1988 Missouri Attorney General election
November 8, 1988
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County results Webster: 50–60% 60–70% 70–80% Wolff: 50–60% 60–70% | |||||||||||||||||
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| Elections in Missouri |
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The 1988 Missouri Attorney General election was held on November 8, 1988, in order to elect the attorney general of Missouri. Republican nominee and incumbent attorney general William L. Webster defeated Democratic nominee Michael A. Wolff.[1]
General election
On election day, November 8, 1988, Republican nominee William L. Webster won re-election by a margin of 428,129 votes against his opponent Democratic nominee Michael A. Wolff, thereby retaining Republican control over the office of attorney general. Webster was sworn in for his second term on January 9, 1989.[2]
Results
| Party | Candidate | Votes | % | |
|---|---|---|---|---|
| Republican | William L. Webster (incumbent) | 1,224,346 | 60.59 | |
| Democratic | Michael A. Wolff | 796,217 | 39.41 | |
| Total votes | 2,020,563 | 100.00 | ||
| Republican hold | ||||
See also
References
- ^ "1988 Attorney General General Election Results - Missouri". uselectionatlas.org. December 20, 2012. Retrieved 2024-10-27.
- ^ "MO Attorney General". ourcampaigns.com. February 12, 2018. Retrieved 2024-10-27.