Aravind_DataAnalytics_Lab6

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Question 5

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

library(ISLR2)

## Warning: package 'ISLR2' was built under R version 4.3.2

library(caTools)

## Warning: package 'caTools' was built under R version 4.3.3

mydefault <- Default
# Fitting logictic regression model on default data
logr_default <- glm(default ~ income + balance, data = mydefault, family = "binomial")
summary(logr_default)

## 
## Call:
## glm(formula = default ~ income + balance, family = "binomial", 
##     data = mydefault)
## 
## 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

1. Logistic regression to classify default using income and balance shows that both of the predictors are significant with Significant level at 0.05.

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

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

set.seed(452)
sample_def = sample.split(mydefault$default, SplitRatio = 0.70)
default_train<-subset(mydefault,sample_def==TRUE)
default_val<-subset(mydefault,sample_def==FALSE)

1. Using 70-30 split for training and validation data sets respectively.

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

# Fitting logictic regression model on default training data
lrdef_train <- glm(default ~ income + balance, data = default_train, family = "binomial")

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.

def_probs=predict(lrdef_train,default_val,type="response")


def_pred=rep("No",length(default_val$default))

def_pred[def_probs > .5] = "Yes"

table(def_pred,default_val$default)

##         
## def_pred   No  Yes
##      No  2889   69
##      Yes   11   31

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

mean(def_pred!=default_val$default) * 100

## [1] 2.666667

Validation Misclassification Rate::((11+69)/3000)=2.66%

1. Validation misclassification rate is 2.66%

3. 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.

Try 1

#Split Try 1
set.seed(297)
sample_def = sample.split(mydefault$default, SplitRatio = 0.60)
default_train<-subset(mydefault,sample_def==TRUE)
default_val<-subset(mydefault,sample_def==FALSE)

#Fit Try 1
# Fitting logictic regression model on default training data
lrdef_train <- glm(default ~ income + balance, data = default_train, family = "binomial")
#Prediction Try 1
def_probs=predict(lrdef_train,default_val,type="response")


def_pred=rep("No",length(default_val$default))

def_pred[def_probs > .5] = "Yes"

table(def_pred,default_val$default)

##         
## def_pred   No  Yes
##      No  3854   85
##      Yes   13   48

#Misclassification rate try 1
mean(def_pred!=default_val$default) * 100

## [1] 2.45

Validation Misclassification Rate(Try 1)::((13+85)/4000)=2.45%

Try 2

#Split Try 1
set.seed(254)
sample_def = sample.split(mydefault$default, SplitRatio = 0.50)
default_train<-subset(mydefault,sample_def==TRUE)
default_val<-subset(mydefault,sample_def==FALSE)

#Fit Try 1
# Fitting logictic regression model on default training data
lrdef_train <- glm(default ~ income + balance, data = default_train, family = "binomial")
#Prediction Try 1
def_probs=predict(lrdef_train,default_val,type="response")


def_pred=rep("No",length(default_val$default))

def_pred[def_probs > .5] = "Yes"

table(def_pred,default_val$default)

##         
## def_pred   No  Yes
##      No  4817  113
##      Yes   16   54

#Misclassification rate try 1
mean(def_pred!=default_val$default) * 100

## [1] 2.58

Validation Misclassification Rate(Try 2)::((16+113)/5000)=2.58%

Try 3

#Split Try 1
set.seed(999)
sample_def = sample.split(mydefault$default, SplitRatio = 0.75)
default_train<-subset(mydefault,sample_def==TRUE)
default_val<-subset(mydefault,sample_def==FALSE)

#Fit Try 1
# Fitting logictic regression model on default training data
lrdef_train <- glm(default ~ income + balance, data = default_train, family = "binomial")
#Prediction Try 1
def_probs=predict(lrdef_train,default_val,type="response")


def_pred=rep("No",length(default_val$default))

def_pred[def_probs > .5] = "Yes"

table(def_pred,default_val$default)

##         
## def_pred   No  Yes
##      No  2406   60
##      Yes   11   23

#Misclassification rate try 1
mean(def_pred!=default_val$default) * 100

## [1] 2.84

Validation Misclassification Rate(Try 3)::((11+60)/2500)=2.84%

1. After multiple train, test splits along with different random seeds, I see that my Try 1 gave a lowest test error rate of 2.45% . This try has a 60-40% split for training and validation sets.

