Sonoma, Chicago and San Diego at the Top: Cheapest Cities for 5-Star Experience
Research highlights:
- Sonoma, Chicago, and San Diego are the top 3 cheapest cities for a 5-star experience.
- Seven of the top ten cities are in California, making it the premier state for experiencing luxury at an affordable price.
- Sonoma, San Diego, and Chicago lead the list as the most affordable cities for luxury dining, while Chicago, Paso Robles, and San Francisco are the top choices for budget-friendly luxury hotels.
After reviewing hundreds of luxury hotels and restaurants, Wethrift reveals the top cities where you can enjoy a 5-star experience in the cheapest way possible.
The study identified 956 Michelin-recognized restaurants and 348 luxury hotels across the U.S. to discover the best cities that offer a low-cost luxury experience.
Sonoma is the cheapest city for a 5-star experience, with an average restaurant price of just $43.75 and an average hotel price of $518.00, earning it a low-cost luxury rating of 1.83. In Sonoma, you can enjoy restaurants such as Valley, then stay a night over at the Cottage Inn & Spa.
Ranking second is Chicago, where you can dine at places like Ghin Khao or Birrieria Zaragoza and enjoy the stay over at Hotel Zachary, Chicago, a Tribute Portfolio Hotel. They offer a 5-star experience with an average restaurant price of $52.88 and a hotel price of $411.25, making it a top choice with a low-cost luxury rating of 1.63.
San Diego ranks third, averaging $49.19 for dining and $579.00 per night, securing its spot with a low-cost luxury rating of 1.51. There, you can visit restaurants like LOLA 55 or Great Maple and then go to The Westgate Hotel.
Coming in fourth is Santa Barbara, where you can dine at Los Agaves and stay at the Simpson House. The average dining experience in this city is $54.46, and the average rating on a hotel per night is $552.00, earning a low-cost luxury rating of 1.34.
In fifth place, Los Angeles offers high-end dining experiences at $55.45 and $559.75 per night, boasting a low-cost luxury rating of 1.29. You can enjoy dining at places like La Azteca or Pollo o la Brasa and stay at the Sunset Marquis.
Ranking sixth, San Francisco offers a deluxe dining experience at La Taqueria and lodging at Mansion on Sutter. The average restaurant price for a 5-star experience in this city is $59.71, and the average hotel price is $487.43, earning a low-cost luxury rating of 1.23.
Seventh on the list, Miami Beach offers a luxurious getaway. You can dine averaging $59.38 at establishments such as Taquiza or stay at The Standard Spa for $526.00, resulting in a low-cost luxury rating of 1.18.
Washington, D.C., takes the eighth place, where luxury comes at an average of $55.12 for restaurants and $707.00 per hotel stay with a low-cost luxury rating of 1.07. This city offers high-end dining at places like La Tejana or Toki Underground and luxury hotels like The Jefferson, Washington, DC.
In ninth place is Paso Robles, with dining averaging $65.00 and hotels at $480.00, giving it a low-cost luxury rating of 1.02. You can enjoy dining at various local restaurants and stay at luxury hotels like the Geneseo Inn for $455.00.
Rounding out the list is Healdsburg, where you can dine at fine restaurants for an average of $64.06 and stay at hotels for $520.00 per night, earning a low-cost luxury rating of 1.00. A good pick would be The Ruse Healdsburg.
Which US cities offer the cheapest 5-star experience?
Rank | City | Average Restaurant Price | Average Hotel Price | Low-cost Luxury Rating |
1 | Sonoma | $43.75 | $518.00 | 1.83 |
2 | Chicago | $52.88 | $411.25 | 1.63 |
3 | San Diego | $49.19 | $579.00 | 1.51 |
4 | Santa Barbara | $54.46 | $552.00 | 1.34 |
5 | Los Angeles | $55.45 | $559.75 | 1.29 |
6 | San Francisco | $59.71 | $487.43 | 1.23 |
7 | Miami Beach | $59.38 | $526.00 | 1.18 |
8 | Washington, D.C. | $55.12 | $707.00 | 1.07 |
9 | Paso Robles | $65.00 | $480.00 | 1.02 |
10 | Healdsburg | $64.06 | $520.00 | 1.00 |
Which are the cheapest US cities for Michelin-recognized restaurants?
Rank | City | No. of Restaurants | Average Price |
1 | Sonoma | 4 | $43.75 |
2 | San Diego | 31 | $49.19 |
3 | Chicago | 152 | $52.88 |
4 | Santa Barbara | 14 | $54.46 |
5 | Miami | 48 | $54.69 |
6 | Washington, D.C. | 122 | $55.12 |
7 | Los Angeles | 140 | $55.45 |
8 | New York City | 396 | $56.63 |
9 | Miami Beach | 16 | $59.38 |
10 | San Francisco | 112 | $59.71 |
Which are the cheapest US cities for luxury hotels?
Rank | City | No. of Hotels | Average Price |
1 | Chicago | 12 | $411.25 |
2 | Paso Robles | 2 | $480.00 |
3 | San Francisco | 7 | $487.43 |
4 | Sonoma | 2 | $518.00 |
5 | Healdsburg | 2 | $520.00 |
6 | Miami Beach | 4 | $526.00 |
7 | Santa Barbara | 2 | $552.00 |
8 | Los Angeles | 8 | $559.75 |
9 | San Diego | 3 | $579.00 |
10 | Washington, D.C. | 6 | $707.00 |
Speaking on the findings, Nick Drewe, discounts expert and founder of online platform Wethrift, says:
“Experiencing luxury doesn’t have to come at a hefty price tag. Our study shows that you can indulge and let loose with these top-tier experiences without breaking the bank, proving that opulence and affordability can go hand in hand.”
“Whether it be savoring gourmet meals or staying in elegant hotels, we believe everyone deserves a taste of luxury. Traveling lies in creating cherished memories, enjoying life’s finer things, and not worrying about financial strain 24/7.”
Methodology:
Wethrift analyzed 956 Michelin-starred and Bib Gourmand restaurants from the Michelin guide and 348 luxury 3 to 5-star rated hotels across the U.S. using data from Booking.com and TripAdvisor. They aimed to identify the top-ranking cities, restaurants, and hotels offering the most affordable luxury experience options.
The restaurant and hotel prices were standardized by applying min-max normalization taken from their respective prices to ensure they had an equal bearing on the overall score.
If you would like to cover this story, please provide proper credit and a link back to wethrift.com.
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