applied.Rmd

Chapter 3: Linear Regression: Exercises: Applied
================================================

8a.
---

```{r}
Auto = read.csv("../data/Auto.csv", header=T, na.strings="?")
Auto = na.omit(Auto)
summary(Auto)
attach(Auto)
lm.fit = lm(mpg ~ horsepower)
summary(lm.fit)
```

### i.
Yes, there is a relationship between horsepower and mpg as deterined by
testing the null hypothesis of all regression coefficients equal to zero.
Since the F-statistic is far larger than 1 and the p-value of the F-statistic
is close to zero we can reject the null hypothesis and state there is a
statistically significant relationship between horsepower and mpg.

### ii.
To calculate the residual error relative to the response we use the mean
of the response and the RSE. The mean of mpg is ``r mean(mpg, na.rm=T)``. 
The RSE of the lm.fit was ``r 4.906`` which indicates a percentage error of
``r 4.906/mean(mpg, na.rm=T) * 100.0``%. The $R^2$ of the lm.fit was
about ``r summary(lm.fit)$r.sq``, meaning ``r summary(lm.fit)$r.sq * 100.0``%
of the variance in mpg is explained by horsepower.

### iii.
The relationship between mpg and horsepower is negative. The more horsepower
an automobile has the linear regression indicates the less mpg fuel efficiency
the automobile will have.

### iv.
```{r}
predict(lm.fit, data.frame(horsepower=c(98)), interval="confidence")
predict(lm.fit, data.frame(horsepower=c(98)), interval="prediction")
```


8b.
---

```{r}
plot(horsepower, mpg)
abline(lm.fit)
```

8c.
---

```{r}
par(mfrow=c(2,2))
plot(lm.fit)
```

Based on the residuals plots, there is some evidence of non-linearity.


9a.
---
```{r}
pairs(Auto)
```

9b.
---
```{r}
cor(subset(Auto, select=-name))
```

9c.
---
```{r}
lm.fit1 = lm(mpg~.-name, data=Auto)
summary(lm.fit1)
```

### i.
Yes, there is a relatioship between the predictors and the response by testing
the null hypothesis of whether all the regression coefficients are zero. The F
-statistic is far from 1 (with a small p-value), indicating evidence against
the null hypothesis.

### ii.
Looking at the p-values associated with each predictor's t-statistic, we see
that displacement, weight, year, and origin have a statistically significant relationship, while cylinders, horsepower, and acceleration do not.

### iii.
The regression coefficient for year, ``r coefficients(lm.fit1)["year"]``,
suggests that for every one year, mpg increases by the coefficient. In other
words, cars become more fuel efficient every year by almost 1 mpg / year.

9d.
---
```{r}
par(mfrow=c(2,2))
plot(lm.fit1)
```
The fit does not appear to be accurate because there is a discernible curve
pattern to the residuals plots. From the leverage plot, point 14 appears to have
high leverage, although not a high magnitude residual.

```{r}
plot(predict(lm.fit1), rstudent(lm.fit1))
```
There are possible outliers as seen in the plot of studentized residuals
because there are data with a value greater than 3.

9e.
---
```{r}
lm.fit2 = lm(mpg~cylinders*displacement+displacement*weight)
summary(lm.fit2)
```
From the correlation matrix, I obtained the two highest correlated pairs and
used them in picking my interaction effects. From the p-values, we can see
that the interaction between displacement and weight is statistically
signifcant, while the interactiion between cylinders and displacement is not.

9f.
---
```{r}
lm.fit3 = lm(mpg~log(weight)+sqrt(horsepower)+acceleration+I(acceleration^2))
summary(lm.fit3)
par(mfrow=c(2,2))
plot(lm.fit3)
```

```{r}
plot(predict(lm.fit3), rstudent(lm.fit3))
```
Apparently, from the p-values, the log(weight), sqrt(horsepower), and
acceleration^2 all have statistical significance of some sort. The residuals
plot has less of a discernible pattern than the plot of all linear regression
terms. The studentized residuals displays potential outliers (>3). The
leverage plot indicates more than three points with high leverage.

However, 2 problems are observed from the above plots:
1) the residuals vs fitted plot indicates heteroskedasticity (unconstant 
variance over mean) in the model.
2) The Q-Q plot indicates somewhat unnormality of the residuals. 

So, a better transformation need to be applied to our model.
From the correlation matrix in 9a., displacement, horsepower and weight show a
similar nonlinear pattern against our response mpg.
This nonlinear pattern is very close to a log form. So in the next attempt, we 
use `log(mpg)` as our response variable.