4. 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.

#including Student as a predictor

set.seed(297)
sample_def = sample.split(mydefault$default, SplitRatio = 0.60)
default_train<-subset(mydefault,sample_def==TRUE)
default_val<-subset(mydefault,sample_def==FALSE)

#Fit Try 1
# Fitting logictic regression model on default training data
lrdef_train <- glm(default ~ income + balance + student, data = default_train, family = "binomial")
#Prediction Try 1
def_probs=predict(lrdef_train,default_val,type="response")


def_pred=rep("No",length(default_val$default))

def_pred[def_probs > .5] = "Yes"

table(def_pred,default_val$default)

##         
## def_pred   No  Yes
##      No  3855   87
##      Yes   12   46

#Misclassification rate try 1
mean(def_pred!=default_val$default) * 100

## [1] 2.475

1. Error/ Misclassification rate with student as additional predictor doesn’t improve the misclassification rate.Best score without student is 2.45% and with Student as predictor is 2.475%

Question 6

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

set.seed(986)
# Fitting logictic regression model on default training data
lrdef_std <- glm(default ~ income + balance, data = mydefault, family = "binomial")

summary(lrdef_std)$coefficients[2:3,]

##             Estimate   Std. Error   z value      Pr(>|z|)
## income  2.080898e-05 4.985167e-06  4.174178  2.990638e-05
## balance 5.647103e-03 2.273731e-04 24.836280 3.638120e-136

2. 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){
  lr_fn = glm(default ~ income + balance, data = data, subset = index,family = "binomial")
  return(summary(lr_fn)$coefficients[2:3,2])
}

3. 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)

## Warning: package 'boot' was built under R version 4.3.3

set.seed(986)

boot(mydefault,boot.fn,R=100)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = mydefault, statistic = boot.fn, R = 100)
## 
## 
## Bootstrap Statistics :
##         original       bias     std. error
## t1* 4.985167e-06 1.076749e-08 1.282837e-07
## t2* 2.273731e-04 8.210079e-07 1.016713e-05

4. Comment on the estimated standard errors obtained using the glm() function and using your bootstrap function. A)Above result indicates that the bootstrap estimate for SE(β^income) is 1.282837e-07, and that the bootstrap estimate for SE(β^balance) is 1.016713e-05. Using the summary() function, we computed the standard errors for the regression coefficients for SE(β^income) is 4.985167e-06 and for SE(β^balance) is 2.273731e-04.

The standard Errors from the bootstrap are lower compared to the standard error estimates obtained from the formulas.As the bootstrap approach does not rely on any of the assumptions that standard formulas take into account. Bootstap estimates for SE(β^income) and SE(β^balance) are likely giving a more accurate estimate of the standard errors that of the summary() function.

Question 7

1. Fit a logistic regression model that predicts Direction using Lag1 and Lag2

set.seed(1)
data("Weekly")
model_1 <- glm(Direction ~ Lag1 + Lag2, data = Weekly, family = "binomial")
summary(model_1)

## 
## Call:
## glm(formula = Direction ~ Lag1 + Lag2, family = "binomial", data = Weekly)
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  0.22122    0.06147   3.599 0.000319 ***
## Lag1        -0.03872    0.02622  -1.477 0.139672    
## Lag2         0.06025    0.02655   2.270 0.023232 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1496.2  on 1088  degrees of freedom
## Residual deviance: 1488.2  on 1086  degrees of freedom
## AIC: 1494.2
## 
## Number of Fisher Scoring iterations: 4

2. Fit a logistic regression model that predicts Direction using Lag1 and Lag2 using all but the first observation.

model_2 <- glm(Direction ~ Lag1 + Lag2, data = Weekly[-1, ], family = "binomial")
summary(model_2)

