Question

A sample containing years to maturity and (percent) yield for 40 corporate bonds is contained in...

A sample containing years to maturity and (percent) yield for 40 corporate bonds is contained in the DATAfile named CorporateBonds (Barron's, April 2, 2012).

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 an estimated quadratic regression equation with years to maturity and squared values of years to maturity as the independent variables. If required, round your answers to four decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300)

y=___+___x+___x^2
How much variation in the sample values of yield is explained by this regression model? If required, round your answer to one decimal places

__%

Test the relationship between each of the independent variables and the dependent variable at a 0.05 level of significance. How would you interpret this model?

Homework Answers

Answer #1

USING minitab>stat>Regression>regression line plot>quadratic

we have

Polynomial Regression Analysis: Yield versus Years

The regression equation is
Yield = 1.017 + 0.4606 Years - 0.01025 Years^2


S = 0.958250 R-Sq = 66.8% R-Sq(adj) = 65.0%


Analysis of Variance

Source DF SS MS F P
Regression 2 68.301 34.1506 37.19 0.000
Error 37 33.975 0.9182
Total 39 102.276


A)an estimated quadratic regression equation with years to maturity and squared values of years to maturity as the independent variables is

y = 1.017 + 0.4606 x - 0.01025 x^2

66.8% variation in the sample values of yield is explained by this regression model.

since p value of f statistics is 0.000 which is less than significance level 0.05 so we can say that there is a relationship between each of the independent variables and the dependent variable at a 0.05 level of significance

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