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.
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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[2] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| Democratic | Andrew Jackson | 139,412 | 51.45% | 20 | |
| National Republican | John Quincy Adams | 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.
- "1828 Presidential General Election Results - New York". U.S. Election Atlas. Retrieved 28 February 2013.
State and district results of the 1828 United States presidential election | ||
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