1960 United States presidential election in Rhode Island

The 1960 United States presidential election in Rhode Island took place on November 8, 1960, as part of the 1960 United States presidential election, which was held throughout all 50 states. Voters chose 4 representatives, or electors to the Electoral College, who voted for president and vice president.

Main article: 1960 United States presidential election

1960 United States presidential election in Rhode Island

November 8, 1960

Nominee John F. Kennedy Richard Nixon Party Democratic Republican Home state Massachusetts California Running mate Lyndon B. Johnson Henry Cabot Lodge, Jr. Electoral vote 4 0 Popular vote 258,032 147,502 Percentage 63.63% 36.36%

County Results Kennedy   50-60%   60-70% Nixon   50-60%

President before election Dwight Eisenhower Republican Elected President John F. Kennedy Democratic

Rhode Island voted for the Democratic nominee, Senator John F. Kennedy of Massachusetts, over the Republican nominee, Vice President Richard Nixon of California. Kennedy ran with Senate Majority Leader Lyndon B. Johnson of Texas, while Nixon's running mate was Ambassador Henry Cabot Lodge, Jr. of Massachusetts.

Kennedy carried Rhode Island by a margin of 27.27%.

Results

1960 United States presidential election in Rhode Island[1]
Party		Candidate	Votes	Percentage	Electoral votes
	Democratic	John F. Kennedy	258,032	63.63%	4
	Republican	Richard Nixon	147,502	36.36%	0
	Write-ins	Write-ins	1	0.01%	0
Totals			405,535	100.00%	4

By county

County	Kennedy%	Kennedy#	Nixon%	Nixon#	Others%	Others#
Providence	67.5%	189,014	32.5%	91,028	0%	0
Bristol	59.6%	11,099	40.4%	7,537	0%	0
Newport	56.8%	15,677	43.2%	11,942	0%	0
Kent	55.7%	30,662	44.3%	24,344	0%	0
Washington	47.8%	11,580	52.2%	12,651	0%	0

References

1. "1960 Presidential General Election Results - Rhode Island". Dave Leip's Atlas of U.S. Presidential Elections. Retrieved 2013-02-07.