## 
## Call:
## glm(formula = Direction ~ Lag1 + Lag2, family = "binomial", data = Weekly[-1, 
##     ])
## 
## Coefficients:
##             Estimate Std. Error z value Pr(>|z|)    
## (Intercept)  0.22324    0.06150   3.630 0.000283 ***
## Lag1        -0.03843    0.02622  -1.466 0.142683    
## Lag2         0.06085    0.02656   2.291 0.021971 *  
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## (Dispersion parameter for binomial family taken to be 1)
## 
##     Null deviance: 1494.6  on 1087  degrees of freedom
## Residual deviance: 1486.5  on 1085  degrees of freedom
## AIC: 1492.5
## 
## Number of Fisher Scoring iterations: 4

3. Use the model from (b) to predict the direction of the first observation. You can do this by predicting that the first observation will go up if P (Direction = “Up”|Lag1, Lag2) > 0.5. Was this observation correctly classified?

predict.glm(model_2, Weekly[1, ], type = "response") > 0.5

##    1 
## TRUE

We can infer that the prediction for the first observation is “Up,” but it was misclassified since the true direction is “Down.”

4. Write a for loop from i = 1 to i = n, where n is the number of observations in the data set, that performs each of the following steps:

1. Fit a logistic regression model using all but the ith observation to predict Direction using Lag1 and Lag2.
2. Compute the posterior probability of the market moving up for the ith observation.
3. Use the posterior probability for the ith observation in order to predict whether or not the market moves up.
4. Determine whether or not an error was made in predicting the direction for the ith observation. If an error was made, then indicate this as a 1, and otherwise indicate it as a 0.

error <- rep(0, dim(Weekly)[1])
for (i in 1:dim(Weekly)[1]) {
    model_3 <- glm(Direction ~ Lag1 + Lag2, data = Weekly[-i, ],  family = "binomial")
    pred.up <- predict.glm(model_3, Weekly[i, ], type = "response") > 0.5
    true.up <- Weekly[i, ]$Direction == "Up"
    if (pred.up != true.up)
        error[i] <- 1
}
error

##    [1] 1 1 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 0 0 1 1 1 1 1 1 0 1 1 1 1 0 1 0 0
##   [38] 0 1 0 1 0 0 1 0 1 1 1 0 1 0 0 0 1 0 0 1 1 0 0 0 0 1 0 1 1 0 0 1 0 1 1 0 0
##   [75] 0 1 0 1 1 0 0 1 1 0 1 1 0 0 1 0 0 1 1 1 0 0 0 0 0 1 0 1 1 0 0 1 0 1 0 0 1
##  [112] 1 0 0 1 0 0 1 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 0 0 1 1 1 0 0 0 1 0 0 0 0 0
##  [149] 0 1 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 0 0 1 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0
##  [186] 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 0 0 1 1 0 1 0 0 1 1
##  [223] 0 0 0 1 1 1 0 1 0 1 0 1 0 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 0 1 0
##  [260] 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 1 0 1 0
##  [297] 0 1 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 1 0 1 1 0 0 1 0 1 0 1 1 0 0 0 1 0 1 0 0
##  [334] 1 1 1 1 0 1 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 0 0 1 1 0 1
##  [371] 1 1 1 1 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 1 0 0 0 0 0 1 0 0 1 1 1 0 1 0 1
##  [408] 0 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 1 0 0 1 0 1 0 0 0 0 0 1 1 1 1
##  [445] 0 1 1 0 1 0 1 1 0 1 0 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 0 1 0 0 0 1 0 0 1 0 0
##  [482] 0 1 1 1 0 1 0 0 0 1 0 1 1 1 0 0 0 0 1 1 1 0 1 1 0 1 0 0 0 1 0 1 0 0 0 1 0
##  [519] 1 1 0 0 1 1 0 0 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 0 0 0 1 1 1 1
##  [556] 1 1 0 1 0 1 0 0 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 0 0 1 0 1
##  [593] 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 1 0 1 0 1
##  [630] 0 1 1 1 1 0 1 1 0 0 0 1 1 1 1 0 1 1 1 0 1 0 0 0 1 1 1 1 1 1 0 1 0 0 1 0 0
##  [667] 0 1 1 0 1 0 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 1 0 1 1
##  [704] 0 0 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 0
##  [741] 1 1 1 1 0 0 0 1 1 1 1 1 1 0 1 0 0 0 0 0 0 1 0 1 1 1 0 0 0 1 0 0 1 0 0 0 1
##  [778] 1 1 0 0 0 1 0 0 1 1 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 1 1 0 0 1 1
##  [815] 0 1 1 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 1 1 0 1 0 1 0 1 1 0 0 1
##  [852] 1 1 0 1 1 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 0 1 1 0 0 1 0 1 0 1 1 0 1 0 1
##  [889] 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1
##  [926] 0 0 0 0 1 0 1 1 1 0 0 1 1 0 1 1 1 1 0 1 0 1 0 1 0 1 0 1 0 0 1 1 1 1 1 0 1
##  [963] 0 0 0 1 1 1 0 1 1 1 1 0 0 0 0 1 1 0 0 0 0 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0
## [1000] 0 1 0 0 1 0 1 0 0 1 1 1 1 0 1 0 0 1 0 0 1 0 0 1 1 0 1 1 1 0 1 1 0 0 0 1 0
## [1037] 1 0 1 1 1 1 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 1 1 1 0 0 0 1 0 1 1 1 0 0
## [1074] 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0

