**Related Video Above: In case you don’t want to go out, here’s how to make hand-crafted cocktails at home.**
(WGHP/WJW) — This Ohio city is among the most expensive place in America for a night out, according to a new study by PriceListo.
That city is not Cleveland or Cincinnati or Dayton, but is, you guessed it, Columbus.
The Top 10 most expensive cities for a night out are as follows:
- San Diego – $225.27
- Charlotte – $224.75
- Austin – $221.23
- Memphis – $216.25
- Columbus – $212.47
- Boston – $211.16
- Seattle – $210.61
- Washington, DC – $210.39
- Houston – $209.54
- Tucson – $203.3
The overall price was calculated by using cost-of-living data from each US city with a population exceeding $500,000 to find out the average price of a cocktail, cab fare (three-mile journey), a bottle of wine and a pint of beer. In addition, data was analyzed from the hotel site Vio to discover the median price for a one-night stay in a budget hotel for each city. The price of each category was then added together to discover an overall cost.
“When planning a fun night out with your friends, the last thing you want to worry about is spending too much money,” PriceListo said in a statement. “This study can help guide those trying to tighten their purse strings in the right direction when enjoying a night out with their friends.”
Inversely, here are the Top 10 most affordable cities for a night out, of which no Ohio cities made the list:
- Las Vegas – $120.76
- San Antonio – $134.56
- Oklahoma City – $136.98
- San Francisco – $142.94
- Fresno – $145.55
- Albuquerque – $157.47
- Jacksonville – $162.42
- Philadelphia – $168.74
- Fort Worth – $172.14
- Chicago – $173.81
