United States presidential election in New Jersey, 1900
|Elections in New Jersey|
The 1900 United States presidential election in New Jersey took place on November 6, 1900. Voters chose 10 representatives, or electors to the Electoral College, who voted for president and vice president.
New Jersey overwhelmingly voted for the Republican nominee, President William McKinley, over the Democratic nominee, former U.S. Representative and 1896 Democratic presidential nominee William Jennings Bryan. McKinley won New Jersey by a margin of 14.17% in this rematch of the 1896 presidential election. The return of economic prosperity and recent victory in the Spanish–American War helped McKinley to score a decisive victory.
|United States presidential election in New Jersey, 1900|
|Party||Candidate||Running mate||Popular vote||Electoral vote|
|Republican||William McKinley of Ohio||Theodore Roosevelt of New York||221,754||55.27%||10||100.00%|
|Democratic||William Jennings Bryan of Nebraska||Adlai Ewing Stevenson I of Illinois||164,879||41.10%||0||0.00%|
|Prohibition||John Granville Woolley of Illinois||Henry Brewer Metcalf of Rhode Island||7,190||1.79%||0||0.00%|
|Socialist||Eugene Victor Debs of Indiana||Job Harriman of California||4,611||1.15%||0||0.00%|
|Socialist Labor||Joseph Francis Malloney of Massachusetts||Valentine Remmel of Pennsylvania||2,081||0.52%||0||0.00%|
|People's||Wharton Barker of Pennsylvania||Ignatius Loyola Donnelly of Minnesota||691||0.17%||0||0.00%|
- "1900 Presidential General Election Results - New Jersey". U.S. Election Atlas. Retrieved 23 December 2013.