Hello,

I wonder if it is possible for you to check what I have done here to check for any obvious errors etc., please?

Kind regards

Question 1

- An investigator was interested in studying the relation between π¦π‘, π₯1π‘ and π₯2π‘. Use R to estimate the linear regression model

π¦π‘= π½0+π½1π₯1π‘+π½2π₯2π‘+π£π‘ (1)

and test the null hypothesis that π½2=1.5. Make sure to show the details of your work.

Given regression model:

Where:

yt = The excess returns on a stock.

And two x variables:

x1t = The change in money supply

x2t = The excess returns on the market

Steps I made:

```
Estimating the linear regression model
Read the data in the R software
Remove missing data
Assign each column a variable
Estimate linear regression model
```

Note: Refined data has 180 observations meaning T=180

#Read in the data

assignment_data1 = read.csv("C:/Uv")

#Remove missing data

assignment_data = na.omit(assignment_data1)

#Assigning variables

x1t = assignment_data[,2]

x2t = assignment_data[,3]

yt = assignment_data[,4]

#Estimation of linear regression model

estlrm = lm(yt~x1t+x2t, anscombe)

#Estimation results

summary(estlrm)

Outcome

Call:

lm(formula = yt ~ x1t + x2t, data = anscombe)

Residuals:

Min 1Q Median 3Q Max

-66.681 -3.519 2.178 7.097 23.886

##
Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -1.7290 1.1138 -1.552 0.1224

x1t -1.7614 0.9433 -1.867 0.0635 .

x2t 1.4441 0.2617 5.518 1.2e-07 ***

Signif. codes: 0 β* β 0.001 ββ 0.01 β*β 0.05 β.β 0.1 β β 1

Residual standard error: 14.8 on 177 degrees of freedom

Multiple R-squared: 0.1655, Adjusted R-squared: 0.156

F-statistic: 17.55 on 2 and 177 DF, p-value: 1.115e-07

Results

The excess returns on stock is described by:

yt = -1.7290 β 1.7614x1t + 1.444x2t + vt

According to adjusted R-squared, 15.6% of the variation in the excess returns on stock is explained by the fitted model. Changes in money supply, x1t, have a β-1.7614β negative effect on the excess returns on the stock where the excess returns on the market, x2t, have a β1.444β positive effect.

```
Hypothesis Test
```

The hypotheses are: H0: π½2 = 1.5, H1: π½2 β 1.5

The model has 3 coefficients so: k = 3

T = 180

T-statistic:

t-statistic = ((Ξ²_2 ) Μ-1.5)/(SE((Ξ²_2 ) Μ))= (1.44-1.5)/0.943= -0.06362672~ t180-3

test = (1.44-1.5)/0.943

test

[1] -0.06362672

T_k=180-3

pvalue=pt(test,T_k)+1 -pt(-test, T_k)

pvalue

[1] 0.9493393

Outcome

p-value equal to 0.9493393

p-value > 5%

Null hypothesis rejected

π½2 β 1.5 at the 5% level

Interpretation: A unit change in money supply will not change the excess returns on a stock by 2 units at the 5% level of significance.

Question 2

The investigator then checked the homoskedasticity in the residuals of model (1). Provide a detailed account of the Whiteβs test for homoskedasticity and use R to illustrate your answer by examining possible homoskedasticity in model (1) at the 1% level of significance.

```
Unrestricted model
```

From question 1: yt = -1.7290 β 1.7614x1t + 1.444x2t + vt

Check assumption

π―π:π½(ππ)=ππ

π―π:π½( ππ)β ππ

```
Calculate the residual series v Μt and v Μt2
```

#Residual series

vt=estlrm$residuals

vt2=vtvtx2t+(x1t^2)+(x2t^2))

Estimation of secondary regression for the Whiteβs test for homoskedasticity

ΞΌ Μ_t^2= Ξ±1 + πΌ2π₯2π‘+πΌ3π₯3π‘+πΌ4π₯2π‘2+πΌ5π₯23π‘+πΌ6π₯2π‘π₯3π‘+ππ‘

kβ = 6

m = kβ β 1 = 5

srm = lm (yt~x1t

summary(srm)

