United States presidential election in Kansas, 1996

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United States presidential election in Kansas, 1996
← 1992 November 5, 1996 2000 →
  Bob Dole, PCCWW photo portrait.JPG Bill Clinton.jpg Ross Perot.jpg
Nominee Bob Dole Bill Clinton Ross Perot
Party Republican Democratic Reform
Home state Kansas Arkansas Texas
Running mate Jack Kemp Al Gore James Campbell [1][2]
Electoral vote 6 0 0
Popular vote 583,245 387,659 92,639
Percentage 54.3% 36.1% 8.6%

County Results

President before election

Bill Clinton

Elected President

Bill Clinton

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,[1] 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%.


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 [1][2] 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


  1. ^ a b c Dave Leip's Atlas of United States Presidential Election Results - 1996 Kansas Results
  2. ^ a b Perot Names Stand-in Veep Candidate http://www-cgi.cnn.com/ALLPOLITICS/1996/news/9608/21/perot.veep/

See also[edit]