# Help in R studio task

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

1. An investigator was interested in studying the relation between 𝑦𝑡, 𝑥1𝑡 and 𝑥2𝑡. Use R to estimate the linear regression model
𝑦𝑡= 𝛽0+𝛽1𝑥1𝑡+𝛽2𝑥2𝑡+𝑣𝑡 (1)
and test the null hypothesis that 𝛽2=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

``````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

#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 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 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 – 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 ‘-1.7614’ negative effect on the excess returns on the stock where the excess returns on the market, x2t, have a ‘1.444’ positive effect.

``````Hypothesis Test
``````

The hypotheses are: H0: 𝛽2 = 1.5, H1: 𝛽2 ≠ 1.5
The model has 3 coefficients so: k = 3
T = 180
T-statistic:
t-statistic = ((β_2 ) ̂-1.5)/(SE((β_2 ) ̂))= (1.44-1.5)/0.943= -0.06362672~ t180-3

test = (1.44-1.5)/0.943
test
 -0.06362672
T_k=180-3
pvalue=pt(test,T_k)+1 -pt(-test, T_k)
pvalue
 0.9493393

Outcome
p-value equal to 0.9493393
p-value > 5%
Null hypothesis rejected
𝛽2 ≠ 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 White’s 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 – 1.7614x1t + 1.444x2t + vt
Check assumption
𝑯𝟎:𝑽(𝒖𝒕)=𝝈𝟐
𝑯𝟏:𝑽( 𝒖𝒕)≠𝝈𝟐

``````Calculate the residual series v ̂t and v ̂t2
``````

#Residual series
vt=estlrm\$residuals
vt2=vtvt
Estimation of secondary regression for the White’s test for homoskedasticity
μ ̂_t^2= α1 + 𝛼2𝑥2𝑡+𝛼3𝑥3𝑡+𝛼4𝑥2𝑡2+𝛼5𝑥23𝑡+𝛼6𝑥2𝑡𝑥3𝑡+𝑒𝑡
k’ = 6
m = k’ – 1 = 5
srm = lm (yt~x1t
x2t+(x1t^2)+(x2t^2))
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 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 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 𝐻0:𝑉(𝑢𝑡)=𝜎2 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 𝑥2𝑡 “Granger-causes” 𝑦𝑡. 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 – 1.7614x1t + 1.444x2t + vt
Unrestricted and restricted models
``````

Unrestricted model:
𝑦𝑡=∑_(j=1)^n▒∝j yt-j + ∑_(i=1)^m▒βi zt-i + 𝐷𝑡+𝜀𝑡
Restricted model:
𝑦𝑡=∑_(j=1)^n▒∝j yt-j + 𝐷𝑡+𝜀𝑡
Granger causality test
𝐻0: x2t does not Granger cause 𝑦𝑡
𝐻1: x2t does Granger cause 𝑦𝑡

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 𝐻0
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: 𝜏=γ ̂/(SE((γ) ̂)')
Estimated model: Δ𝑦𝑡𝑑=𝛾𝑦𝑡-1d+Σpi=1 𝛿iΔ𝑦 d 𝑡−𝑖 +𝜀𝑡
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 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 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
``````

𝐻0: yt has a unit root, i.e. not stationary
𝐻1: yt does not have a unit root, i.e. stationary
Results
The value of test statistic τ = -3.602
Critical values of DF-GLS test at 5% significance is -2.93
-3.602 > -2.93 therefore τ < -2.93, τ 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 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 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
``````

𝐻0: ∆yt has a unit root, i.e. not stationary
𝐻1: ∆yt does not have a unit root, i.e. stationary
Results
The value of test statistic τ = -7.8321
Critical values of DF-GLS test at 5% significance is -1.94
-1.94 > -7.8321 therefore τ < -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.

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