United States presidential election in Arizona, 1988

United States presidential election in Arizona, 1988

← 1984 November 8, 1988 1992 →
Nominee George H. W. Bush Michael Dukakis Party Republican Democratic Home state Texas Massachusetts Running mate Dan Quayle Lloyd Bentsen Electoral vote 7 0 Popular vote 702,541 454,029 Percentage 60.0% 38.7%
County Results   Dukakis—60-70%   Dukakis—50-60%   Bush—<50%   Bush—50-60%   Bush—60-70%
President before election Ronald Reagan Republican Elected President George H. W. Bush Republican

Ronald Reagan
Republican

George H. W. Bush
Republican

The 1988 United States presidential election in Arizona took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Arizona voters chose 7 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 6% 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. Nearly every county turned out for Bush, with the exception of Native American Apache County and heavily unionized Greenlee County voting primarily for Dukakis.