United States presidential election in Oregon, 1996

United States presidential election in Oregon, 1996
Oregon
November 5, 1996
 
Nominee Bill Clinton Bob Dole Ross Perot
Party Democratic Republican Reform
Home state Arkansas Kansas Texas
Running mate Al Gore Jack Kemp James Campbell
Electoral vote 7 0 0
Popular vote 649,641 538,152 121,221
Percentage 47.15% 39.06% 8.80%

County Results
  Clinton—50-60%
  Clinton—<50%
  Dole—<50%
  Dole—50-60%
  Dole—60-70%
President before election

Bill Clinton
Democratic

 Elected President

Bill Clinton
Democratic

The 1996 United States presidential election in Oregon took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose seven representatives, or electors to the Electoral College, who voted for President and Vice President.

Oregon was won by President Bill Clinton (D) over Senator Bob Dole (R-Kansas), with Clinton winning 47.2% to 39.1% by a margin of 8.1%. Billionaire businessman Ross Perot (Reform-Texas) finished in third with 8.8% of the popular vote.[1]

Oregon was one of thirteen states where on the election ballot, James Campbell of California, Perot's former boss at IBM, was listed as a stand-in Vice-Presidential candidate.[2]

Results

United States presidential election in Oregon, 1996[1]
Party Candidate Votes Percentage Electoral votes
Democratic Bill Clinton 649,641 47.15% 7
Republican Bob Dole 538,152 39.06% 0
Reform Ross Perot 121,221 8.80% 0
Pacific Green Ralph Nader 49,415 3.59% 0
Libertarian Harry Browne 8,903 0.65% 0
Independent Howard Phillips 3,379 0.25% 0
Natural Law John Hagelin 2,798 0.20% 0
Write-ins 2,329 0.17% 0
Socialist Mary Cal Hollis 1,922 0.14% 0
Totals 1,377,760[3] 100.0% 7

References

  1. 1 2 "1992 Presidential General Election Results - Oregon". U.S. Election Atlas. Retrieved June 11, 2012.
  2. ↑ "Perot Names Stand-in Veep Candidate". CNN.com. August 21, 1996. Retrieved June 11, 2012.
  3. ↑ 1996 Presidential General Election Results - Oregon". U.S. Election Atlas.
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