1908 United States presidential election in Rhode Island
|Elections in Rhode Island|
The 1908 United States presidential election in Rhode Island took place on November 3, 1908. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president.
Rhode Island voted for the Republican nominee, Secretary of War William Howard Taft, over the Democratic nominee, former U.S. Representative William Jennings Bryan. Taft won the state by a margin of 26.6%.
|1908 United States presidential election in Rhode Island|
|Party||Candidate||Running mate||Popular vote||Electoral vote|
|Republican||William Howard Taft of Ohio||James Schoolcraft Sherman of New York||43,942||60.76%||4||100.00%|
|Democratic||William Jennings Bryan of Nebraska||John Worth Kern of Indiana||24,706||34.16%||0||0.00%|
|Socialist||Eugene Victor Debs of Indiana||Benjamin Hanford of New York||1,365||1.89%||0||0.00%|
|Independence||Thomas Louis Hisgen of Massachusetts||John Temple Graves of Georgia||1,105||1.53%||0||0.00%|
|Prohibition||Eugene Wilder Chafin of Illinois||Aaron Sherman Watkins of Ohio||1,016||1.40%||0||0.00%|
|Socialist Labor||August Gillhaus of New York||Donald L. Munro of Virginia||183||0.25%||0||0.00%|
- "1908 Presidential Election Statistics". Dave Leip’s Atlas of U.S. Presidential Elections. Retrieved 2018-03-05.
- "1908 Presidential General Election Results - Rhode Island". U.S. Election Atlas. Retrieved 23 December 2013.