United States presidential election in Iowa, 1988
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County Results
Dukakis—60-70%
Dukakis—50-60%
Dukakis—<50%
Bush—<50%
Bush—50-60%
Bush—60-70%
Bush—70-80%
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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.
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.
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