1988 United States presidential election in Rhode Island

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1988 United States presidential election in Rhode Island
Flag of Rhode Island.svg
  1984 November 8, 1988 1992  
Turnout70.2% [1] Decrease2.svg 7.8 pp
  Dukakis campaign portrait 3x4.jpg VP George Bush crop.jpg
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote40
Popular vote225,123177,761
Percentage55.64%43.93%

Rhode Island Presidential Election Results 1988.svg
Rhode Island Presidential Election Municipality Results 1988.svg

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.

Contents

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 is the last time until the 2024 that a Republican candidate won over 40% of the vote in Rhode Island.

To date, this is the last time that the towns of Barrington, Charlestown, Little Compton, Middletown, North Kingstown, and Portsmouth voted Republican.

Results

1988 United States presidential election in Rhode Island [2]
PartyCandidateVotesPercentageElectoral votes
Democratic Michael Dukakis 225,12355.64%4
Republican George H. W. Bush 177,76143.93%0
Libertarian Ron Paul 8250.20%0
New Alliance Lenora Fulani 2800.07%0
Peace and Freedom Herbert G. Lewin 1950.05%0
America First David Duke 1590.04%0
Socialist Workers James Warren 1300.03%0
Socialist Willa Kenoyer 960.02%0
Write-ins Write-ins 510.01%0
Totals404,620100.00%4
Voter Turnout (Voting age/Registered)53%/74%

By county

CountyMichael Dukakis
Democratic
George H.W. Bush
Republican
Various candidates
Other parties
MarginTotal votes cast
# %# %# %# %
Bristol 11,16851.04%10,62648.56%890.40%5422.48%21,883
Kent 37,22151.84%34,31447.79%2660.37%2,9074.05%71,801
Newport 17,59750.76%16,92348.82%1440.42%6741.94%34,664
Providence 135,92758.80%94,24840.77%9840.43%41,67918.03%231,159
Washington 23,21051.51%21,65048.04%2020.45%1,5603.47%45,062
Totals225,12355.64%177,76143.93%1,7360.43%47,36211.71%404,620

Counties that flipped from Republican to Democratic

See also

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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 February 7, 2013.