Regression Statistics 4 MultipleR 5 R Square 6 Adjusted R Square 7 Standard Erro

Question

Regression Statistics 4 MultipleR 5 R Square 6 Adjusted R Square 7 Standard Error 8 Observations 9 10 ANOVA 0.245257 0.060151 0.043947 0.00515 60 F Significance F 1 9.84603E-05 9.85E-05 3.712028 0.058926051 df MS 12 Regression 3sidul 14 Total 15 16 17 Intercept 58 0.00153843 2.65E-05 59 0.00163689 Coefficients Standard Error-, t Stat-, p-value Lower 95% Upper 95% 0.104695 0.001654144 63.29272 3.17E-55 0.101384134 0.10800639 18 Interest rate differencial 19 0.05951 0.030889062 -1.92666 0.058926 -0.121343972 0.00231838 20 21 22 RESIDUAL OUTPUT

Explanation / Answer

As I can see the P value for interest rate differential, it is 0.058926 which is more than 0.05 for 95% confidence interval, hence one cannot reject the null hypothesis.

Also t value for 59 degree of freedom shall be 2.0010 which is again numerically larger than 1.9266 achieved hence one cannot reject the null.

This goes on to say that there is no valid linear relation between interest rate differential and currency rates at 95% confidence level

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