3.We now review k-fold cross-validation.

(a) Explain how k-fold cross-validation is implemented.

K fold cross validations, seperates the data into a training data set and test data. Then it is split into however many observations we wish for,the seperates them into groups. after running anylsis on these groups the test error is estimated by the Mean of the MSE.

(b) What are the advantages and disadvantages of k-fold crossvalidation relative to:

i. The validation set approach?

Disadvantage:

  1. This apporach produces a differnet MSE when repeatedto to the randomness.

Advantages:

1.Not computationally intensive, their for it is quicker.

2.We only repeat the anylsis K number of times(k=10,k=5), instead of performing the entire data set.

(ii. LOOCV?)

Disadvantage:

1.We use the entire data set, their for it is computationally intensive.

  1. has higher Varience when k<n

Advantages:

  1. has less bias, almost the entire data set is being used.

  2. produces a less varible MSE.

5. In Chapter 4, we used logistic regression to predict the probability of default using income and balance on the Default data set. We will now estimate the test error of this logistic regression model using the validation set approach. Do not forget to set a random seed before beginning your analysis.

library(ISLR)

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

set.seed(1)
glm.fit<-glm(default~income+balance, data = Default, family = binomial)
glm.fit
## 
## Call:  glm(formula = default ~ income + balance, family = binomial, 
##     data = Default)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.154e+01    2.081e-05    5.647e-03  
## 
## Degrees of Freedom: 9999 Total (i.e. Null);  9997 Residual
## Null Deviance:       2921 
## Residual Deviance: 1579  AIC: 1585

(b) Using the validation set approach, estimate the test error of this model. In order to do this, you must perform the following steps:

i. Split the sample set into a training set and a validation set.

train<- sample(nrow(Default), .5*nrow(Default))

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

glm.fit<-glm(default~income+balance, data = Default, family = binomial, subset = train)
glm.fit
## 
## Call:  glm(formula = default ~ income + balance, family = binomial, 
##     data = Default, subset = train)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.208e+01    1.858e-05    6.053e-03  
## 
## Degrees of Freedom: 4999 Total (i.e. Null);  4997 Residual
## Null Deviance:       1457 
## Residual Deviance: 734.4     AIC: 740.4

iii. 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.

glm.probs <- predict(glm.fit, newdata=Default[-train, ], type = "response")
glm.pred <- rep("N", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Y"

iv. Compute the validation set error, which is the fraction of the observations in the validation set that are misclassi???ed.

error<-mean(glm.pred!=Default[-train, ]$default)

print(“our error is 2.86”)

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

Split 1
set.seed(1)

glm.fit<-glm(default~income+balance, data = Default, family = "binomial")
glm.fit
## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.154e+01    2.081e-05    5.647e-03  
## 
## Degrees of Freedom: 9999 Total (i.e. Null);  9997 Residual
## Null Deviance:       2921 
## Residual Deviance: 1579  AIC: 1585
train<- sample(nrow(Default), .75*nrow(Default))
glm.fit.train<-glm(default~income+balance, data = Default, family = "binomial" ,subset = train)
glm.fit.train
## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.167e+01    2.115e-05    5.738e-03  
## 
## Degrees of Freedom: 7499 Total (i.e. Null);  7497 Residual
## Null Deviance:       2226 
## Residual Deviance: 1178  AIC: 1184
glm.probs <- predict(glm.fit.train, data = Default[-train, ], type = "response")
glm.pred <- rep("N", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Y"

error1<-mean(glm.pred!=Default[-train, ]$default)
SPlit 2
set.seed(1)

glm.fit<-glm(default~income+balance, data = Default, family = "binomial")
glm.fit
## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.154e+01    2.081e-05    5.647e-03  
## 
## Degrees of Freedom: 9999 Total (i.e. Null);  9997 Residual
## Null Deviance:       2921 
## Residual Deviance: 1579  AIC: 1585
train<- sample(nrow(Default), .25*nrow(Default))

glm.fit.train<-glm(default~income+balance, data = Default, family = "binomial" ,subset = train)
glm.fit.train
## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.233e+01    2.330e-05    6.233e-03  
## 
## Degrees of Freedom: 2499 Total (i.e. Null);  2497 Residual
## Null Deviance:       735.2 
## Residual Deviance: 367   AIC: 373
glm.probs <- predict(glm.fit.train, data = Default[-train, ], type = "response")
glm.pred <- rep("N", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Y"


error2<-mean(glm.pred!=Default[-train, ]$default)
Split 3
set.seed(1)

glm.fit<-glm(default~income+balance, data = Default, family = "binomial")
glm.fit
## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.154e+01    2.081e-05    5.647e-03  
## 
## Degrees of Freedom: 9999 Total (i.e. Null);  9997 Residual
## Null Deviance:       2921 
## Residual Deviance: 1579  AIC: 1585
train<- sample(nrow(Default), .80*nrow(Default))

glm.fit.train<-glm(default~income+balance, data = Default, family = "binomial" ,subset = train)
glm.fit.train
## 
## Call:  glm(formula = default ~ income + balance, family = "binomial", 
##     data = Default, subset = train)
## 
## Coefficients:
## (Intercept)       income      balance  
##  -1.155e+01    1.859e-05    5.708e-03  
## 
## Degrees of Freedom: 7999 Total (i.e. Null);  7997 Residual
## Null Deviance:       2354 
## Residual Deviance: 1255  AIC: 1261
glm.probs <- predict(glm.fit.train, data = Default[-train, ], type = "response")
glm.pred <- rep("N", length(glm.probs))
glm.pred[glm.probs > 0.5] = "Y"


error3<-mean(glm.pred!=Default[-train, ]$default)

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

train<- sample(nrow(Default), .80*nrow(Default))
glm.fit <- glm(default~income+balance+student, data = Default, family = "binomial", subset = train)
glm.prob <- predict(glm.fit,Default[-train, ],type = "response")
glm.pred <- rep("No",length(glm.prob))
glm.pred[glm.prob > .5] ="Yes"
mean(glm.pred!=Default[-train,]$default)
## [1] 0.036

6. We continue to consider the use of a logistic regression model to predict the probability of default using income and balance on the Default data set. In particular, we will now compute estimates for the standard errors of the income and balance logistic regression coe???cients in two di???erent ways: (1) using the bootstrap, and (2) using the standard formula for computing the standard errors in the glm() function. Do not forget to set a random seed before beginning your analysis.

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

library(ISLR)
str(Default)
## 'data.frame':    10000 obs. of  4 variables:
##  $ default: Factor w/ 2 levels "No","Yes": 1 1 1 1 1 1 1 1 1 1 ...
##  $ student: Factor w/ 2 levels "No","Yes": 1 2 1 1 1 2 1 2 1 1 ...
##  $ balance: num  730 817 1074 529 786 ...
##  $ income : num  44362 12106 31767 35704 38463 ...
set.seed(1)
glm.fit<-glm(default~income+balance,data = Default, family = binomial)
summary(glm.fit)
## 
## 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) 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 coeffcient estimates for income and balance in the multiple logistic regression model.

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

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

library(boot)
boot(Default,boot.fn,100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Default, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##          original        bias     std. error
## t1* -1.154047e+01  9.699111e-02 4.101121e-01
## t2*  2.080898e-05  6.715005e-08 4.127740e-06
## t3*  5.647103e-03 -5.733883e-05 2.105660e-04

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

Both the glm and the boot strap appears to be the same.

9. We will now consider the Boston housing data set, from the MASS library.

(a) Based on this data set, provide an estimate for the population mean of medv. Call this estimate ^ ??.

library(MASS)
summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08204   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00
mu <-mean(Boston$medv)
mu
## [1] 22.53281

(b) Provide an estimate of the standard error of ^ ??. Interpret this result. Hint: We can compute the standard error of the sample mean by dividing the sample standard deviation by the square root of the number of observations.

error<-sd(Boston$medv)/sqrt(length(Boston$medv))
error
## [1] 0.4088611

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

library(boot)
boot.fn = function(data,index)
{
  mu = mean(data[index])
  return(mu)
}
bootstrap<-boot(Boston$medv, boot.fn, 1000)
bootstrap
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original       bias    std. error
## t1* 22.53281 -0.005122332   0.4117619

(d) Based on your bootstrap estimate from (c), provide a 95% con???dence interval for the mean of medv. Compare it to the results obtained using t.test(Boston$medv). Hint: You can approximate a 95% con???dence interval using the formula [^ ?????2SE(^ ??), ^ ?? +2SE(^ ??)].

t.test(Boston$medv)
## 
##  One Sample t-test
## 
## data:  Boston$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
confidence.interval<-c(22.533-2*.4119,22.533+2*.4119)
confidence.interval
## [1] 21.7092 23.3568

(e) Based on this data set, provide an estimate, ^ ??med, for the median value of medv in the population.

medi<-median(Boston$medv)
medi
## [1] 21.2

(f) We now would like to estimate the standard error of ^ ??med. 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 ???ndings.

boot.fn<- function(data,index)
{
  mu<-median(data[index])
  return(mu)
}
boot(Boston$medv,boot.fn,100)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0395   0.3988174

(g) Based on this data set, provide an estimate for the tenth percentile of medv in Boston suburbs. Call this quantity ^ ??0.1. (You can use the quantile() function.)

Tenth<-quantile(Boston$medv,.1)
Tenth
##   10% 
## 12.75

(h) Use the bootstrap to estimate the standard error of ^ ??0.1. Comment on your ???ndings.

boot.fn<-function(data,index)
{
  mu<quantile(data[index],.1)
  return(mu)
}
boot(Boston$medv,boot.fn,10000)
## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = Boston$medv, statistic = boot.fn, R = 10000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1* 22.53281       0           0