Results

Q3

a.Explain how k-fold cross-validation is implemented?

The k-fold cross-validation is implemented by taking a certain number of observations defined by n and randomly splitting them into groups defined by K. The groups that are split act as a validation set and the remainder act as a validation set. The test error of the data is then calculated by averaging the k resulting MSE estimates.

b.the advantages and disadvantages:

1.The validation set approach

There are two drawbacks to using the variable set approach, the first is that the estimate of the test error could be highly variablke based on the data that is inputed into the training and vailidation data sets. Secondly, the vailidation set error rate may tend to overestimate the test error rate for the model fit of the entire data set.

  1. LOOCV

The LOOCV approach has an advatage as it will produce less bias. The validation approach would produce diffrent MSE when applied multiple times due to the splitting of the data, while performing LOOCV this process will yield the same results. Due to the fact that we are splitting the data based on 1 observation. Another disadvantage this model has is that is computationally intensive.

Q5

a.Fit a logistic regression model that uses income and balance to predict default.

library(ISLR)
attach(Default)
set.seed(1)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit.glm)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4725  -0.1444  -0.0574  -0.0211   3.7245  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1579.0  on 9997  degrees of freedom
## AIC: 1585
## 
## Number of Fisher Scoring iterations: 8

b.Using the validation set approach, estimate the test error of this model

  1. Split the sample set into a training set and a validation set.
train = sample(dim(Default)[1], dim(Default)[1]/2)

2.Fit a multiple logistic regression model using only the training observations.

fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
summary(fit.glm)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.5830  -0.1428  -0.0573  -0.0213   3.3395  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.194e+01  6.178e-01 -19.333  < 2e-16 ***
## income       3.262e-05  7.024e-06   4.644 3.41e-06 ***
## balance      5.689e-03  3.158e-04  18.014  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1523.8  on 4999  degrees of freedom
## Residual deviance:  803.3  on 4997  degrees of freedom
## AIC: 809.3
## 
## Number of Fisher Scoring iterations: 8

3.Obtain a prediction of default status for each individual in the validation set by computing the posterior probability of default for that individual, and classifying the individual to the default category if the posterior probability is greater than 0.5.

probs = predict(fit.glm, newdata = Default[-train,], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > .5] = "yes"

4.Compute the validation set error, which is the fraction of the observations in the validation set that are misclassified.

mean(pred.glm != Default[-train,]$default)
## [1] 0.0352

The validation set approach gave me a test rate error if 3.52%

c.Repeat the process in (b) three times, using three different splits of the observations into a training set and a validation set. Comment on the results obtained.

train = sample(dim(Default)[1], dim(Default)[1]/2)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train,], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > .5] = "yes"
mean(pred.glm != Default[-train,]$default)
## [1] 0.0374
train = sample(dim(Default)[1], dim(Default)[1]/2)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train,], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > .5] = "yes"
mean(pred.glm != Default[-train,]$default)
## [1] 0.0346
train = sample(dim(Default)[1], dim(Default)[1]/2)
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train,], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > .5] = "yes"
mean(pred.glm != Default[-train,]$default)
## [1] 0.0352

These three diffrent splits produced a diffrent resulting error rate which shows that the rate does in fact vary based on which observations are in the training/validation sets.

d.Now consider a logistic regression model that predicts the prob- ability of default using income, balance, and a dummy variable for student. Estimate the test error for this model using the val- idation set approach. Comment on whether or not including a dummy variable for student leads to a reduction in the test error rate.

train = sample(dim(Default)[1], dim(Default)[1]/2)
fit.glm = glm(default ~ income + balance + student, data = Default, family = "binomial", subset = train)
probs = predict(fit.glm, newdata = Default[-train,], type = "response")
pred.glm = rep("No", length(probs))
pred.glm[probs > .5] = "yes"
mean(pred.glm != Default[-train,]$default)
## [1] 0.0384

When including the dummy variable student it does not seem to lower the test error rate.