The outputs show that log transform of mpg yield better model fitting
(better R^2, normality of residuals).

```{r}
lm.fit2<-lm(log(mpg)~cylinders+displacement+horsepower+weight+acceleration+year+origin,data=Auto)
summary(lm.fit2)
```

```{r}
par(mfrow=c(2,2)) 
plot(lm.fit2)
plot(predict(lm.fit2),rstudent(lm.fit2))
```

10a.
----

```{r}
library(ISLR)
summary(Carseats)
attach(Carseats)
lm.fit = lm(Sales~Price+Urban+US)
summary(lm.fit)
```

10b.
----
### Price
The linear regression suggests a relationship between price and sales given the
low p-value of the t-statistic. The coefficient states a negative
relationship between Price and Sales: as Price increases, Sales decreases.

### UrbanYes
The linear regression suggests that there isn't a relationship between the
location of the store and the number of sales based on the high p-value of the
t-statistic.

### USYes
The linear regression suggests there is a relationship between whether the store
is in the US or not and the amount of sales. The coefficient states a positive relationship between USYes and Sales: if the store is in the US, the sales will
increase by approximately 1201 units.

10c.
----
Sales = 13.04 + -0.05 Price + -0.02 UrbanYes + 1.20 USYes

10d.
----
Price and USYes, based on the p-values, F-statistic, and p-value of the
F-statistic.

10e.
----
```{r}
lm.fit2 = lm(Sales ~ Price + US)
summary(lm.fit2)
```

10f.
----
Based on the RSE and R^2 of the linear regressions, they both fit the data
similarly, with linear regression from (e) fitting the data slightly better.

10g.
----
```{r}
confint(lm.fit2)
```

10h.
----
```{r}
plot(predict(lm.fit2), rstudent(lm.fit2))
```
All studentized residuals appear to be bounded by -3 to 3, so not potential
outliers are suggested from the linear regression.

```{r}
par(mfrow=c(2,2))
plot(lm.fit2)
```
There are a few observations that greatly exceed $(p+1)/n$ (``r 3/397``) on the leverage-statistic plot that suggest that the corresponding points have high leverage.

11.
---

```{r}
set.seed(1)
x = rnorm(100)
y = 2*x + rnorm(100)
```

11a.
----
```{r}
lm.fit = lm(y~x+0)
summary(lm.fit)
```
The p-value of the t-statistic is near zero so the null hypothesis is rejected.

11b.
----
```{r}
lm.fit = lm(x~y+0)
summary(lm.fit)
```
The p-value of the t-statistic is near zero so the null hypothesis is rejected.

11c.
----
Both results in (a) and (b) reflect the same line created in 11a. In other
words, $y = 2x + \epsilon$ could also be written $x = 0.5 * (y - \epsilon)$.

11d.
----
$$
\begin{array}{cc}
t = \beta / SE(\beta) &
\beta = \frac {\sum{x_i y_i}} {\sum{x_i^2}} &
SE(\beta) = \sqrt{\frac {\sum{(y_i - x_i \beta)^2}} {(n-1) \sum{x_i^2}}}
\end{array}
\\
t = {\frac {\sum{x_i y_i}} {\sum{x_i^2}}}
    {\sqrt{\frac {(n-1) \sum{x_i^2}} {\sum{(y_i - x_i \beta)^2}}}}
\\
\frac {\sqrt{n-1} \sum{x_i y_i}}
      {\sqrt{\sum{x_i^2} \sum{(y_i - x_i \beta)^2}}}
\\
\frac {\sqrt{n-1} \sum{x_i y_i}}
      {\sqrt{\sum{x_i^2} \sum{(y_i^2 - 2 \beta x_i y_i  + x_i^2 \beta^2)}}}
\\
\frac {\sqrt{n-1} \sum{x_i y_i}}
      {\sqrt{\sum{x_i^2} \sum{y_i^2} - 
            \sum{x_i^2} \beta (2 \sum{x_i y_i} - \beta \sum{x_i^2})}}
\\
\frac {\sqrt{n-1} \sum{x_i y_i}}
      {\sqrt{\sum{x_i^2} \sum{y_i^2} - 
            \sum{x_i y_i} (2 \sum{x_i y_i} - \sum{x_i y_i})}}
\\
t = \frac {\sqrt{n-1} \sum{x_i y_i}} 
          {\sqrt{\sum{x_i^2} \sum{y_i^2} - (\sum{x_i y_i})^2 }}
$$

```{r}
(sqrt(length(x)-1) * sum(x*y)) / (sqrt(sum(x*x) * sum(y*y) - (sum(x*y))^2))
```
This is same as the t-statistic shown above.

11e.
----
If you swap t(x,y) as t(y,x), then you will find t(x,y) = t(y,x).

11f.
----
```{r}
lm.fit = lm(y~x)
lm.fit2 = lm(x~y)
summary(lm.fit)
summary(lm.fit2)
```
You can see the t-statistic is the same for the two linear regressions.

