1988 United States presidential election in Rhode Island

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1988 United States presidential election in Rhode Island

← 1984 November 8, 1988 1992 →
Turnout70.2%[1] Decrease 7.8 pp
  1988 Dukakis (cropped 3x4).jpg 1988 Bush (cropped).jpg
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 4 0
Popular vote 225,123 177,761
Percentage 55.64% 43.93%

Rhode Island Presidential Election Results 1988.svg
County Results
Dukakis
  50–60%


Rhode Island Presidential Election Municipality Results 1988.svg
Municipality results

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

The 1988 United States presidential election in Rhode Island took place on November 8, 1988, as part of the 1988 United States presidential election, which was held throughout all 50 states and D.C. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.

Rhode Island voted for the Democratic nominee, Massachusetts Governor Michael Dukakis, over Republican Vice President George H. W. Bush. Dukakis took 55.64% of the vote to Bush's 43.93%, a margin of 11.71%. This made it one of 10 states (plus the District of Columbia) to vote for Dukakis, while Bush won a convincing electoral victory nationwide.

A liberal New England state, Rhode Island gave Dukakis his strongest state victory in the nation, with only the District of Columbia voting more Democratic. It was one of just two states (along with Iowa) to vote Democratic by a double-digit margin, and one of only two states (along with Hawaii) to have all of its counties go to Dukakis. Despite this, it was still a relatively strong Republican performance compared to how the state has trended since. The state has voted Democratic in every presidential election that followed. This remains the last time a Republican candidate won over 40% of the vote in Rhode Island.

As of 2020, this was the last presidential election in which the towns of Barrington, Charlestown, Little Compton, Middletown, North Kingstown, and Portsmouth voted for the Republican candidate.

Results[]

1988 United States presidential election in Rhode Island[2]
Party Candidate Votes Percentage Electoral votes
Democratic Michael Dukakis 225,123 55.64% 4
Republican George H. W. Bush 177,761 43.93% 0
Libertarian Ron Paul 825 0.20% 0
New Alliance Lenora Fulani 280 0.07% 0
Peace and Freedom Herbert G. Lewin 195 0.05% 0
America First David Duke 159 0.04% 0
Socialist Workers James Warren 130 0.03% 0
Socialist Willa Kenoyer 96 0.02% 0
Write-ins Write-ins 51 0.01% 0
Totals 404,620 100.00% 4
Voter Turnout (Voting age/Registered) 53%/74%

By county[]

Michael Stanley Dukakis

Democratic

George Herbert Walker Bush

Republican

Various candidates

Other parties

County % # % # % #
Providence 58.8% 135,927 40.8% 94,248 0.4% 984
Kent 51.8% 37,221 47.8% 34,314 0.4% 266
Washington 51.5% 23,210 48.0% 21,650 0.4% 202
Bristol 51% 11,168 48.6% 10,626 0.4% 89
Newport 50.8% 17,597 48.8% 16,923 0.4% 144

See also[]

References[]

  1. ^ This figure is calculated by dividing the total number of votes cast in 1988 (385,027) by an estimate of the number of registered voters in Rhode Island in 1988 (548,758). See "Voter Turnout, 1988". Rhode Island Board of Elections. Retrieved February 6, 2018.
  2. ^ "1988 Presidential General Election Results - Rhode Island". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved 2013-02-07.
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