United States presidential election in Arizona, 1948
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All 4 Arizona votes to the Electoral College |
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The 1948 United States presidential election in Arizona took place on November 2, 1948, as part of the 1948 United States presidential election. Arizona voters chose four[2] representatives, or electors, to the Electoral College, who voted for President and Vice President.
Arizona was won by incumbent President Harry S. Truman (D–Missouri), running with Senator Alben W. Barkley, with 53.79% of the popular vote, against Governor Thomas Dewey (R–New York), running with Governor Earl Warren, with 43.82% of the popular vote.[3][4]
Results
| United States presidential election in Arizona, 1948 | ||||
|---|---|---|---|---|
| Party | Candidate | Votes | % | |
| Democratic | Harry S. Truman (inc.) | 95,251 | 53.79% | |
| Republican | Thomas Dewey | 77,597 | 43.82% | |
| Progressive | Henry A. Wallace | 3,310 | 1.87% | |
| Prohibition | Claude A. Watson | 786 | 0.44% | |
| Socialist Labor | Edward A. Teichert | 121 | 0.07% | |
| Total votes | 177,065 | 100% | ||
References
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| General articles |
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| Local results |
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| Other 1948 elections |
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