1888 United States presidential election in Massachusetts explained

See main article: 1888 United States presidential election.

Election Name:1888 United States presidential election in Massachusetts
Country:Massachusetts
Type:presidential
Ongoing:no
Previous Election:1884 United States presidential election in Massachusetts
Previous Year:1884
Next Election:1892 United States presidential election in Massachusetts
Next Year:1892
Turnout:71.7%[1] 2.4 pp
Election Date:November 6, 1888
Image1:Benjamin Harrison 1896.jpg
Nominee1:Benjamin Harrison
Party1:Republican Party (United States)
Home State1:Indiana
Running Mate1:Levi P. Morton
Electoral Vote1:14
Popular Vote1:183,892
Percentage1:53.42%
Nominee2:Grover Cleveland
Party2:Democratic Party (United States)
Home State2:New York
Running Mate2:Allen G. Thurman
Electoral Vote2:0
Popular Vote2:151,590
Percentage2:44.04%
Map Size:350px
President
Before Election:Grover Cleveland
Before Party:Democratic Party (United States)
After Election:Benjamin Harrison
After Party:Republican Party (United States)

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.

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%.

See also

Notes and References

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