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| Elections in Rhode Island |
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The 1908 United States presidential election in Rhode Island took place on November 3, 1908, as part of the 1908 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
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Rhode Island voted for the Republican nominees, Secretary of War William Howard Taft of Ohio and his running mate James S. Sherman of New York. They defeated the Democratic nominees, former U.S. Representative William Jennings Bryan of Nebraska and his running mate John W. Kern of Indiana. Taft won the state by a margin of 26.6%.
With 60.76% of the popular vote, Rhode Island would be Taft's fifth strongest victory in terms of percentage in the popular vote after Vermont, Maine, Michigan and North Dakota. [1]
Bryan had previously lost Rhode Island to William McKinley in both 1896 and 1900.
