1988 United States presidential election in Rhode Island explained

See main article: 1988 United States presidential election.

Election Name:1988 United States presidential election in Rhode Island
Country:Rhode Island
Type:presidential
Ongoing:no
Previous Election:1984 United States presidential election in Rhode Island
Previous Year:1984
Next Election:1992 United States presidential election in Rhode Island
Next Year:1992
Turnout:70.2%[1] 7.8 pp
Election Date:November 8, 1988
Image1:File:Dukakis campaign portrait 3x4.jpg
Nominee1:Michael Dukakis
Party1:Democratic Party (United States)
Home State1:Massachusetts
Running Mate1:Lloyd Bentsen
Electoral Vote1:4
Popular Vote1:225,123
Percentage1:55.64%
Nominee2:George H. W. Bush
Party2:Republican Party (United States)
Home State2:Texas
Running Mate2:Dan Quayle
Electoral Vote2:0
Popular Vote2:177,761
Percentage2:43.93%
President
Before Election:Ronald Reagan
Before Party:Republican Party (United States)
After Election:George H. W. Bush
After Party:Republican Party (United States)

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. As of the 2020 presidential election, this is the last time 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
DemocraticMichael Dukakis225,12355.64%4
RepublicanGeorge H. W. Bush177,76143.93%0
LibertarianRon Paul825 0.20%0
New AllianceLenora Fulani2800.07%0
Peace and FreedomHerbert G. Lewin1950.05%0
America FirstDavid Duke1590.04%0
Socialist WorkersJames Warren1300.03%0
Socialist960.02%0
Write-insWrite-ins510.01%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
%%%%
Bristol11,16851.04%10,62648.56%890.40%5422.48%21,883
Kent37,22151.84%34,31447.79%2660.37%2,9074.05%71,801
Newport17,59750.76%16,92348.82%1440.42%6741.94%34,664
Providence135,92758.80%94,24840.77%9840.43%41,67918.03%231,159
Washington23,21051.51%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

Notes and 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 Web site: Voter Turnout, 1988. February 6, 2018. Rhode Island Board of Elections.
  2. Web site: 1988 Presidential General Election Results - Rhode Island. 2013-02-07 . Dave Leip's Atlas of U.S. Presidential Elections.