1988 United States presidential election in Hawaii

1988 United States presidential election in Hawaii

← 1984 November 8, 1988 1992 →
 
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 4 0
Popular vote 192,364 158,625
Percentage 54.27% 44.75%

County Results
Dukakis
  50–60%


President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

The 1988 United States presidential election in Hawaii took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Hawaii voters chose 4 electors to the Electoral College, which selected the president and vice president.

Hawaii 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. The election was very partisan for Hawaii, with 99% of the electorate voting for either the Democratic or Republican parties.[1] All four of the Hawaiian island counties voted in majority for Dukakis. Hawaii weighed in for this election as 17% more Democratic than the national average. Dukakis won the election in Hawaii with a solid 10-point win. This is the last time any candidate flipped every county in any state from the previous election, as Dukakis won every county in the state after Ronald Reagan did so in 1984.

  1. ^ "Dave Leip's Atlas of U.S. Presidential Elections". Uselectionatlas.org. Retrieved July 21, 2013.