Newsweek poll assumes lower GOP turnout than in 2008

What do you make of a poll that has Republicans losing on the generic ballot by 5 points among registered voters, and by 8 points among “definite” voters, even though the GOP leads among independent voters by 47-30 percent?

I am referring, of course, to the new Newsweek poll that’s been released. How do you suppose Republicans can be losing so badly if they’re winning independents by a ratio better than three-to-two? Here’s how: If you look at how each party does by the voters’ partisan ID, and do the three-by-three equation (handy tool for that here),

```
(0.93)r+ (.02)d + (0.47)i = 0.43 
```
```
(0.04)r + (0.96)d + (0.30)i = 0.48
```
```
r + d + i = 1 
```

you’ll find their registered voter sample is weighted as follows:

39 percent Democrat
29 percent Republican,
32 percent independent.

They give different numbers at the end of their report, but this is how the poll is actually weighted, according to the math. It’s possible they made an error and confused the independent and GOP numbers, but more likely they took a Democrat-leaning sample and weighted it to be a bit more Democratic. There’s not necessarily anything wrong with that, but how likely is this turnout model? Well, in 2008, as Obama won his near-landslide victory, national turnout looked like this:

39 percent Democrat
32 percent Republican
29 percent independent

So Newsweek’s poll assumes a lower GOP turnout than in 2008 — nearly 10 percent lower — and equal Democratic turnout to 2008. How likely do you suppose that is? (The likely — or “definite” — sample produces an even more Democratic result, and so although there isn’t enough information to determine its sample, we can probably assume that it is no more Republican than the registered voter sample.)

For reference, here is the 2004 turnout by party ID: