1988 United States presidential election in Iowa

The 1988 United States presidential election in Iowa took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president.

1988 United States presidential election in Iowa
Flag of Iowa (variant).svg
← 1984 November 8, 1988 1992 →
  1988 Dukakis.jpg 1988 Bush.jpg
Nominee Michael Dukakis George H.W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 8 0
Popular vote 670,557 545,355
Percentage 54.71% 44.50%

Iowa Presidential Election Results by County, 1988.svg
County Results

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

Iowa was won by Democratic Governor Michael Dukakis of Massachusetts with 54.71% of the popular vote over Republican Vice President George H.W. Bush's 44.50%, a victory margin of 10.22%. This made it one of 10 states (plus the District of Columbia) to vote for Dukakis, while Bush won a convincing electoral victory nationwide.

The farm crisis of the 1980s under the incumbent Republican administration made the Midwest one of the targets for the Dukakis campaign in 1988, which ultimately proved successful in the region with Democrats performing strongly in many farm states. Nowhere was this more evident than in Iowa, which was the second most Democratic state in the nation in 1988 in terms of both vote percentage and victory margin; it was one of just two states (along with Rhode Island) to vote Democratic by a double-digit margin. This Democratic support was spread widely across the state, with Dukakis winning 75 of the state's 99 counties. Iowa was also the only state in the nation which Dukakis won by a larger margin than fellow Democrat Bill Clinton would win it by four years later in 1992; Dukakis won 13 counties in Iowa in 1988 which would vote for Bush in 1992.

While Dukakis personally overperformed in Iowa due to the farm crisis, 1988 was also the beginning of a long-term re-alignment of the state toward the Democratic Party, as the historically-Republican state has been a Democratic-leaning state ever since; prior to this election, Iowa had voted Republican in the five preceding elections. Beginning in 1988, Iowa has voted Democratic in six of the eight elections that have since occurred. Despite this, as of the 2016 presidential election, this is the last election in which Marion County, Buena Vista County, Sac County, and Humboldt County voted for the Democratic candidate. Additionally, this was also the last time a candidate from either party would win the state of Iowa without winning the popular vote, until Donald Trump did so in 2016 (although he won the general election).

ResultsEdit

1988 United States presidential election in Iowa[1]
Party Candidate Votes Percentage Electoral votes
Democratic Michael Dukakis 670,557 54.71% 8
Republican George H.W. Bush 545,355 44.50% 0
Independent Lyndon LaRouche 3,526 0.29% 0
Libertarian Ron Paul 2,494 0.20% 0
N/A Write-ins 1,613 0.13% 0
America First David Duke 755 0.06% 0
New Alliance Lenora Fulani 540 0.04% 0
Socialist Willa Kenoyer 334 0.03% 0
Socialist Equality Edward Winn 235 0.02% 0
Socialist Workers James Warren 205 0.02% 0
Totals 1,225,614 100.00% 8
Voter Turnout (Voting age/Registered) 59%/73%

ReferencesEdit

  1. ^ "1988 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 28 April 2013.