page_title: 1988 United States presidential election in Kentucky
section_title: Partisan background
hierarchical_section_title: 1988 United States presidential election in Kentucky / Partisan background
caption_attribution_description: George Bush campaigning for President in 1988 in Owensboro, Kentucky.
context_page_description: The 1988 United States presidential election in Kentucky took place on November 8, 1988. All 50 states and the District of Columbia, were part of the 1988 United States presidential election. Kentucky voters chose nine electors to the Electoral College, which selected the president and vice president.
  Kentucky 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.
  Kentucky weighed in for this election as 4 points more Republican than the national average. As of the 2016 presidential election, this is the last time that Jefferson County voted for a Republican presidential candidate. The 1988 presidential election is also the last time that Kentucky voted for a different presidential candidate from neighbouring West Virginia.
context_section_description: The presidential election of 1988 was a very partisan election for Kentucky, with more than 99 percent of the electorate voting for either the Democratic or Republican parties, and only five parties appearing on the ballot. The vast majority of counties turned out for Bush, including the highly populated center of Louisville's Jefferson County, the last time that county voted for the Republican candidate. Conversely, several small clumps of counties in the Eastern Coal Field (surrounding Pike County) and in the Jackson Purchase (surrounding McCracken County) regions of the state voted primarily Democratic.
caption: George H. W. Bush campaigning in Owensboro, Kentucky
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