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| Elections in Wisconsin |
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The 1988 United States presidential election in Wisconsin took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. State voters chose 11 electors to the Electoral College, which selected the president and vice president.
Wisconsin was won by Massachusetts Governor Michael Dukakis who was running against incumbent United States Vice President George H. W. Bush of Texas. Dukakis ran with Texas Senator Lloyd Bentsen as Vice President, and Bush ran with Indiana Senator Dan Quayle. Dukakis won the election in Wisconsin with a four point margin. The state has since consistently voted for the Democratic Party, until the narrow victory of Republican Donald Trump in 2016.
The election was very partisan, with over 99 percent of the electorate voting for either the Republican or Democratic parties, although five additional candidates were on the ballot.[1] Dukakis and Bush almost evenly split Wisconsin's seventy-two counties – Dukakis won 37 and Bush won 35. Dukakis won the large urban counties containing Madison (Dane County), Milwaukee, Racine and Kenosha, alongside almost entirely Native American Menominee County and the heavily unionized Scandinavian-American counties of the northwest. Bush won the suburban "WOW counties" and the more conservative, historically German Catholic, counties of the rural eastern half of the state.[2] Over the state as a whole, Dukakis did best, as usual, in Menominee County, and Bush did best in Ozaukee County.
- ^ "1988 Presidential General Election Results – Wisconsin". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved July 21, 2013.
- ^ Phillips, Kevin P.; The Emerging Republican Majority, pp. 381-382, 414 ISBN 978-0-691-16324-6
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