1996 United States House of Representatives elections in Kansas
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All 4 Kansas seats to the United States House of Representatives | ||||||||||||||||||||||||||||
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The 1996 United States House of Representatives elections in Kansas were held on November 5, 1996 to determine who will represent the state of Kansas in the United States House of Representatives. Kansas has four seats in the House, apportioned according to the 1990 United States census. Representatives are elected for two-year terms.
Overview
| United States House of Representatives elections in Kansas, 1996[1] | |||||
|---|---|---|---|---|---|
| Party | Votes | Percentage | Seats | +/– | |
| Republican | 591,146 | 56.36% | 4 | - | |
| Democratic | 424,984 | 40.52% | 0 | - | |
| Libertarian | 23,253 | 2.22% | 0 | - | |
| Reform | 9,495 | 0.91% | 0 | - | |
| Totals | 1,048,878 | 100.00% | 4 | - | |
References
- ^ Clerk of the U.S. House of Representatives. "Statistics of the Presidential and Congressional Election of November 5, 1996" (PDF). U.S. Government Printing Office. p. 25.
