Question

A sample containing years to maturity and yield () for corporate bonds are contained in the table below.

Company Ticker	Years		Yield
GE	1.00		0.767
MS	1.10		1.716
WFC	1.35		0.897
TOTAL	1.75		1.378
TOTAL	3.25		1.748
GS	3.75		3.558
MS	4.00		4.413
JPM	4.25		2.310
C	4.75		3.332
RABOBK	4.75		2.805
TOTAL	5.00		2.069
MS	5.00		4.739
AXP	5.00		2.181
MTNA	5.00		4.366
BAC	5.00		3.699
VOD	5.00		1.855
SHBASS	5.00		2.861
AIG	5.00		3.452
HCN	7.00		4.184
MS	9.25		5.798
GS	9.25		5.365
GE	9.50		3.778
GS	9.75		5.367
C	9.75		4.414
BAC	9.75		4.949
RABOBK	9.75		4.203
WFC	10.00		3.682
TOTAL	10.00		3.270
MTNA	10.00		6.046
LNC	10.00		4.163
FCX	10.00		4.030
NEM	10.00		3.866
PAA	10.25		3.856
HSBC	12.00		4.079
GS	25.50		6.913
C	25.75		8.204
GE	26.00		5.130
GE	26.75		5.138
T	28.50		4.930
BAC	29.75		5.903

a. Choose the correct scatter diagram of the data using  years to maturity as the independent variable.

A	$appletImage?dbid=188426822$
B	$appletImage?dbid=1276957862$
C	$appletImage?dbid=748657751$
D	$appletImage?dbid=1991259189$

- Select your answer -ABCDItem 1

Does a simple linear regression model appear to be appropriate?

- Select your answer -A simple linear regression model appears to be appropriate.A simple linear regression model does not appear to be appropriate.Item 2

b. Develop an estimated regression equation with  years to maturity and  as the independent variables (to 4 decimals). Enter negative values as negative numbers.

      

What is the value of the coefficient of determination? Note: Report  between 0 and 1.

(to 3 decimals)

What is the value of the  test statistic?

(to 2 decimals)

What is the -value? If your answer is zero enter "0".

(to 4 decimals)

c. As an alternative to fitting a second-order model, fit a model using the natural logarithm of price as the independent variable; that is, (to 3 decimals).

    

What is the value of the coefficient of determination? Note: Report  between 0 and 1.

(to 3 decimals)

What is the value of the  test statistic?

(to 2 decimals)

What is the -value? If your answer is zero enter "0".

(to 4 decimals)

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

- Select your answer -Yes, it doesNo, it does notItem 14 Because  .

Answer 1

a. Draw a scatter plot using Minitab, the results are given below: Scatterplot of Yield vs Year Yield 10 15 Year 20 25 30 From the above graph, the correct graph is D. b. By observing the graph, a simple linear regression model does not appear to be appropriate. Perform a regression analysis using Minitab, the results are given below:

WORKSHEET 1 Regression Analysis: Yield versus Year, year2 Regression Equation Yield = 1.003 - 04629 Year -0.01032 year2 Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 1.003 0.437 2.30 0.027 Year 0.4629 0.0816 5.67 0.000 18.06 year2 -0.01032 0.00259 -3.98 0.000 18.06 Model Summary S R-sq R-sq(adj) R-sq(pred) 0.955911 66.88% 65.09% 61.24% Analysis of Variance Source DF Adj SS Adj MS F-Value P-value Regression 2 68.28 34.141537.36 0.000 Year 1 29.42 29.4226 32.200.000 year2 1 14.48 14.4795 15.850.000 Error 37 33.81 0.9138 Lack-of-Fit 20 19.80 0.9901 1.200.354 Pure Error 17 14.01 0.8240 Total 39 102.09

The estimated regression equation is, Yield=1.003+0.4629 Year -0.0103 Year. The value of the coefficient of determination is 0.6688. The value of F test statistic is 37.36. The p-value is 0. c. Perform the regression analysis using Minitab, the results are given below:

Regression Analysis: Yield versus log(year) Regression Equation Yield = 0.781 - 1.583 log(year) Coefficients Term Coef SE Coef T-Value P-Value VIF Constant 0.781 0.381 2.05 0.047 loglyear) 1.583 0.179 8.85 0.000 1.00 Model Summary S R-sq R-sq(adj) R-sq(pred) 0.937136 67.31% 66,45% 63.76% Analysis of Variance Source DF Adj SS Adj MS F-Value P-value Regression 1 68.72 68.7198 78.250.000 log year) 1 68.72 68.7198 78.250.000 Error 3 8 33.37 0.8782 Lack-of-Fit 21 19.37 0.9221 1.12 0.411 Pure Error 17 14.01 0.8240 Total 39 102.09

The estimated regression equation is, Yield = 0.781+1.583 log (year). The value of the coefficient of determination is 0.6731. The value of F test statistic is 78.25. The p-value is 0. Yes, the estimated regression equation using natural logarithm of x provides a better fit than estimated regression equation developed in part (b).