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1966 Iowa gubernatorial election

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1966 Iowa gubernatorial election

← 1964 November 8, 1966 1968 →
 
Nominee Harold Hughes William G. Murray
Party Democratic Republican
Popular vote 494,259 394,518
Percentage 55.34% 44.17%

County results
Hughes:      50–60%      60–70%      70–80%
Murray:      40–50%      50–60%      60–70%

Governor before election

Harold Hughes
Democratic

Elected Governor

Harold Hughes
Democratic

The 1966 Iowa gubernatorial election was held on November 8, 1966. Incumbent Democrat Harold Hughes defeated Republican nominee William G. Murray with 55.34% of the vote.

Primary elections

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Primary elections were held on September 6, 1966.[1]

Democratic primary

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Candidates

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Results

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Democratic primary results[1]
Party Candidate Votes %
Democratic Harold Hughes (incumbent) 80,198 100.00
Total votes 80,198 100.00

Republican primary

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Candidates

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Results

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Republican primary results[1]
Party Candidate Votes %
Republican William G. Murray 87,371 50.5
Republican Robert K. Beck 85,733 49.5
Total votes 173,109 100.00

General election

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Candidates

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Major party candidates

  • Harold Hughes, Democratic
  • William G. Murray, Republican

Other candidates

  • David B. Quiner, Independent
  • Charles Sloca, Independent

Results

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1966 Iowa gubernatorial election[2]
Party Candidate Votes % ±%
Democratic Harold Hughes (incumbent) 494,259 55.34%
Republican William G. Murray 394,518 44.17%
Independent David B. Quiner 3,680 0.41%
Independent Charles Sloca 715 0.08%
Majority 99,741
Turnout 893,175
Democratic hold Swing

References

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  1. ^ a b c "Summary of Official Canvass of Votes Cast in Iowa Primary Election" (PDF). Secretary of State of Iowa. 1966. Retrieved March 20, 2020.
  2. ^ "Summary of Official Canvass of Votes Cast in Iowa General Election" (PDF). Secretary of State of Iowa. 1966. Retrieved March 20, 2020.