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SUMMARY OUTPUTRegression StatisticsMultiple R0.5992984R Square0.3591586Adjusted R Square0.3430165Standard Error0.0440354Observations408ANOVA dfSSMSFSignificance FRegression100.43140.043122.24985.12098E-33Residual3970.76980.0019Total4071.2013    CoefficientsStandard Errort StatP-valueLower 95%Upper 95%Intercept0.12570.06581.91010.0568-0.00370.2550D/P0.03480.02921.19110.2343-0.02260.0922E/P0.05130.02492.06350.03970.00240.1002b/m-0.16900.0380-4.44300.0000-0.2437-0.0942ntis0.05090.13590.37450.7083-0.21640.3182LIQUIDITY-0.33960.0376-9.03410.0000-0.4135-0.2657tbl-0.00710.1778-0.04010.9681-0.35670.3425lty0.18330.27000.67880.4977-0.34760.7142DFY7.59021.05597.18860.00005.51449.6660infl-2.22210.7056-3.14910.0018-3.6093-0.8349IP1.43140.81211.76250.0788-0.16523.0280Volatility = β0+β1 (D/Pi) +β2 (E/Pi) + β3 (b/mi) + β4 (ntisi) + β5 (LIQUIDITYi) + β6 (tbli) + β7 (ltyi) + β8 (DFYi) + β9 (infli) + β10 (IPi)We assess the fitness of the model in three ways. They are standard error of estimate, the coefficient of determination and F-test. The Sε (0.0440354) accounts 163% for mean of y, so Sε is very large. R square is 0.3591586 which means about 35.92% of the total variation in volatility can be explained by those 10 variables, while about 64.08% remains unexplained. Using the F-test, we can find the rejection region is F< F0.05, 10, 397 = 1.855. As we can see from the table F = 22.2498 which is large than F0.05, 10, 397. At least one βi is not equal to zero, so we can conclude this model is valid.  We also need to check the model assumptions.

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Summary Outputregression Statisticsmultiple R And Β0. (June 8, 2021). Retrieved from https://www.freeessays.education/summary-outputregression-statisticsmultiple-r-and-%ce%b20-essay/