IT and Business Application lab : Assignment 8

 

 

Assignment 8 : Panel Data Analysis

 

Problem I: Do Panel Data Analysis of "Produc" data analyzing on three types of model :

1. Pooled affect model
2. Fixed affect model
3. Random affect model

Problem II: Determine which model is the best by using functions:

1. pFtest : Fixed vs Pooled
2. plmtest : Pooled vs Random
3. phtest: Random vs Fixed
 
 

Solution:

 

To load the dataCommands:

> data("Produc" , package ="plm")
> head(Produc)  

 

 

Pooled Affects Model
Commands:

> pool <- plm(log(pcap)~ log(hwy) + log(water) + log(util) + log(pc) + log(gsp) + log(emp) + log(unemp) , data =Produc, model=("pooling"), index = c("state","year"))
> summary(pool)
 

 

 

Fixed Affects ModelCommands:

> fixed <- plm(log(pcap)~ log(hwy) + log(water) + log(util) + log(pc) + log(gsp) + log(emp) + log(unemp) , data =Produc, model=("within"), index = c("state","year"))
> summary(fixed)

 

 

Random Affects ModelCommands:

> random <- plm(log(pcap)~ log(hwy) + log(water) + log(util) + log(pc) + log(gsp) + log(emp) + log(unemp) , data =Produc, model=("random"), index = c("state","year"))
> summary(random)

 

 

 

Tests:
Pooled vs Fixed:
H0: Pooled Affects Model
H1: Fixed Affects Model
 

 

As the p-value is very small, we reject the null hypothesis and accept the alternate hypothesis.
=> Fixed affects model is accepted.

Pooled vs Random:
H0: Pooled Affects Model
H1: Random Affects Model

 

As the p-value is very small, we reject the null hypothesis and accept the alternate hypothesis.
=> Random affects model is accepted.

Random vs Fixed:
H0: Random Affects Model
H1: Fixed Affects Model 

 

 

As the p-value is very small, we reject the null hypothesis and accept the alternate hypothesis.
=> Fixed affects model is accepted.

Conclusion:
By conducting all the above tests, we come to a conclusion that Fixed Affects Model is the best to do the panel data analysis for "Produc".

IT and Business Application lab : Assignment 6

Problem 1


1) Create log of returns data (from 01.01.2012 to 01.01.2013) and calculate historical volatility

2) Create ACF plot for the log returns data ,perform adf test and interpret.
Commands:

> stockprice<-read.csv(file.choose(),header=T)
> head(stockprice)
> closingprice<-stockprice[,5]
> closingprice.ts<-ts(closingprice,frequency=252)
> returns<-(closingprice.ts-lag(closingprice.ts,k=-1))/lag(closingprice.ts,k=-1)
> z<-scale(returns)+10
> logreturns<-log(z)
> logreturns
> acf(logreturns)

 

 

From the above graph, we can see that the measurements lie with in the 95% confidence interval. Therefore, the time series is stationary.

> T=252^0.5
> historicalvolatility<-sd(logreturns)*T
> historicalvolatility
> adf.test(logreturns)

 

From the test results, we can see that p-value=0.01 (<0.05).
Therefore, we reject the null hypothesis and accept the alternate hypothesis which states that the time series is stationary.

 

 

 

 

IT and Business Application lab : Assignment 5

Problem I:

1. Find returns of NSE data of greater than 6 months having selected the 10th data point as start and 95th data point as end.

2. Find plot of that return

Solutions:

Commands:

z<-read.csv(file.choose(),header=T)

head(z)

open<-z$Open[10:95]

open.ts<-ts(open,deltat=1/252)

open.ts

summary(open.ts)

z.diff<-diff(open.ts)

z.diff

returns<-cbind(open.ts,z.diff,lag(open.ts,k=-1))

returns

returns<-z.diff/lag(open.ts,k=-1)

returns

plot(returns)

Problem II

Do logit analysis for 700 data points and predict the probability for the next 150 data points.

