1996 United States presidential election in Rhode Island

1996 United States presidential election in Rhode Island

← 1992 November 5, 1996 2000 →
Turnout64.8%[1] Decrease 11.8 pp
 
Nominee Bill Clinton Bob Dole Ross Perot
Party Democratic Republican Reform
Home state Arkansas Kansas Texas
Running mate Al Gore Jack Kemp Patrick Choate
Electoral vote 4 0 0
Popular vote 233,050 104,683 43,723
Percentage 59.71% 26.82% 11.20%


President before election

Bill Clinton
Democratic

Elected President

Bill Clinton
Democratic

The 1996 United States presidential election in Rhode Island took place on November 5, 1996, as part of the 1996 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island was won by President Bill Clinton (D) over Senator Bob Dole (R-KS), with Clinton winning 59.71% to 26.82% by a margin of 32.89%. Billionaire businessman Ross Perot (Reform Party of the United States of America-TX) finished in third, with 11.20% of the popular vote.[2]

As of 2020, this was the most recent presidential election in which the town of Scituate voted for a Democrat.

  1. ^ This figure is calculated by dividing the total number of votes cast in 1996 (390,247) by an estimate of the number of registered voters in Rhode Island in 1996 (602,692). See "Voter Turnout, 1996". Rhode Island Board of Elections. Retrieved February 6, 2018.
  2. ^ "Dave Leip's Atlas of U.S. Presidential Elections".

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