United States presidential election in Kansas, 1996

Main article: United States presidential election, 1996

United States presidential election in Kansas, 1996

November 5, 1996

Nominee Bob Dole Bill Clinton Ross Perot Party Republican Democratic Reform Home state Kansas Arkansas Texas Running mate Jack Kemp Al Gore James Campbell Electoral vote 6 0 0 Popular vote 583,245 387,659 92,639 Percentage 54.3% 36.1% 8.6%

370px County Results   Clinton—60-70%   Clinton—40-50%   Dole—40-50%   Dole—50-60%   Dole—60-70%   Dole—70-80%

President before election Bill Clinton Democratic Elected President Bill Clinton Democratic

The 1996 United States presidential election in Kansas took place on November 5, 1996 as part of the 1996 United States presidential election. Voters chose 6 representatives, or electors to the Electoral College, who voted for President and Vice President.

Kansas was won by Senator Bob Dole (R-KS) over President Bill Clinton (D), with Dole winning 54.29% to 36.08% by a margin of 18.21%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third with 8.62% of the popular vote, a sharp decline from 1992, when Perot captured 27% of the state's votes and held George H.W. Bush's victory margin to just over 5%.

Results

United States presidential election in Kansas, 1996
Party		Candidate	Running mate	Votes	Percentage	Electoral votes
	Republican	Bob Dole	Jack Kemp	583,245	54.29%	6
	Democratic	Bill Clinton (incumbent)	Al Gore	387,659	36.08%	0
	Reform	Ross Perot	James Campbell	92,639	8.62%	0
	Libertarian	Harry Browne	Jo Jorgensen	4,557	0.42%	0
	Independent	Howard Phillips	Herbert Titus	3,519	0.33%	0
	Independent	Dr. John Hagelin	Dr. V. Tompkins	1,655	0.15%	0
	Write-in	Ralph Nader	Winona LaDuke	914	0.09%	0
	Write-in	Charles Collins	Rosemary Giumarra	112	0.01%	0