1828 United States presidential election in Connecticut
The 1828 United States presidential election in Connecticut took place between October 31 and December 2, 1828, as part of the 1828 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for President and Vice President.
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| Elections in Connecticut |
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Connecticut voted for the National Republican candidate, John Quincy Adams, over the Democratic candidate, Andrew Jackson. Adams won Connecticut by a margin of 48.41%.
With 71.36% of the popular vote, Connecticut would prove to be Adams' fourth strongest state in the 1828 election after Rhode Island, Massachusetts and Vermont.[1]
Results
| 1828 United States presidential election in Connecticut[2] | |||||
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| Party | Candidate | Votes | Percentage | Electoral votes | |
| National Republican | John Quincy Adams | 13,829 | 71.36% | 8 | |
| Democratic | Andrew Jackson | 4,448 | 22.95% | 0 | |
| N/A | Other | 1,101 | 5.68% | 0 | |
| Totals | 19,378 | 100.0% | 8 | ||
References
- "1828 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1828 Presidential General Election Results - Connecticut". U.S. Election Atlas. Retrieved 28 February 2013.
State and district results of the 1828 United States presidential election | ||
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