The Democrats in 2006 compared to the Republicans in 1994

Update: The numbers changed a bit since this entry was posted the other day. The Democrats got 55% of the average district vote, not 56%. (The confusion came because we used numbers from the New York Times that counted third-party votes in a different way than we did, I think; see note at the end of this entry.)

Back to the main story:
The Democrats’ victory in the 2006 election has been compared to the Republicans’ in 2004. But the Democrats actually did a lot better in terms of the vote. The Democrats received 54.8% of the average district vote for the two parties in 2006, whereas the Republicans only averaged 51.6% in 1994.

There was a big jump in 2006. Here’s the time series:

postelection.png

(Bigger version is here.) The shaded areas on the graph show the periods where Republicans have controlled the House. The 2006 outcome of 55% for the Democrats is comparable to their typical vote shares as the matjority party in the decades preceding the 1994 realignment.

54.8% of the vote, 53.3% of the seats

Even with their large vote majority, the Democrats only received 53.3% of the seats in the House. This is as we and Bafumi et al. anticipated. More info on the seats-votes relationship is in our recent paper. (For example, had the Republicans received 54.8% of the vote in 2006, we estimate they would’ve won about 245 seats.)

By the way, the Democrats’ 54.8% share of the two-party vote tracks closely with the “generic congressional vote” in which they were getting 56% in the polls (that is, 52.1%/(52.1% + 40.6%)).

Technical notes

We actually calculate average district votes by imputing 75% for uncontested races (to represent the strength that the party might have had if the district had been contested; the 75% comes from computing the average vote in districts just before and just after being uncontested, based on historical data), so we needed to make corrections for uncontesteds. In 2004, we have 31 uncontested Democrats and 37 uncontested Republicans: the average district vote for the Democrats was 50.5% using our correction (or 50.0% if you simply plug in 100% for uncontested races). In 2006, there were 45 uncontested Democrats and 10 uncontested Republicans, yielding an average district vote of 54.8% using our correction (or 56.8% if you simply plug in 100%). The 56.8% number is more dramatic but I think it overstates the Democrats’ strength in giving them 100% in all those districts.

Another option is to use total vote (see here) rather than average district vote. We discuss this in Section 3.3 of our paper (in particular, see Figure 4). The short answer is that we use average district vote because it represents total support for the parties across the country. The Democrats tend to do better in lower-turnout districts and so their total vote is typically slightly lower than their average district vote. See the scatterplot here.

P.S. Here’s a picture of the estimated seats-votes curve for the 2006 election:

sv2006_from_2004.png

More discussion here.

P.P.S. The first version of this blog entry said that the Democrats received 56% of the votes. When doing this calculation, I didn’t have a spreadsheet with all the 2006 data, but the New York Times reported that the average swing was 7.6%. We then corrected for uncontesteds (using the NYT numbers on how many districts were uncontested for each party) to get an average district vote of 56.1%.

Actually, though, it appears based on John Kastellec’s district-by-district tabulations that 56% wasn’t right either. We think the discrepancy arose because of how the media’s count reflects districts in which only one major party candidate ran but third party candidates received some votes. We just want to work with the two-party vote, and we consider an election as uncontested unless it is contested by both the Democrats and the Republicans. The 7.6% average swing reported in the press was counting third party votes in some way, we think.

3 thoughts on “The Democrats in 2006 compared to the Republicans in 1994

  1. Thanks. Good work. I used a slightly different methodology based on aggregating all votes to come up with a slightly smaller advantage for the Democrats.

    http://isteve.blogspot.com/2006/11/democrats-win-

    Democrats tend to do well in Congressional Districts that are "rotten boroughs" where there aren't many voters living because most of the population are aliens, legal and illegal. The Constitutionality of drawing district boundaries liek that has been disputed, with Judge Posner of the 7th Circuit court of appeals ruling that counting aliens in forming districts deprives citizens in other districts of equal represenation. But the issue has yet to go to the Supreme Courth.

  2. Steve,

    The Democrats do better in low-turnout districts: the nonvoters include noncitizens but also children, prisoners, and people who simply choose not to vote. (Differences also arise because, near the end of the decade, districts become more unequal in population.) My impression is that most of the difference between avg district vote and total vote comes from differences in turnout of eligible voters, but I haven't looked at this in detail.

  3. Steve's also wrong about the legal issues. Posner's decision had nothing to do with the Constitution. It basically gave the states a little leeway to disadvantage non-citizens under the Voting Rights Act. The Constitution doesn't ban non-citizens from voting, and apportionment has always been on the basis of population, not the electorate. Women, children, and slaves couldn't vote back in the founding fathers' day, but they were all counted for purposes of apportionment (though slaves were downweighted to 3/5, of course).

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