1988 United States presidential election in Delaware
The 1988 United States presidential election in Delaware took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Delaware voters chose 3 electors to the Electoral College, which selected the president and vice president.
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County Results
Bush 50–60%
60–70% | ||||||||||||||||||||||||||
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Delaware was won by incumbent United States Vice President George H. W. Bush of Texas, who was running against Massachusetts Governor Michael Dukakis. Bush ran with Indiana Senator Dan Quayle as Vice President, and Dukakis ran with Texas Senator Lloyd Bentsen.
Delaware weighed in for this election as about 4% more Republican than the national average. As of the 2016 presidential election, this is the last time that the Republican nominee carried Delaware.
Partisan background
The presidential election of 1988 was a very partisan election for Delaware, with nearly 99.5% of the electorate voting for either the Democratic or Republican parties, and only 4 political parties on the ballot, statewide.[1] All counties in Delaware voted in majority for Bush. As of the 2016 presidential election, this is the last election in which New Castle County voted for the Republican candidate.
Elections in Delaware | ||||||||||||
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Results
United States presidential election in Delaware, 1988 | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Republican | George H. W. Bush | 139,639 | 55.88% | 3 | |
Democratic | Michael Dukakis | 108,647 | 43.48% | 0 | |
Libertarian | Ron Paul | 1,162 | 0.47% | 0 | |
New Alliance Party | Lenora Fulani | 443 | 0.18% | 0 | |
Totals | 249,891 | 100.00% | 3 |
See also
References
- "Dave Leip's Atlas of U.S. Presidential Elections". Uselectionatlas.org. Retrieved 2013-07-21.