1996 United States presidential election in Hawaii
Appearance
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![]() County Results
Clinton—70-80%
Clinton—50-60% | |||||||||||||||||||||||||||||||||
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Elections in Hawaii |
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The 1996 United States presidential election in Hawaii took place on November 5, 1996 as part of the 1996 United States presidential election. Voters chose 4 representatives, or electors to the Electoral College, who voted for President and Vice President.
Hawaii was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 56.93% to 31.64% by a margin of 25.29%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third with 7.6% of the popular vote .[1]
Results
United States presidential election in Hawaii, 1996 | ||||||
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Party | Candidate | Running mate | Votes | Percentage | Electoral votes | |
Democratic | Bill Clinton (incumbent) | Al Gore | 205,012 | 56.93% | 4 | |
Republican | Bob Dole | Jack Kemp | 113,943 | 31.64% | 0 | |
Reform | Ross Perot | Patrick Choate | 27,358 | 7.60% | 0 | |
Green | Ralph Nader | Winona LaDuke | 10,386 | 2.88% | 0 | |
Libertarian | Harry Browne | Jo Jorgensen | 2,493 | 0.69% | 0 | |
Natural Law | Dr. John Hagelin | Dr. V. Tompkins | 570 | 0.16% | 0 | |
U.S. Taxpayers' Party | Howard Phillips | Herbert Titus | 358 | 0.10% | 0 | |
Totals | 360,120 | 100.0% | 4 |