United States presidential election in Montana, 1892
Main article: United States presidential election, 1892
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| Elections in Montana |
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The 1892 United States presidential election in Montana took place on November 8, 1892, as part of the 1892 United States presidential election. Voters chose 3 representatives, or electors to the Electoral College, who voted for President and Vice President.
Montana voted for the Republican nominee, incumbent President Benjamin Harrison, over the Democratic nominee, former President Grover Cleveland, who was running for a second, non-consecutive term and over the People's Party (Populists) nominee James B. Weaver. Harrison won Montana by a margin of 2.65%.
This was the first time that Montana voted in a national election.
Results
| United States presidential election in Montana, 1892[1] | |||||
|---|---|---|---|---|---|
| Party | Candidate | Votes | Percentage | Electoral votes | |
| Republican | Benjamin Harrison (incumbent) | 18,871 | 42.44% | 3 | |
| Democratic | Grover Cleveland | 17,690 | 39.79% | 0 | |
| People's | James Weaver | 7,338 | 16.50% | 0 | |
| Prohibition | John Bidwell | 562 | 1.26% | 0 | |
| Totals | 44,461 | 99.99% | 3 | ||
References
- ↑ "1892 Presidential General Election Results - Montana". U.S. Election Atlas. Retrieved 1 August 2012.
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