Q6

a.Using the summary() and glm() functions, determine the esti- mated standard errors for the coefficients associated with income and balance in a multiple logistic regression model that uses both predictors.

set.seed(1)
attach(Default)
## The following objects are masked from Default (pos = 3):
## 
##     balance, default, income, student
fit.glm = glm(default ~ income + balance, data = Default, family = "binomial")
summary(fit.glm)
## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Deviance Residuals: 
##     Min       1Q   Median       3Q      Max  
## -2.4725  -0.1444  -0.0574  -0.0211   3.7245  
## 
## Coefficients:
##               Estimate Std. Error z value Pr(>|z|)    
## (Intercept) -1.154e+01  4.348e-01 -26.545  < 2e-16 ***
## income       2.081e-05  4.985e-06   4.174 2.99e-05 ***
## balance      5.647e-03  2.274e-04  24.836  < 2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 2920.6  on 9999  degrees of freedom
## Residual deviance: 1579.0  on 9997  degrees of freedom
## AIC: 1585
## 
## Number of Fisher Scoring iterations: 8

The standard error for the income coeffiecent is 4.985e-06 and the standard error for the balance coeffiecent was 2.274e-04.

b.Write a function, boot.fn(), that takes as input the Default data set as well as an index of the observations, and that outputs the coefficient estimates for income and balance in the multiple logistic regression model.

boot.fn = function(data, index) {
  fit = glm(default ~ income + balance, data = data, family = "binomial", subset = index)
  return(coef(fit))
}

c.Use the boot() function together with your boot.fn() function to estimate the standard errors of the logistic regression coefficients for income and balance.

library(boot)
boot(Default, boot.fn, 1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01 -3.945460e-02 4.344722e-01
## t2*  2.080898e-05  1.680317e-07 4.866284e-06
## t3*  5.647103e-03  1.855765e-05 2.298949e-04

The standard errors for the coefficients are 4.866284e-06 and 2.298949e-04.

d.Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function.

The standard error for both functions are relatively similar.

Q9

a. Based on this data set, provide an estimate for the population mean of medv. Call this estimate μˆ.

library(MASS)
attach(Boston)
sample_mu = mean(medv)
sample_mu
## [1] 22.53281

This displays the sample mean of 22.53281

b.Provide an estimate of the standard error of μˆ. Interpret this result.

sample_se = sd(medv)/sqrt(nrow(Boston))
sample_se
## [1] 0.4088611

The standard error estimate is .4088611. This standard error will give us the accuracy of the estimate how much the mean will vary if a different sample was chosen.

c.Now estimate the standard error of μˆ using the bootstrap. How does this compare to your answer from (b)?

library(boot)
set.seed(1)
boot.fn = function(data,index){
  mu = mean(data[index])
  return(mu)
}
set.seed(1)
boot(medv,boot.fn, 100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 0.009027668   0.3482331

The standard error obtained from bootstrapping is slightly less than the one obtained above.

d.Based on your bootstrap estimate from (c), provide a 95 % con- fidence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv).

t.test(medv)
## 
##  One Sample t-test
## 
## data:  medv
## t = 55.111, df = 505, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  21.72953 23.33608
## sample estimates:
## mean of x 
##  22.53281
ci.mu = c(sample_mu-2*0.3482331,sample_mu+2*0.3482331)
ci.mu
## [1] 21.83634 23.22927

The confidence interval obtained is very similar to the one from the t-test.

e.Based on this data set, provide an estimate, μˆmed, for the median value of medv in the population.

mu_median = median(medv)
mu_median
## [1] 21.2

The estimate median is 21.2

f.We now would like to estimate the standard error of μˆmed.Unfortunately, there is no simple formula for computing the standard error of the median. Instead, estimate the standard error of the median using the bootstrap. Comment on your findings.

library(boot)
set.seed(1)
boot.fn = function(data,index){
  mu.median = median(data[index])
  return(mu.median)
}

set.seed(1)
boot(medv,boot.fn, 100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2  -0.029   0.3461316

Here we computed the standard error of the median using the boostrapping method.

g.Based on this data set, provide an estimate for the tenth per- centile of medv in Boston suburbs. Call this quantity μˆ0.1. (You can use the quantile() function.)

mu.quantile = quantile(medv,c(.1))
mu.quantile
##   10% 
## 12.75

The estimate for the tenth percentile of medv in boston suburbs is 12.75.

h.Use the bootstrap to estimate the standard error of μˆ0.1. Comment on your findings.

boot.fn = function(data,index){
  mu.quantile = quantile(data[index], c(.1))
  return(mu.quantile)
}

boot(medv, boot.fn,1000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0285   0.4861472

This is showing the standard error of the tenth percentile using the bootstrapping method. I observed that the standard error was 0.4917843 which is very low telling us that this method is very accurate.