In both situations, the dummy indicators resulted in a slightly higher R squared and F significance but a lowered overall F value. Also, when the dummy indicators were included with the original Independent variables none of them had great T stats or p values. Nothing would indicate that any of these dummy variables add a large amount of accuracy to these models.
To predict the turnout and results of the 2014 Senate elections in four Colorado Counties (Pueblo, El Paso, Yuma and Arapahoe) I collected previous data for all counties on Senate elections dating back to 1998. The 5 years I used data from were ’98, ’02, ’04, ’08 and ’10. I then found the congressional approval rating for each of those years. I compared the Pearson Correlation Coefficient I got from comparing everything both by year and by congressional approval rating. Whichever comparison had a higher ‘r’ was given a weight of 2 and then added to the other prediction with the total being divided by 3.
Turnout(of pop): 41.81%, Dem Vote(total %): 49.35%, Rep Vote(total%): 44.92, Unaffiliated(total%): 5.73%
After getting my weighted predictions, I computed how far each county deviated from the average for each category in 2010. I then adjusted each county to have the same deviation from my prediction. This adjustment seemed reasonable for the turnout numbers because this is an off year and it resulted in a lower number. However, for the total party votes this widened the gap in between votes for democrats and votes for Republicans. With many of my counties, particularly Yuma which is my strong democratic county, they weren’t voting consistently year after year, so assuming that their predisposition to vote increasingly democratic in the past few years would continue didn’t seem accurate. Thus I went with just my predicted voting percentages for the parties and my prediction adjusted by typical standard deviation for the county in order to predict turnout.
My results are as follows:
This overall model is insignificant since using critical value methods to assess it has shown that we fail to reject the null hypothesis that the slope is equal to 0.
This overall model is significant and we are clearly able to reject the null hypothesis that the slope is equal to 0.
This overall model is insignificant and you can see that we fail to reject the null hypothesis that the slope is equal to 0.