Question

The Capital Asset Price Model (CAPM) is a financial model that attempts to predict the rate of return on a financial instrument, such as a common stock, in such a way that it is linearly related to the rate of return on the overal market. Specifically,

RStockA,i = β0 + β1RMarket,i + ei

You are to study the relationship between the two variables and estimate the above model:

iRStockA,i - rate of return on Stock A for month i, i=1,2,⋯59.

iRMarket,i - market rate of return for month ii, i=1,2,⋯,59

β1 represent's the stocks 'beta' value, or its systematic risk. It measure's the stocks volatility related to the market volatility. β0 represents the risk-free interest rate.

The data in the  file contains the data on the rate of return of a large energy company which will be referred to as Acme Oil and Gas and the corresponding rate of return on the Toronto Composite Index (TSE) for 59 randomly selected months.

TSERofReturn	AcmeRofReturn
2.29651	-0.34793
-1.61176	-1.75424
0.8957	0.24095
-0.46309	-0.52434
1.17586	-1.39147
0.36339	-0.89941
-0.09888	0.62191
1.54007	0.21203
1.20388	0.89063
0.40541	-0.31979
-0.50512	-0.26566
-2.94253	-0.48511
0.39141	-1.22745
2.9549	2.35981
-2.39621	-0.02795
-0.16892	-0.63943
-0.09888	-0.69269
-0.60317	-0.57024
-1.8639	-1.26911
1.79222	-0.16832
-0.16892	-0.73469
2.08639	0.33578
-1.31759	-0.99294
1.17586	0.06602
-0.1409	-0.02439
-1.56973	1.75941
5.16818	3.23171
-0.00082	1.19321
-1.24755	0.74471
-0.4771	-0.28887
-0.86933	0.4171
-0.46309	-1.21974
0.5595	1.06245
-0.32301	-0.14503
-0.50512	1.69671
-0.00082	0.58354
0.34938	-2.45484
-0.68722	0.452
4.08955	0.93878
-3.01257	-1.62261
-3.71298	0.25316
-0.29499	-0.51118
0.93772	1.53503
1.63813	0.82144
0.71359	0.61567
-3.22269	-0.22444
0.5455	1.42175
-0.60317	-1.03702
1.91829	0.51314
-0.15491	0.07771
-1.91994	0.10144
-0.23896	0.22354
-1.59775	1.36347
0.23732	-0.61873
-1.19151	-0.96878
-1.30358	0.00046
2.87085	1.67688
2.05837	-2.55599
-1.10747	-0.01911

Therefore RAcme,i represents the monthly rate of return for a common share of Acme Oil and Gas stock; RTSE,i represents the monthly rate of return (increase or decrease) of the TSE Index for the same month, month ii. The first column in this data file contains the monthly rate of return on Acme Oil and gas stock; the second column contains the monthly rate of return on the TSE index for the same month.

(a) Use software to estimate this model. Use four-decimals in each of your least-squares estimates your answer.


RAcme,i^ = ____+____RTSE,i

(b) Find the coefficient of determination. Expresses as a percentage, and use two decimal places in your answer.

r2=

(c) In the context of the data, interpret the meaning of the coefficient of determination.

A. There is a strong, positive linear relationship between the monthly rate of return of Acme stock and the monthly rate of return of the TSE Index.
B. There is a weak, positive linear relationship between the monthly rate of return of Acme stock and the monthly rate of return of the TSE Index.
C. The percentage found above is the percentage of variation in the monthly rate of return of the TSE Index that can be explained by its linear dependency with the monthly rate of return of Acme stock.
D. The percentage found above is the percentage of variation in the monthly rate of return of Acme stock that can be explained by its linear dependency with the monthly rate of return of the TSE Index.


(d) Find the standard deviation of the prediction/regression, using two decimals in your answer.

Se =

(e, i) You wish to test if the data collected supports the statistical model listed above. That is, can the monthly rate of return on Acme stock be expressed as a linear function of the monthly rate of return on the TSE Index? Select the correct statistical hypotheses which you are to test.

A. H0:β0=0HA:β0≠0H0:β0=0HA:β0≠0
B. H0:β1=0HA:β1≠0H0:β1=0HA:β1≠0
C. H0:β1=0HA:β1>0H0:β1=0HA:β1>0
D. H0:β1=0HA:β1<0H0:β1=0HA:β1<0
E. H0:β0=0HA:β0>0H0:β0=0HA:β0>0
F. H0:β1≠0HA:β1≠0H0:β1≠0HA:β1≠0
G. H0:β0=0HA:β0<0H0:β0=0HA:β0<0
H. H0:β0≠0HA:β0≠0H0:β0≠0HA:β0≠0

(e, ii) Use the FF-test, test the statistical hypotheses determined in (e, i). Find the value of the test statistic, using three decimals in your answer.

Fcalc =

(e, iii) Find the P-value of your result in (e, ii). Use three decimals in your answer.

P-value =

(f) Find a 95% confidence interval for the slope term of the model, β1.

Lower Bound =

(use three decimals in your answer)

Upper Bound =

(use three decimals in your answer)


(h) Find a 95% confidence interval for the β0 term of the model.

Lower Bound =

(use three decimals in your answer)

Upper Bound =

(use three decimals in your answer)


(k) Last month, the TSE Index's monthly rate of return was 1.5%. This is, at the end of last month the value of the TSE Index was 1.5% higher than at the beginning of last month. With 95% confidence, find the last month's rate of return on Acme Oil and Gas stock.

Lower Bound =

(use three decimals in your answer)

Upper Bound =

(use three decimals in your answer)

Answer 1

using excel>data>data analysis>Regression

we have

Regression Analysis
Regression Statistics
Multiple R	0.358999297
R Square	0.128880495
Adjusted R Square	0.113597697
Standard Error	1.03846292
Observations	59
ANOVA
	df	SS	MS	F	Significance F
Regression	1	9.09423771	9.09423771	8.433042984	0.005234035
Residual	57	61.46909845	1.078405236
Total	58	70.56333616
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	0.021969143	0.135197959	0.162496118	0.871489886	-0.248760099	0.292698385
TSERofReturn	0.233072633	0.080259994	2.903970211	0.005234035	0.072354766	0.393790499

(a)the regression equation is

RAcme,i^ = 0.0220+0.2331*RTSE,i

(b)the coefficient of determination

r2=12.89%

(c) the meaning of the coefficient of determination.

D. The percentage found above is the percentage of variation in the monthly rate of return of Acme stock that can be explained by its linear dependency with the monthly rate of return of the TSE Index.


(d) the standard deviation of the prediction/regression is

Se =1.04

(e, i) the correct null and alternative hypothesis is
B. H0:β1=0

HA:β1≠0

(e, ii)

Fcalc =8.433

(e, iii) the P-value of your result
P-value =0.005

(f) a 95% confidence interval for the slope term of the model, β1.

Lower Bound =0.072

Upper Bound =0.394

(h) a 95% confidence interval for the β0 term of the model.

Lower Bound =-0.249

Upper Bound =0.292