United States presidential election in Maine, 1988
|
||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
|
||||||||||||||||||||||||||
|
||||||||||||||||||||||||||
County Results
Bush—50-60%
Bush—60-70%
|
||||||||||||||||||||||||||
|
||||||||||||||||||||||||||
The 1988 United States presidential election in Maine took place on November 8, 1988 as part of the 1988 United States presidential election, which was held throughout all 50 states and D.C. Voters chose 8 representatives, or electors to the Electoral College, who voted for President and Vice President.
Maine voted for the Republican nominee, Vice President George H. W. Bush, over the Democratic nominee, Massachusetts Governor Michael Dukakis, by a margin of 11.45%. Bush took 55.34% of the vote to Dukakis's 43.88%.
Bush's comfortable win was expected, as the state was at the time Republican-leaning, and the Bush family also had ties to Maine, with Bush owning a house in Kennebunkport, Maine.
Like the rest of liberal and secular New England, however, Maine would soon begin to drift to the Democratic Party as the Republican Party became increasingly dominated by conservatives, Southerners, and Evangelical Christians. Consequently, this would be the last time that a Republican would carry the state of Maine in a presidential election, and in the following two decades it would become regarded as a safe blue state. Maine's Second Congressional District, however, gave its electoral vote to the Republicans in 2016.
...
Wikipedia