1988 United States Senate election in Ohio
The 1988 United States Senate election in Ohio was held on November 8, 1988. Incumbent Democratic U.S. Senator Howard Metzenbaum won re-election.[1] As of 2020, this is the last time in a presidential election year Ohio voted differently for president and for senate.
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County Results Metzenbaum: 50–60% 60–70% 70–80% Voinovich: 50-60% 60-70% | |||||||||||||||||
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Elections in Ohio | ||||||||||||||||||||
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Major candidates
Democratic
- Howard Metzenbaum, incumbent U.S. Senator
Republican
- George Voinovich, Mayor of Cleveland and former Lieutenant Governor of Ohio.
Results
Party | Candidate | Votes | % | |
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Democratic | Howard Metzenbaum (incumbent) | 2,480,038 | 56.97% | |
Republican | George Voinovich | 1,872,716 | 42.31% | |
Independent | David Marshall | 151 | 0.00% | |
Majority | 607,322 | 8.68% | ||
Turnout | 4,352,905 | 100.00% |
See also
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