If you think San Francisco is the most expensive city in the Golden State, think again. That spot belongs to a quiet beach neighborhood just south of Los Angeles: Huntington Beach. And just when you are about to make reservations for a meeting in Silicon Valley, you may want to reconsider San Francisco for that task. That’s because, according to new research from LosAngelesHotels.org, it turns out that San Jose, located in Santa Clara county and perhaps better known for its complicated Winchester House than tourism hospitality, is the second most expensive city in California for hotel stays.
The survey compared hotel rates in 30 city destinations across California. For each destination, the average price for the cheapest available double room during September and October 2019 — the months when hotel rates in most Californian cities tend to be at their highest — was recorded (only centrally located hotels rated at least 3 stars were considered for the survey).
With an average of $198 per night for the least expensive room, San Jose takes 2nd spot in the rankings, and Huntington Beach, known for its world-class surfing beaches, is slightly pricier at an average rate of $223 per night. Sunnyvale, also located in Santa Clara county, completes the podium with an average rate of $193.
With Oakland, San Francisco and Fremont also in the Top 10, the Bay Area is certainly well represented. Significantly more affordable are Los Angeles and San Diego, where travelers can expect to find average rates of just $126 and $114 per night, respectively.
The following table shows the 10 most expensive city destinations in California. The prices shown reflect the average rate for each destination’s cheapest available double room for the period spanning September and October of this year.
1. Huntington Beach $223
2. San Jose $198
3. Sunnyvale $193
4. Oakland $186
5. San Franciso $184
6. Santa Clarita $165
7. Sacramento $158
8. Riverside $153
9. San Bernadino $153
10. Fremont $152