Question

In: Statistics and Probability

A financial analyst tested the performance of Portfolio A against the S&P500 Index during 150 months...

A financial analyst tested the performance of Portfolio A against the S&P500 Index during 150 months of transactions. The figure below shows part of the output from a regression between the two sets of returns.

ANOVA
df SS MS
Regression 1 0.080 0.080
Residual 120 0.107 0.000
Total 121 0.187
Coefficients Standard Error
Intercept -0.001 0.003
S&P500 0.346 0.037

What is the estimated equation that corresponds to these results?Find the R2 and provide a brief explanation about the meaning of yourresult.Find the t-statistic for the coefficient of the intercept and of the variableS&P500.Give a 95% confidence interval for the coefficient for the variable S&P500. What return for the Portfolio A would be expected if S&P500’s returnwas 2.5%?

Solutions

Expert Solution

estimated equation is Y^ = -0.001 + 0.346*S&P500

R² = SSR/SST = 0.080/0.187 = 0.4278

about 42.78% of variation in observation of return for portfolio A is explained by variable S&P500

t-statistic for the coefficient of the intercept = -0.001/0.003= -0.33333

t-statistic for the coefficient of the variableS&P500 = 0.346/0.037= 9.351351351

----------------

confidence interval for slope                  
                  
n =   122              
alpha,α =    0.05              
estimated slope=   0.346              
std error =    0.037              
                  
Df = n-2 =   120              
t critical value =    1.9799   [excel function: =t.inv.2t(α,df) ]          
                  
margin of error ,E = t*std error =    1.9799   *   0.037   =   0.07326
                  
95%   confidence interval is ß1 ± E               
lower bound = estimated slope - margin of error =    0.346   -   0.0733   =   0.273
upper bound = estimated slope + margin of error =    0.346   +   0.0733   =   0.419

---------------

Y^ = -0.001 + 0.346*S&P500

S&P500’s return =  2.5%

return for the Portfolio A = -0.001 + 0.346*2.5 = 0.864 %


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