Last night, Scott Brown unleashed an eyebrow-raiser: he votes as often with Democrats as with Republicans. (Because, you know, he’s an Independent or an independent or something.) Later Brown was more precise: he votes with Republicans only 54% of the time. The claim was one of a related set that also include his claim that he was the second-most bipartisan senator (based on voting record) and that, in contrast to his independence, most GOP senators voted with their party 93% percent of the time or more.
How factually challenged? Pretty.
According to OpenCongress, Brown is indeed the second-least likely to vote with his party at 67.1% (behind Susan Collins at 65.9%). The least likely Democrat to vote with his party is Joe Manchin at 81.0%. The two Independents (Lieberman and Sanders) each voted closest to higher discipline Democrats, so we’ll give Brown one point for being #2.
But, that 67.1% is nowhere near 54%. Not only is he off by more than 10 percentage points, but the spread takes him from a claim of nearly 1-to-1 party voting to the reality that he’s twice as likely to vote R than D. I’ll let a statistician characterize the difference more precisely, but 2-to-1 is significantly more often than 1-to-1. He’s misleading Massachusetts voters. Is it a sign of desperation that he feels 67% is not independent enough?
As for the other Republican senators, a majority of the delegation does not vote with the party 93% of the time. In fact only Mitch McConnell is at or over 93%. (Interestingly, more than half the Democrats vote with their party more than 93%.) On its own, this would not be a egregious error, except that Brown used it to try and distance himself from national Republicans. It aggravates his other misstatement.
Video snippet of Brown’s 50/50 reference.
Video snippet of Brown saying he’s the least partisan senator in the US Senate
Video snippet of Brown saying he’s the second least partisan senator in the US Senate