12a.
----
When the sum of the squares of the observed y-values are equal to the sum of the 
squares of the observed x-values.

12b.
----
```{r}
set.seed(1)
x = rnorm(100)
y = 2*x
lm.fit = lm(y~x+0)
lm.fit2 = lm(x~y+0)
summary(lm.fit)
summary(lm.fit2)
```
The regression coefficients are different for each linear regression.

12c.
----
```{r}
set.seed(1)
x <- rnorm(100)
y <- -sample(x, 100)
sum(x^2)
sum(y^2)
lm.fit <- lm(y~x+0)
lm.fit2 <- lm(x~y+0)
summary(lm.fit)
summary(lm.fit2)
```
The regression coefficients are the same for each linear regression. 
So long as sum sum(x^2) = sum(y^2) the condition in 12a. will be satisfied.
Here we have simply taken all the $x_i$ in a different order and made them
negative.

13a.
----
```{r}
set.seed(1)
x = rnorm(100)
```

13b.
----
```{r}
eps = rnorm(100, 0, sqrt(0.25))
```

13c.
----
```{r}
y = -1 + 0.5*x + eps
```
y is of length 100. $\beta_0$ is -1, $\beta_1$ is 0.5.

13d.
----
```{r}
plot(x, y)
```
I observe a linear relationship between x and y with a positive slope, with a 
variance as is to be expected.

13e.
----
```{r}
lm.fit = lm(y~x)
summary(lm.fit)
```
The linear regression fits a model close to the true value of the coefficients
as was constructed. The model has a large F-statistic with a near-zero p-value
so the null hypothesis can be rejected.

13f.
----
```{r}
plot(x, y)
abline(lm.fit, lwd=3, col=2)
abline(-1, 0.5, lwd=3, col=3)
legend(-1, legend = c("model fit", "pop. regression"), col=2:3, lwd=3)
```

13g.
----
```{r}
lm.fit_sq = lm(y~x+I(x^2))
summary(lm.fit_sq)
```
There is evidence that model fit has increased over the training data given the
slight increase in $R^2$ and $RSE$. Although, the p-value of the t-statistic
suggests that there isn't a relationship between y and $x^2$.

13h.
----
```{r}
set.seed(1)
eps1 = rnorm(100, 0, 0.125)
x1 = rnorm(100)
y1 = -1 + 0.5*x1 + eps1
plot(x1, y1)
lm.fit1 = lm(y1~x1)
summary(lm.fit1)
abline(lm.fit1, lwd=3, col=2)
abline(-1, 0.5, lwd=3, col=3)
legend(-1, legend = c("model fit", "pop. regression"), col=2:3, lwd=3)
```
As expected, the error observed in $R^2$ and $RSE$ decreases considerably.

13i.
----
```{r}
set.seed(1)
eps2 = rnorm(100, 0, 0.5)
x2 = rnorm(100)
y2 = -1 + 0.5*x2 + eps2
plot(x2, y2)
lm.fit2 = lm(y2~x2)
summary(lm.fit2)
abline(lm.fit2, lwd=3, col=2)
abline(-1, 0.5, lwd=3, col=3)
legend(-1, legend = c("model fit", "pop. regression"), col=2:3, lwd=3)
```
As expected, the error observed in $R^2$ and $RSE$ increases considerably.

13j.
----
```{r}
confint(lm.fit)
confint(lm.fit1)
confint(lm.fit2)
```
All intervals seem to be centered on approximately 0.5, with the second fit's
interval being narrower than the first fit's interval and the last fit's
interval being wider than the first fit's interval.

14a.
----
```{r}
set.seed(1)
x1 = runif(100)
x2 = 0.5 * x1 + rnorm(100)/10
y = 2 + 2*x1 + 0.3*x2 + rnorm(100)
```
$$
Y = 2 + 2 X_1 + 0.3 X_2 + \epsilon \\
\beta_0 = 2, \beta_1 = 2, \beta_3 = 0.3
$$

14b.
----
```{r}
cor(x1, x2)
plot(x1, x2)
```

14c.
----
```{r}
lm.fit = lm(y~x1+x2)
summary(lm.fit)
```
$$\beta_0 = 2.0533, \beta_1 = 1.6336, \beta_3 = 0.5588$$
The regression coefficients are close to the true coefficients, although with
high standard error. We can reject the null hypothesis for $\beta_1$ because
its p-value is below 5%. We cannot reject the null hypothesis for $\beta_2$
because its p-value is much above the 5% typical cutoff, over 60%.

14d.
----
```{r}
lm.fit = lm(y~x1)
summary(lm.fit)
```
Yes, we can reject the null hypothesis for the regression coefficient given the
p-value for its t-statistic is near zero.

14e.
----
```{r}
lm.fit = lm(y~x2)
summary(lm.fit)
```
Yes, we can reject the null hypothesis for the regression coefficient given the
p-value for its t-statistic is near zero.

14f.
----
No, because x1 and x2 have collinearity, it is hard to distinguish their effects
when regressed upon together. When they are regressed upon separately, the
linear relationship between y and each predictor is indicated more clearly.

14g.
----
```{r}
x1 = c(x1, 0.1)
x2 = c(x2, 0.8)
y = c(y, 6)
lm.fit1 = lm(y~x1+x2)
summary(lm.fit1)
lm.fit2 = lm(y~x1)
summary(lm.fit2)
lm.fit3 = lm(y~x2)
summary(lm.fit3)
```
In the first model, it shifts x1 to statistically insignificance and shifts
x2 to statistiscal significance from the change in p-values between the two
linear regressions.

```{r}
par(mfrow=c(2,2))
plot(lm.fit1)
```

```{r}
par(mfrow=c(2,2))
plot(lm.fit2)
```

```{r}
par(mfrow=c(2,2))
plot(lm.fit3)
```
In the first and third models, the point becomes a high leverage point. 

```{r}
plot(predict(lm.fit1), rstudent(lm.fit1))
plot(predict(lm.fit2), rstudent(lm.fit2))
plot(predict(lm.fit3), rstudent(lm.fit3))
```
Looking at the studentized residuals, we don't observe points too far from the
|3| value cutoff, except for the second linear regression: y ~ x1.

15a.
----
```{r}
library(MASS)
summary(Boston)
Boston$chas <- factor(Boston$chas, labels = c("N","Y"))
summary(Boston)
attach(Boston)
lm.zn = lm(crim~zn)
summary(lm.zn) # yes
lm.indus = lm(crim~indus)
summary(lm.indus) # yes
lm.chas = lm(crim~chas) 
summary(lm.chas) # no
lm.nox = lm(crim~nox)
summary(lm.nox) # yes
lm.rm = lm(crim~rm)
summary(lm.rm) # yes
lm.age = lm(crim~age)
summary(lm.age) # yes
lm.dis = lm(crim~dis)
summary(lm.dis) # yes
lm.rad = lm(crim~rad)
summary(lm.rad) # yes
lm.tax = lm(crim~tax)
summary(lm.tax) # yes
lm.ptratio = lm(crim~ptratio)
summary(lm.ptratio) # yes
lm.black = lm(crim~black)
summary(lm.black) # yes
lm.lstat = lm(crim~lstat)
summary(lm.lstat) # yes
lm.medv = lm(crim~medv)
summary(lm.medv) # yes
```
All, except chas. Plot each linear regression using "plot(lm)" to see
residuals.

15b.
----
```{r}
lm.all = lm(crim~., data=Boston)
summary(lm.all)
```
zn, dis, rad, black, medv

15c.
----
```{r}
x = c(coefficients(lm.zn)[2],
      coefficients(lm.indus)[2],
      coefficients(lm.chas)[2],
      coefficients(lm.nox)[2],
      coefficients(lm.rm)[2],
      coefficients(lm.age)[2],
      coefficients(lm.dis)[2],
      coefficients(lm.rad)[2],
      coefficients(lm.tax)[2],
      coefficients(lm.ptratio)[2],
      coefficients(lm.black)[2],
      coefficients(lm.lstat)[2],
      coefficients(lm.medv)[2])
y = coefficients(lm.all)[2:14]
plot(x, y)
```
Coefficient for nox is approximately -10 in univariate model and 31 in
multiple regression model.

15d.
----
```{r}
lm.zn = lm(crim~poly(zn,3))
summary(lm.zn) # 1, 2
lm.indus = lm(crim~poly(indus,3))
summary(lm.indus) # 1, 2, 3
# lm.chas = lm(crim~poly(chas,3)) : qualitative predictor
lm.nox = lm(crim~poly(nox,3))
summary(lm.nox) # 1, 2, 3
lm.rm = lm(crim~poly(rm,3))
summary(lm.rm) # 1, 2
lm.age = lm(crim~poly(age,3))
summary(lm.age) # 1, 2, 3
lm.dis = lm(crim~poly(dis,3))
summary(lm.dis) # 1, 2, 3
lm.rad = lm(crim~poly(rad,3))
summary(lm.rad) # 1, 2
lm.tax = lm(crim~poly(tax,3))
summary(lm.tax) # 1, 2
lm.ptratio = lm(crim~poly(ptratio,3))
summary(lm.ptratio) # 1, 2, 3
lm.black = lm(crim~poly(black,3))
summary(lm.black) # 1
lm.lstat = lm(crim~poly(lstat,3))
summary(lm.lstat) # 1, 2
lm.medv = lm(crim~poly(medv,3))
summary(lm.medv) # 1, 2, 3
```
See inline comments above, the answer is yes for most, except for black and
chas.