06-01-2015, 12:57 PM
Warning, statistics ahead...
Running a regression in excel's Data Analysis toolpack, I can replicate your regression equation plus output an ANOVA table to get a better idea of this relationship.
Using your exact set up, I get:
R square = 0.1779 (not very predictive)
Significance F = very low (this is good)
p-value for the payout ratio variable = very low (this is good)
p-value of the intercept variable = 0.053 (this fails the 0.05 threshold)
Since the intercept has a high p-value, I reran the regression using no intercept. I actually got better results with the y = 0.54085*x equation.
R square = 0.2559 (not great, but better)
Significance F = very low (this is good)
p-value for the payout ratio variable = very low (this is good)
Wrap up:
Even with the improved equation, the predictive value of the payout ratio isn't that strong. These two ratios are most likely loosely correlated, with no causation. Use with caution.
Running a regression in excel's Data Analysis toolpack, I can replicate your regression equation plus output an ANOVA table to get a better idea of this relationship.
Using your exact set up, I get:
R square = 0.1779 (not very predictive)
Significance F = very low (this is good)
p-value for the payout ratio variable = very low (this is good)
p-value of the intercept variable = 0.053 (this fails the 0.05 threshold)
Since the intercept has a high p-value, I reran the regression using no intercept. I actually got better results with the y = 0.54085*x equation.
R square = 0.2559 (not great, but better)
Significance F = very low (this is good)
p-value for the payout ratio variable = very low (this is good)
Wrap up:
Even with the improved equation, the predictive value of the payout ratio isn't that strong. These two ratios are most likely loosely correlated, with no causation. Use with caution.