1998 United States Senate election in Iowa
(Redirected from United States Senate election in Iowa, 1998)
The 1998 United States Senate election in Iowa was held November 3, 1998. Incumbent Republican United States Senator Chuck Grassley sought re-election to a fourth term in the United States Senate, running against former State Representative David Osterberg, who won the Democratic nomination unopposed. Grassley had not faced a competitive election since 1980; this year proved no different, and Grassley defeated Osterberg in a landslide.
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 County results Grassley: 50-60% 60-70% 70-80% 80-90% >90% | |||||||||||||||||
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Democratic primary
editCandidates
edit- David Osterberg, former State Representative
Results
edit| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Democratic | David Osterberg | 86,064 | 99.45% | |
| Democratic | Write-ins | 476 | 0.55% | |
| Total votes | 86,540 | 100.00% | ||
Republican primary
editCandidates
edit- Chuck Grassley, incumbent United States Senator
Results
edit| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | Chuck Grassley (Incumbent) | 149,943 | 99.72% | |
| Republican | Write-ins | 419 | 0.28% | |
| Total votes | 150,362 | 100.00% | ||
General election
editResults
edit| Party | Candidate | Votes | % | ±% | |
|---|---|---|---|---|---|
| Republican | Chuck Grassley (Incumbent) | 648,480 | 68.41% | −1.20% | |
| Democratic | David Osterberg | 289,049 | 30.49% | +3.29% | |
| Natural Law | Susan Marcus | 7,561 | 0.80% | −0.47% | |
| Socialist Workers | Margaret Trowe | 2,542 | 0.27% | +0.16% | |
| Write-ins | 275 | 0.03% | |||
| Majority | 359,431 | 37.92% | −4.50% | ||
| Turnout | 947,907 | ||||
| Republican hold | Swing | ||||
See also
editReferences
edit- ^ a b "Results" (PDF). sos.iowa.gov. 1998. Retrieved October 15, 2019.
- ^ "Election info" (PDF). clerk.house.gov. 1998. Retrieved October 15, 2019.