Call:

lm(formula = yt ~ x1t * x2t + (x1t^2) + (x2t^2))

Residuals:

Min 1Q Median 3Q Max

-65.387 -2.861 1.623 6.825 25.571

##
Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -1.7199 1.1126 -1.546 0.1240

x1t -2.0984 0.9845 -2.131 0.0344 *

x2t 1.5513 0.2767 5.605 7.9e-08 ***

x1t:x2t 0.3002 0.2542 1.181 0.2392

Signif. codes: 0 β* β 0.001 ββ 0.01 β*β 0.05 β.β 0.1 β β 1

Residual standard error: 14.79 on 176 degrees of freedom

Multiple R-squared: 0.172, Adjusted R-squared: 0.1579

F-statistic: 12.19 on 3 and 176 DF, p-value: 2.758e-07

```
Calculate test statistic
```

R2= 0.172

T= 177 + 3 = 180

T * R2 = 180 * 0.172 = 30.96 ~Xm2

m = 9 , T = 180

p-value = 2.758e-07 > 1%

Do not reject π»0:π(π’π‘)=π2 at 1% level of significance

Meaning: ut is a constant, the assumption of constant variance of ut is satisfied.

Question 3

The investigator also checked whether π₯2π‘ βGranger-causesβ π¦π‘. Use R to perform the Granger causality test (with order = 4) to complete this task and show the detailed work of your analysis at the 5% level of significance.

```
Estimated original model from question 1: yt = -1.7290 β 1.7614x1t + 1.444x2t + vt
Unrestricted and restricted models
```

Unrestricted model:

π¦π‘=β_(j=1)^nββj yt-j + β_(i=1)^mβΞ²i zt-i + π·π‘+ππ‘

Restricted model:

π¦π‘=β_(j=1)^nββj yt-j + π·π‘+ππ‘

Granger causality test

π»0: x2t does not Granger cause π¦π‘

π»1: x2t does Granger cause π¦π‘

grangertest(yt~x2t,order=4)

Granger causality test

Model 1: yt ~ Lags(yt, 1:4) + Lags(x2t, 1:4)

Model 2: yt ~ Lags(yt, 1:4)

Res.Df Df F Pr(>F)

1 167

2 171 -4 1.3048 0.2703

```
Results
Test statistic = 1.3048
p-value = 0.2703 > 5% => not rejected π»0
There is no unidirectional causality from x2t to yt
Meaning:
```

Changes in the excess returns on the market does not Granger Cause changes of the excess return on the stock. Past values of the excess returns on the market cannot be used to predict the excess return on the stock.

Question 4

In addition to the methods used in the first three questions, you have also learnt several other methods in this module, which can also be used to analyse the data. Choose one of these methods, provide a detailed account of the selected method, explain why the selected method is suitable for analysing the data, and use software R to illustrate your results. (Note: It is not necessary to use all variables in the data set for this question.)

```
Method used: DF-GLS test
```

Test statistic: π=Ξ³ Μ/(SE((Ξ³) Μ)')

Estimated model: Ξπ¦π‘π=πΎπ¦π‘-1d+Ξ£pi=1 πΏiΞπ¦ d π‘βπ +ππ‘

Conduct test on the excess returns on a stock data for possible unit root and non-stationary

DF-GLS test

testyr=ur.ers(yt,model="trend",lag.max=12)

summary(testyr)

###############################################

# Elliot, Rothenberg and Stock Unit Root Test

###############################################

Test of type DF-GLS

detrending of series with intercept and trend

Call:

lm(formula = dfgls.form, data = data.dfgls)

Residuals:

