The cheapest cities in the USA have just been unveiled in the TripIndex Cities cost comparison from TripAdvisor® which is now in its third year.
The survey compares the costs of dinner and cocktails for two, a return taxi trip and a one night stay at a 4 star hotel. In total 20 US cities were included, which is 5 more than last year. The hotel stay is calculated for a stay between June and August this summer, which gives a good insight into summer 2013 holiday prices.
The cheapest city again this summer is Las Vegas with this year’s package only costing a mere 9 USD more than last year! Summer 2013 total cost is 272.94 USD and last summer was 263.46 USD. This is despite a general 18% increase in 4 star hotel stays. In fact a hotel stay in Vegas is almost a third of a comparable hotel in Honolulu in Hawaii, which was revealed as the most expensive US destination in the same survey! The only downside is that the dinner for two element in Las Vegas comes out as most expensive out of all 20 cities. Generally though Las Vegas continues to offer superb value, providing you don’t hit the casinos too heavily of course!
The second cheapest city in the USA is Miami in Florida which wasn’t included in last summer’s survey.
The third cheapest destination is Dallas in Texas which was last year’s number two city. The total cost for the four items has risen by 45 USD over last summer.
In summary, for the third year in a row, indications are that cities in southern USA are generally cheaper, with Miami, Dallas and Atlanta all in the top four spots.
Cheapest Cities in USA for Summer 2013
The national average is $398.95 based on hotel stay, taxi trip, dinner and cocktails for two;
1. Las Vegas, Nevada $272.94
2. Miami, Florida $318.62
3. Dallas, Texas $328.55
4. Atlanta, Georgia $331.14
5. Portland, Oregon $347.67
6. New Orleans, Louisiana $348.66
7. Houston, Texas $355.23
8. Minneapolis, Minnesota $372.00
9. Denver, Colorado $390.11
10. Philadelphia, Pennsylvania $393.91
Thanks to TripAdvisor for the data.
