Question

A statistical program is recommended.

A sample containing years to maturity and yield (%) for 40 corporate bonds is contained in the data file named CorporateBonds.

Company Ticker	Years	Yield
GE	1	0.767
MS	1	1.816
WFC	1.25	0.797
TOTAL	1.75	1.378
TOTAL	3.25	1.748
GS	3.75	3.558
MS	4	4.413
JPM	4.25	2.31
C	4.75	3.332
RABOBK	4.75	2.805
TOTAL	5	2.069
MS	5	4.739
AXP	5	2.181
MTNA	5	4.366
BAC	5	3.699
VOD	5	1.855
SHBASS	5	2.861
AIG	5	3.452
HCN	7	4.184
MS	9.25	5.798
GS	9.25	5.365
GE	9.5	3.778
GS	9.75	5.367
C	9.75	4.414
BAC	9.75	4.949
RABOBK	9.75	4.203
WFC	10	3.682
TOTAL	10	3.27
MTNA	10	6.046
LNC	10	4.163
FCX	10	4.03
NEM	10	3.866
PAA	10.25	3.856
HSBC	12	4.079
GS	25.5	6.913
C	25.75	8.204
GE	26	5.13
GE	26.75	5.138
T	28.5	4.93
BAC	29.75	5.903

(a)

Develop a scatter diagram of the data using x = years to maturity as the independent variable.

a

b

c

d

Does a simple linear regression model appear to be appropriate?

Given the downward trend of the data on the left side of the plot, a linear regression model would predict lower values for the data on the right side of the plot. So, a curvilinear regression model appears to be more appropriate.

Since the data on the left and right sides of the plot both trend upward at about the same rate, a linear model is appropriate.    

Since the data on the left and right sides of the plot both trend downward at about the same rate, a linear model is appropriate.

Given the upward trend of the data on the left side of the plot, a linear regression model would predict higher values for the data on the right side of the plot. So, a curvilinear regression model appears to be more appropriate.

(b)

Develop an estimated regression equation with x = years to maturity and x² as the independent variables. (Round your numerical values to two decimal places.)

ŷ =

(c)

As an alternative to fitting a second-order model, fit a model using the natural logarithm of years to maturity as the independent variable; that is, ŷ = b₀ + b₁ ln(x). (Round your numerical values to two decimal places.)ŷ =

Does the estimated regression using the natural logarithm of x provide a better fit than the estimated regression developed in part (b)? Explain.

The regression equation developed in part (b) provides a better fit since its R² value is higher and it predicts that yield will begin to decrease after a certain point with respect to years to maturity.

The regression equation developed in part (c) provides a better fit because it has less influential observations than the equation developed in part (b).    

The regression equation developed in part (c) provides a better fit since its R² value is higher and it predicts that yield will always increase with respect to years to maturity.

The regression equation developed in part (b) provides a better fit since it uses more independent variables than the equation developed in part (c).

Answer 1

Solution

a)

Does a simple linear regression model appear to be appropriate?

Given the upward trend of the data on the left side of the plot, a linear regression model would predict higher values for the data on the right side of the plot. So, a curvilinear regression model appears to be more appropriate.

(b)

Develop an estimated regression equation with x = years to maturity and x² as the independent variables

we will solve it by using excel and the steps are

Enter the Data into excel

Click on Data tab

Click on Data Analysis

Select Regression

Select input Y Range as Range of dependent variable.

Select Input X Range as Range of independent variable

click on labels if your selecting data with labels

click on ok.

So this is the output of Regression in Excel.

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.82
R Square	0.67
Adjusted R Square	0.65
Standard Error	0.96
Observations	40.00
ANOVA
	df	SS	MS	F	Significance F
Regression	2.00	68.30	34.15	37.19	0.00
Residual	37.00	33.97	0.92
Total	39.00	102.28
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	1.02	0.44	2.34	0.03	0.13	1.90	0.13	1.90
Years	0.46	0.08	5.66	0.00	0.30	0.63	0.30	0.63
Years^2	-0.01	0.00	-3.96	0.00	-0.02	-0.01	-0.02	-0.01

Yield = 1.02+0.46*Years-0.01*Years^2

(c)

As an alternative to fitting a second-order model, fit a model using the natural logarithm of years to maturity as the independent variable

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.82
R Square	0.67
Adjusted R Square	0.66
Standard Error	0.94
Observations	40.00
ANOVA
	df	SS	MS	F	Significance F
Regression	1.00	68.48	68.48	76.98	0.00
Residual	38.00	33.80	0.89
Total	39.00	102.28
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	0.83	0.38	2.18	0.04	0.06	1.60	0.06	1.60
log(Years)	1.56	0.18	8.77	0.00	1.20	1.92	1.20	1.92

Yield = 0.83+1.56*log(Years)

Does the estimated regression using the natural logarithm of x provide a better fit than the estimated regression developed in part (b)? Explain

The regression equation developed in part (c) provides a better fit because it has less influential observations than the equation developed in part (b).