United States presidential election in Oregon, 1996

United States presidential election in Oregon, 1996

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. "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.