Solutions:

Commands

z<-read.csv(file.choose(),header=T)
Readdata<-z[1:700,1:9]
Readdata$ed<-factor(Readdata$ed)
Readdata.est<-glm(default~age+ed+employ+address+income+debtinc+creddebt+othdebt,data=Readdata,family="binomial")
summary(Readdata.est)
Forecast<-z[701:850,1:8]
Forecast$ed<-factor(Forecast$ed)
Forecast$prob<-predict(Readdata.est,newdata=Forecast,type="response")
head(Forecast)

IT and Business Application lab : Assignment 3

Problem 1: Using Mileage-Groove data, fit 'lm' and 'comment on the applicability of 'lm'.

Solution:

Commands:

>test1<-read.csv(file.choose(),header=T)

> a<-test1$mileage

> b<-test1$groove

> reg1<- lm(a~b)

> reg1

> res_var<-resid(reg1)

> res_var

> plot(b,res_var)

> qqnorm(res_var)

>qqline(res_var)

Linear Regression is not applicable as can be seen from the graph

Problem 2:

a) Plot residual and standard residual of independent variable

Solution: 

Commands:

 > test2<-read.csv(file.choose(),header=T)

> a<-test2$Pluto

> b<-test2$alpha

> reg1<- lm(a~b)

> reg1

> res_var<-resid(reg1)

> sres<-rstandard(reg1)

B) Plot qqplot and add a qqline
Solution:

Commands:

> plot(b,res_var)

> qqnorm(res_var)

> qqline(res_var)

Problem 3: Justify null hypothesis using Anova

Solution:

Commands:

> anova1<-read.csv(file.choose(),header=T)

> anova2<-aov(anova1$Comfort_level~anova1$Chair)     

p value=0.687

Since p-value is greater than 5%, we can't reject the null hypothesis.

 



IT and Business Application lab : Assignment 2

Problem 1

a) Create two matrices

Solution:            

Command

> a<- c(1,2,3,4,5,6,7,8,9)

> b<- c(32,48,1,5,10,12,15,18,23)

> dim(a)<- c(3,3)

> dim(b)<- c(3,3)

> a

     [,1] [,2] [,3]

[1,]    1    4    7

[2,]    2    5    8

[3,]    3    6    9

> b

     [,1] [,2] [,3]

[1,]   32    5   15

[2,]   48   10   18

[3,]    1   12   23  

b) Select 3rd column from first matrix {a} and 1st column from second matrix {b}

Solution:

Command

> a1<-a[ ,3]

> a1

[1] 7 8 9

> b1<-b[ ,1]

> b1

[1] 32  48 1

c) Combine the two matrices

Solution:

Command

> combine<-cbind(a1,b1)

> combine

     a1 b1

[1,]  7 32

[2,]  8 48

[3,]  9 1

Problem 2: Find the product of two matrices

Solution:

Command

> multiple<- a%*%b

> multiple

     [,1] [,2] [,3]

[1,]  231  129  248

[2,]  312  156  304

[3,]  393  183  360

Problem 3: Find the regression using nifty data from 01 Dec 2012 to 31 Dec 2012 ?

Solution:

Command

> nifty<-read.csv(file.choose(),header=T)

> open<- nifty[ ,2]

> high<- nifty[ ,3]

> data<- cbind(open,high)

> reg1<-lm(high~open,data=nifty)

> reg1

Call:

lm(formula = high ~ open, data = nifty)

Coefficients:

(Intercept)        open

  1578.3358       0.7355

Problem 4: Generate and plot a normal distribution curve

Solution:

Command

> x<-seq(0,100)

> y<-dnorm(x,mean=50,sd=5)

> plot(x,y,type="l")

IT and Business Application lab : Assignment 1

Problem 1:

a) Create line plot of x, where x is a variable.

Solution:            

Command:          

> x<-c(1,2,3)

> plot(x,type="l")

b) Create histogram of NSE data extracted from 1st October 2012 to 4th January 2013.

Solution:            

Command:       

>z<-read.csv(file.choose(),header=T)

>zcol1<-z[ ,3]

> plot(zcol1,type="h")

Problem 2:

Plot both lines and points. Also name the plot, x axis and y axis

Command:

>plot(zcol1,type="b",main="NSEGraph",xlab="Time",ylab="nifty")

Problem 3:

Create scatter plot by selecting the low values.

Command:

> zcol4<- z[,4]

> plot(zcol1,zcol4)

Problem 4:

Find the volatility of data?

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