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Statistic

Essay by   •  April 1, 2016  •  Term Paper  •  821 Words (4 Pages)  •  1,020 Views

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SUMMARY OUTPUT

Regression Statistics

Multiple R

0.5992984

R Square

0.3591586

Adjusted R Square

0.3430165

Standard Error

0.0440354

Observations

408

ANOVA

 

df

SS

MS

F

Significance F

Regression

10

0.4314

0.0431

22.2498

5.12098E-33

Residual

397

0.7698

0.0019

Total

407

1.2013

 

 

 

 

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

0.1257

0.0658

1.9101

0.0568

-0.0037

0.2550

D/P

0.0348

0.0292

1.1911

0.2343

-0.0226

0.0922

E/P

0.0513

0.0249

2.0635

0.0397

0.0024

0.1002

b/m

-0.1690

0.0380

-4.4430

0.0000

-0.2437

-0.0942

ntis

0.0509

0.1359

0.3745

0.7083

-0.2164

0.3182

LIQUIDITY

-0.3396

0.0376

-9.0341

0.0000

-0.4135

-0.2657

tbl

-0.0071

0.1778

-0.0401

0.9681

-0.3567

0.3425

lty

0.1833

0.2700

0.6788

0.4977

-0.3476

0.7142

DFY

7.5902

1.0559

7.1886

0.0000

5.5144

9.6660

infl

-2.2221

0.7056

-3.1491

0.0018

-3.6093

-0.8349

IP

1.4314

0.8121

1.7625

0.0788

-0.1652

3.0280

Volatility = β01 (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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