly that the variability is strong, with values
consistenly lying outside 1 standardized residual unit, but
still within a tight range that doesn't extend past 3, which
is often considered an approximate threshold to indicate
values that aren't explained well by the model. There are
many points with high leverage, and these values have less
residual by default, of course, and in both of these graphs
we see a few points (323, 330) that are rearing their ugly
heads. These seems to be the bit of "uptick" toward the
higher end of the horsepower scale that would probably be
picked up by a quadratic fit.