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1892 United States House of Representatives election in Wyoming

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1892 United States House of Representatives election in Wyoming

← 1890 November 8, 1892 (1892-11-08) 1894 →
 
Nominee Henry A. Coffeen Clarence D. Clark
Party Democratic Republican
Popular vote 8,855 8,394
Percentage 51.34% 48.66%

County results
Coffeen:      50–60%      60–70%
Clark:      50–60%
     No Data

U.S. Representative before election

Clarence D. Clark
Republican

Elected U.S. Representative

Henry A. Coffeen
Democratic

The Wyoming United States House election for 1892 was held on November 3, 1892. Democratic Henry A. Coffeen defeated Republican incumbent Clarence D. Clark with 51.34% of the vote making Clark the first incumbent Representative from Wyoming to lose reelection.

Results

[edit]
United States House of Representatives election in Wyoming, 1892[1]
Party Candidate Votes %
Democratic Henry A. Coffeen 8,855 51.34%
Republican Clarence D. Clark (inc.) 8,394 48.66%
Total votes 17,249 100%

References

[edit]
  1. ^ "United States House of Representatives election in Wyoming, 1892".