1828 United States presidential election in New York

The 1828 United States presidential election in New York took place on November 4, 1828, as part of the 1828 United States presidential election.^[2] 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 choose its electoral votes through its State Legislature.^[3]

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1828 United States presidential election in New York

← 1824 November 4, 1828 1832 â†’
Turnout	80.2%[1]
Nominee Andrew Jackson John Quincy Adams Party Democratic National Republican Alliance - Anti-Masonic Home state Tennessee Massachusetts Running mate John C. Calhoun Richard Rush Electoral vote 20 16 Popular vote 139,412 131,563 Percentage 51.45% 48.55%
County results Jackson   50–60%   60–70%   70–80% Adams   50–60%   60–70%   70–80%

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Thurlow Weed, Adams' campaign manager in New York, was sympathetic to the anti-masonic movement and support for Adams in New York was aligned with the Anti-Masonic Party.^[4]

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%.