1828 United States presidential election in New York
The 1828 United States presidential election in New York took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose 36 representatives, or electors to the Electoral College, who voted for President and Vice President. For this election, New York used the Congressional District Method of choosing electors, with 34 of its electors being chosen by the winner in each of the state's congressional districts, and the remaining two being chosen by the first 34 electors. Jackson won 18 congressional districts while Adams won 16. This election marks the first time New York did not chose its electoral votes through its State Legislature.[2]
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Turnout | 80.2[1] | |||||||||||||||||||||||||
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Elections in New York State |
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New York voted for the Democratic candidate, Andrew Jackson, over the National Republican candidate, John Quincy Adams. Jackson won New York by a margin of 2.9%.
Results
1828 United States presidential election in New York[3] | |||||
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Party | Candidate | Votes | Percentage | Electoral votes | |
Democratic | Andrew Jackson | 139,412 | 51.45% | 20 | |
National Republican | John Quincy Adams (incumbent) | 131,563 | 48.55% | 16 | |
Totals | 270,975 | 100.0% | 36 | ||
References
- Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- Moore, John L., ed. (1985). Congressional Quarterly's Guide to U.S. Elections (2nd ed.). Washington, D.C.: Congressional Quarterly, Inc. pp. 254–56.
- "1828 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 28 February 2013.