1988 United States presidential election in New York

1988 United States presidential election in New York

← 1984 November 8, 1988 1992 →
 
Nominee Michael Dukakis George H. W. Bush
Party Democratic Republican
Alliance Liberal Conservative
Home state Massachusetts Texas
Running mate Lloyd Bentsen Dan Quayle
Electoral vote 36 0
Popular vote 3,347,882 3,081,871
Percentage 51.62% 47.52%

County Results

President before election

Ronald Reagan
Republican

Elected President

George H. W. Bush
Republican

International policy with the buckling Soviet Union was a critical component of the political landscape in the late 1980s. Vice President Bush can be seen here standing with United States President Ronald Reagan and Soviet General Secretary Mikhail Gorbachev, on the New York waterfront, 1988.

The 1988 United States presidential election in New York took place on November 8, 1988, as part of the 1988 United States presidential election. Voters chose 36 representatives, or electors to the Electoral College, who voted for president and vice president.

New York was won by Democratic Governor Michael Dukakis of Massachusetts with 51.62% of the popular vote over Republican Vice President George H. W. Bush of Texas, who took 47.52%, a victory margin of 4.10%.[1] This result made New York roughly 12% more Democratic than the nation-at-large. Dukakis’ statewide victory is largely attributable to winning four of five boroughs of New York City overall with 66.2% of the vote.

Bush became the first Republican to win the White House without carrying Broome County and the first to win the White House without carrying Montgomery County since Rutherford B. Hayes in 1876.

  1. ^ "1988 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved October 13, 2012.

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