United States presidential election in Arkansas, 1992
Elections in Arkansas |
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The 1992 United States presidential election in Arkansas took place on November 3, 1992, as part of the 1992 United States presidential election. Voters chose six representatives, or electors to the Electoral College, who voted for President and Vice President.
Arkansas was won by the state's Governor Bill Clinton (D) with 53.21% of the popular vote over incumbent President George H.W. Bush (R-Texas) with 35.48%. Businessman Ross Perot (I-Texas) finished in third with 10.43% of the popular vote.[1] Clinton ultimately won the national vote, defeating incumbent President Bush. Clinton was Governor of Arkansas at the time, and as he was popular within the state, he easily won his home state by a margin of 17.73%, making it the first time that Arkansas had voted Democratic since 1976 when it voted for Jimmy Carter. Arkansas was the only state in which Clinton got over 50% of the popular vote in, besides Washington, D.C.
George H.W. Bush won this state by a substantial margin four years earlier, but his campaign statements that year and his statement during the debates that Arkansas was a backwater state caused furor among the state's residents. Ross Perot did the same thing. The backlash with the two candidates, made it all the more possible for the native son presidential candidate Bill Clinton to win this state by double digits and in nearly every county.
After the 1996 election, Arkansas would then return to being solidly Republican.[2]
Results
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