5. Take the average of the n numbers obtained in (d)iv in order to obtain the LOOCV estimate for the test error. Comment on the results.

mean(error)

## [1] 0.4499541

Question 9

1. Based on this data set, provide an estimate for the population mean of medv. Call this estimate μ^.

myboston <- Boston
mu_est <- mean(myboston$medv)
mu_est

## [1] 22.53281

A)μ^=22.5328

2. Provide an estimate of the standard error of μ^ .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.

se_mu = sd(myboston$medv)/sqrt(length(myboston$medv))
se_mu

## [1] 0.4088611

1. Standard error of μ^=0.40886

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

#boot strap function
set.seed(123)
boot_mu_fn <- function(data,index)
              return(mean(data[index]))

boot_result<- boot(myboston$medv,boot_mu_fn, R = 1000)

boot_result

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = myboston$medv, statistic = boot_mu_fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original      bias    std. error
## t1* 22.53281 -0.01607372   0.4045557

1. Standard error from Bootstrap for μ^ is 0.4045557 , this is slightly lower than that from (b),which is 0.40886

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

Hint: You can approximate a 95 % confidence interval using the formula [ μ^− 2∗SE(μ^), μ^+ 2∗SE(μ^)].

#confidence intervals using the SE from bootstrap is:

lower_bd <- mu_est - (2*0.4045557) #mu hat - 2 * SE(from bootstrap)
upper_db <- mu_est + (2*0.4045557) #mu hat + 2 * SE(from bootstrap)
lower_bd

## [1] 21.72369

upper_db

## [1] 23.34192

Lower bound is 21.72369 Upper bound is 23.34192

using One Sampe t-Test

t.test(myboston$medv)

## 
##  One Sample t-test
## 
## data:  myboston$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

1. Confidence intervals calculated using the bootstrap estimate of SE and one sample t-Test are approximately same.

5. Based on this data set, provide an estimate, μ^median for the median value of medv in the population.

median(myboston$medv)

## [1] 21.2

6. We now would like to estimate the standard error of μ^median .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.

#boot strap function
set.seed(123)
boot_med_fn <- function(data,index)
              return(median(data[index]))

boot(myboston$medv,boot_med_fn, R = 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = myboston$medv, statistic = boot_med_fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*     21.2 -0.0203   0.3676453

SE(μ^median) using the bootstrap is 0.37(rounded to 2 decimals). This is small compared to the SE(μ^mean). this indicates that the mediam estimate for the population is mostly accurate.

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

#boot strap function

quantile(myboston$medv,0.1)

##   10% 
## 12.75

μ^0.1=12.75

8. Use the bootstrap to estimate the standard error of μ^0.1. Comment on your findings

set.seed(1)
boot_10q_fn <- function(data,index)
              return(quantile(data[index],0.1))

boot(myboston$medv,boot_10q_fn, R = 1000)

## 
## ORDINARY NONPARAMETRIC BOOTSTRAP
## 
## 
## Call:
## boot(data = myboston$medv, statistic = boot_10q_fn, R = 1000)
## 
## 
## Bootstrap Statistics :
##     original  bias    std. error
## t1*    12.75  0.0339   0.4767526

Bootstrap μ^0.1=0.48 this is relatively small considering the 10th percentile of 12.75.