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| Elections in Arizona |
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The 1988 United States presidential election in Arizona took place on November 8, 1988. All fifty states and the District of Columbia, were part of the 1988 United States presidential election. State voters chose seven electors to the Electoral College, which selected the president and vice president.
Arizona 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.
Arizona weighed in for this election as 14 points more Republican than the national average. The presidential election of 1988 was a very partisan election for Arizona, with nearly 99% of the electorate voting for either the Democratic or Republican parties.[1] Nearly every county turned out for Bush, with the exception of Native American Apache County and heavily unionized Greenlee County voting primarily for Dukakis.
As of the 2020 presidential election[update], this is the last occasion when the counties of Coconino, Pima and Santa Cruz have voted for the Republican presidential candidate.[2]
Bush won the election in the traditionally conservative and Republican state of Arizona with a solid 21-point margin.
- ^ "1988 Presidential General Election Results – Arizona". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved July 21, 2013.
- ^ Sullivan, Robert David; ‘How the Red and Blue Map Evolved Over the Past Century’; America Magazine in The National Catholic Review; June 29, 2016