1928 United States presidential election in Delaware

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.

1928 United States presidential election in Delaware

← 1924November 6, 19281932 →
 
NomineeHerbert HooverAl Smith
PartyRepublicanDemocratic
Home stateCaliforniaNew York
Running mateCharles CurtisJoseph T. Robinson
Electoral vote30
Popular vote68,86036,643
Percentage65.03%34.60%

County Results
Hoover
  50-60%
  60-70%


President before election

Calvin Coolidge
Republican

Elected President

Herbert Hoover
Republican

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]

Results

1928 United States presidential election in Delaware[3]
PartyCandidateVotesPercentageElectoral votes
RepublicanHerbert Hoover68,86065.03%3
DemocraticAlfred E. Smith36,64334.60%0
SocialistNorman Thomas3290.31%0
CommunistWilliam Z. Foster560.06%0
Totals105,891100.0%3

Results by county

CountyDemRepOth
Kent5,7278,33527
New Castle22,46447,641307
Sussex7,16312,88454
STATEWIDE35,35468,860388

See also

References