Min 1Q Median 3Q Max

-63.544 -6.369 0.601 7.631 25.733

##
Coefficients:

Estimate Std. Error t value Pr(>|t|)

yd.lag -1.52852 0.42435 -3.602 0.000425 ***

yd.diff.lag1 0.35778 0.40630 0.881 0.379916

yd.diff.lag2 0.28591 0.38231 0.748 0.455691

yd.diff.lag3 0.28585 0.35561 0.804 0.422735

yd.diff.lag4 0.07563 0.32656 0.232 0.817151

yd.diff.lag5 -0.05515 0.30005 -0.184 0.854418

yd.diff.lag6 -0.08919 0.27239 -0.327 0.743775

yd.diff.lag7 -0.05843 0.24162 -0.242 0.809253

yd.diff.lag8 -0.06147 0.20828 -0.295 0.768283

yd.diff.lag9 -0.20371 0.17524 -1.162 0.246855

yd.diff.lag10 -0.19682 0.14889 -1.322 0.188134

yd.diff.lag11 -0.19340 0.11596 -1.668 0.097388 .

yd.diff.lag12 -0.05587 0.07541 -0.741 0.459926

Signif. codes: 0 β* β 0.001 ββ 0.01 β*β 0.05 β.β 0.1 β β 1

Residual standard error: 15.35 on 154 degrees of freedom

Multiple R-squared: 0.6139, Adjusted R-squared: 0.5813

F-statistic: 18.83 on 13 and 154 DF, p-value: < 2.2e-16

Value of test-statistic is: -3.602

Critical values of DF-GLS are:

1pct 5pct 10pct

critical values -3.46 -2.93 -2.64

```
Null hypothesis
```

π»0: yt has a unit root, i.e. not stationary

π»1: yt does not have a unit root, i.e. stationary

Results

The value of test statistic Ο = -3.602

Critical values of DF-GLS test at 5% significance is -2.93

-3.602 > -2.93 therefore Ο < -2.93, Ο is in the rejection region

The null hypothesis is rejected meaning data of the excess returns on a stock is stationary and does not have a unit root

Differenced data test of the excess returns on stock

dyt=diff(yt)

testdyt=ur.ers(dyt,model = "constant", lag.max=12)

summary(testdyt)

###############################################

# Elliot, Rothenberg and Stock Unit Root Test

###############################################

Test of type DF-GLS

detrending of series with intercept

Call:

lm(formula = dfgls.form, data = data.dfgls)

Residuals:

Min 1Q Median 3Q Max

-71.651 -9.224 -3.291 4.059 22.163

##
Coefficients:

Estimate Std. Error t value Pr(>|t|)

yd.lag -10.50764 1.34161 -7.832 7.46e-13 ***

yd.diff.lag1 8.45162 1.30406 6.481 1.18e-09 ***

yd.diff.lag2 7.42681 1.23377 6.020 1.24e-08 ***

yd.diff.lag3 6.53005 1.14288 5.714 5.60e-08 ***

yd.diff.lag4 5.53392 1.03931 5.325 3.55e-07 ***

yd.diff.lag5 4.56873 0.92527 4.938 2.05e-06 ***

yd.diff.lag6 3.69254 0.79687 4.634 7.63e-06 ***

yd.diff.lag7 2.95325 0.66066 4.470 1.51e-05 ***

yd.diff.lag8 2.30471 0.52597 4.382 2.17e-05 ***

yd.diff.lag9 1.61140 0.40006 4.028 8.84e-05 ***

yd.diff.lag10 1.04724 0.28198 3.714 0.000285 ***

yd.diff.lag11 0.53731 0.16990 3.163 0.001886 **

yd.diff.lag12 0.21746 0.07417 2.932 0.003885 **

Signif. codes: 0 β* β 0.001 ββ 0.01 β*β 0.05 β.β 0.1 β β 1

Residual standard error: 16.05 on 153 degrees of freedom

Multiple R-squared: 0.864, Adjusted R-squared: 0.8524

F-statistic: 74.74 on 13 and 153 DF, p-value: < 2.2e-16

Value of test-statistic is: -7.8321

Critical values of DF-GLS are:

1pct 5pct 10pct

critical values -2.58 -1.94 -1.62

```
Null hypothesis
```

π»0: βyt has a unit root, i.e. not stationary

π»1: βyt does not have a unit root, i.e. stationary

Results

The value of test statistic Ο = -7.8321

Critical values of DF-GLS test at 5% significance is -1.94

-1.94 > -7.8321 therefore Ο < -1.93

The null hypothesis is rejected meaning data of the differenced excess returns on a stock does not have a unit root and it is stationary.