1888 United States presidential election in Massachusetts
(Redirected from United States presidential election in Massachusetts, 1888)
The 1888 United States presidential election in Massachusetts took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose 14 representatives, or electors to the Electoral College, who voted for president and vice president.
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Turnout | 71.7%[1] 2.4 pp | |||||||||||||||||||||||||
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Massachusetts voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 9.38%.
Results
edit1888 United States presidential election in Massachusetts[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Republican | Benjamin Harrison of Indiana | Levi Parsons Morton of New York | 183,892 | 53.42% | 14 | 100.00% | ||
Democratic | Grover Cleveland of New York (incumbent) | Allen Granberry Thurman of Ohio | 151,590 | 44.04% | 0 | 0.00% | ||
Prohibition | Clinton Bowen Fisk of New Jersey | John Anderson Brooks of Missouri | 8,701 | 2.53% | 0 | 0.00% | ||
N/A | Others | Others | 60 | 0.02% | 0 | 0.00% | ||
Total | 344,243 | 100.00% | 14 | 100.00% |
See also
editReferences
edit- ^ Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.
- ^ "1888 Presidential General Election Results - Massachusetts". U.S. Election Atlas. Retrieved December 23, 2013.