1884 United States presidential election in Massachusetts explained

See main article: 1884 United States presidential election.

Election Name:1884 United States presidential election in Massachusetts
Country:Massachusetts
Type:presidential
Ongoing:no
Previous Election:1880 United States presidential election in Massachusetts
Previous Year:1880
Next Election:1888 United States presidential election in Massachusetts
Next Year:1888
Turnout:69.3%[1] 1.9 pp
Election Date:November 4, 1884
Image1:Unsuccessful 1884.jpg
Nominee1:James G. Blaine
Party1:Republican Party (United States)
Home State1:Maine
Running Mate1:John A. Logan
Electoral Vote1:14
Popular Vote1:146,724
Percentage1:48.36%
Nominee2:Grover Cleveland
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Thomas A. Hendricks
Electoral Vote2:0
Popular Vote2:122,352
Percentage2:40.33%
Image3:Bfbutler.jpg
Nominee3:Benjamin Butler
Party3:Greenback Party
Home State3:Massachusetts
Running Mate3:Absolom West
Electoral Vote3:0
Popular Vote3:24,382
Percentage3:8.04%
Map Size:350px
President
Before Election:Chester A. Arthur
Before Party:Republican Party (United States)
After Election:Grover Cleveland
After Party:Democratic Party (United States)

The 1884 United States presidential election in Massachusetts took place on November 4, 1884, as part of the 1884 United States presidential election. Voters chose 14 representatives, or electors to the Electoral College, who voted for president and vice president.

Massachusetts voted for the Republican nominee, James G. Blaine, over the Democratic nominee, Grover Cleveland. Blaine won the state by a margin of 8.03%. Greenback nominee and former Massachusetts governor Benjamin Butler won 8.04% of the vote.

See also

Notes and References

  1. Bicentennial Edition: Historical Statistics of the United States, Colonial Times to 1970, part 2, p. 1072.