1918 United States Senate election in Wyoming
November 5, 1918
 
Nominee Francis E. Warren John E. Osborne
Party Republican Democratic
Popular vote 23,975 17,528
Percentage 57.77% 42.23%
U.S. senator before election

Francis E. Warren
Republican

 Elected U.S. Senator

Francis E. Warren
Republican

The 1918 United States Senate election in Wyoming took place on November 5, 1918. Incumbent Republican Senator Francis E. Warren ran for re-election to his fifth consecutive term in the Senate, in his first popular election. He was opposed by former Governor John E. Osborne, the Democratic nominee, who had until recently served as the U.S. Assistant Secretary of State. Owing in part to Warren's long record of service, and also to the strong Republican performance throughout the country in the 1918 elections, Warren defeated Osborne by a wide margin.

Democratic primary

Candidates

Results

Democratic primary[1]
Party Candidate Votes %
Democratic John E. Osborne 4,186 100.00%
Total votes 4,186 100.00%

Republican primary

Candidates

Results

Republican primary[1]
Party Candidate Votes %
Republican Francis E. Warren (inc.) 12,709 100.00%
Total votes 12,709 100.00%

General election

Results

1918 United States Senate election in Wyoming[1]
Party Candidate Votes %
Republican Francis E. Warren (inc.) 23,975 57.77%
Democratic John E. Osborne 17,528 42.23%
Total votes 41,503 100.00%
Republican hold

References

  1. 1 2 3 Chaplin, W. E. (1919). 1919 Official Directory of Wyoming and Election Returns for 1918. pp. 35–37.
This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.