1948 United States presidential election in Connecticut

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1948 United States presidential election in Connecticut

← 1944 November 2, 1948 1952 →
  ThomasDewey.png Harry S. Truman.jpg
Nominee Thomas E. Dewey Harry S. Truman
Party Republican Democratic
Home state New York Missouri
Running mate Earl Warren Alben William Barkley
Electoral vote 8 0
Popular vote 437,754 423,297
Percentage 49.55% 47.91%

President before election

Harry S. Truman

Elected President

Harry S. Truman

The 1948 United States presidential election in Connecticut took place on November 7, 1944, as part of the 1948 United States presidential election. State voters chose eight electors to the Electoral College, which selected the president and vice president.

Connecticut was won by Republican candidate New York governor Thomas E. Dewey over Democratic candidate, incumbent President Harry S. Truman.

Dewey won the state by a very narrow margin of 1.64 percent.


1948 United States presidential election in Connecticut[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Thomas Edmund Dewey of New York Earl Warren of Ohio 437,754 49.55% 8 100.00%
Democratic Harry S. Truman of Missouri Alben William Barkley of Kentucky 423,297 47.91% 0 0.00%
Progressive Henry Agard Wallace of Iowa Glen Hearst Taylor of Idaho 13,713 1.55% 0 0.00%
Socialist Norman Thomas of New York Tucker Powell Smith of Michigan 6,964 0.79% 0 0.00%
Socialist Labor Edward A. Teichert of Pennsylvania Stephen Emery of New York 1,184 0.13% 0 0.00%
Socialist Workers Farrell Dobbs of Minnesota Grace Carlson of Minnesota 606 0.07% 0 0.00%
Total 883,518 100.00% 8 100.00%


  1. ^ "1948 Presidential General Election Results - Connecticut". U.S. Election Atlas. Retrieved 23 December 2013.