1928 United States presidential election in Delaware

1928 United States presidential election in Delaware

← 1924 November 6, 1928 1932 →
 
Nominee Herbert Hoover Al Smith
Party Republican Democratic
Home state California New York
Running mate Charles Curtis Joseph T. Robinson
Electoral vote 3 0
Popular vote 68,860 36,643
Percentage 65.03% 34.60%

County Results
Hoover
  50-60%
  60-70%


President before election

Calvin Coolidge
Republican

Elected President

Herbert Hoover
Republican

The 1928 United States presidential election in Delaware took place on November 6, 1928. All contemporary 48 states were part of the 1928 United States presidential election. State voters chose three electors to the Electoral College, which selected the president and vice president.

Delaware was won by Republican former Secretary of Commerce Herbert Hoover of California, who was running against Democratic Governor of New York Alfred E. Smith. Hoover's running mate was Senate Majority Leader Charles Curtis of Kansas, while Smith's running mate was Senator Joseph Taylor Robinson of Arkansas.

Hoover won with a majority of 65.03% of the vote to Smith's 34.60%, a margin of 30.43%. Socialist candidate Norman Thomas finished a distant third, with 0.31%. Hoover’s performance is the best by any presidential candidate in Delaware, surpassing its nearest rival from Barack Obama in 2008 by 3.12%.[1] In this election, Delaware voted 13% to the right of the nation at-large.[2]

  1. ^ "Presidential General Election Results Comparison – Delaware". Dave Leip’s U.S. Election Atlas.
  2. ^ "Dave Leip's Atlas of U.S. Presidential Elections". uselectionatlas.org. Retrieved March 31, 2023.

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