1916 United States presidential election in Rhode Island
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Elections in Rhode Island |
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The 1916 United States presidential election in Connecticut took place on November 7, 1916 as part of the 1916 United States Presidential Election which was held throughout all contemporary 48 states. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.
Rhode Island was won by the Republican nominee, U.S. Supreme Court Justice Charles Evans Hughes of New York, and his running mate Senator Charles W. Fairbanks of Indiana. They defeated Democratic nominees, incumbent Democratic President Woodrow Wilson and Vice President Thomas R. Marshall.
Hughes won Rhode Island by a narrow margin of 5.08 percent.
Results
United States presidential election in Rhode Island, 1916[1] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Charles Evans Hughes of New York | Charles Warren Fairbanks of Indiana | 44,858 | 51.08% | 5 | 100.00% | ||
Democratic | Woodrow Wilson of New Jersey | Thomas Riley Marshall of Indiana | 40,394 | 46.00% | 0 | 0.00% | ||
Socialist | Allan Louis Benson of New York | George Ross Kirkpatrick of New Jersey | 1,914 | 2.18% | 0 | 0.00% | ||
Prohibition | James Franklin Hanly of Indiana | Ira Landrith of Tennessee | 470 | 0.54% | 0 | 0.00% | ||
Socialist Labor | Arthur Elmer Reimer of Massachusetts | Caleb Harrison of Illinois | 180 | 0.20% | 0 | 0.00% | ||
Total | 87,816 | 100.00% | 5 | 100.00% |
References
- ^ "1916 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.