Christmas is a fantastic time of the year, but some places for celebrating it are better than others. That’s because America’s favorite holiday is also one of the most expensive. From decorations and food to gifts and travel, there’s a long list of expenses to prepare for during the year.
In 2018, Americans are expected to spend up to $720 billion over the holidays, according to the National Retail Federation. The average person plans to spend $1,007.24. If you don’t have a reasonable holiday budget – or enough self-control – then your Plan B should include celebrating in the cheapest Christmas destinations. Overspending is a common mistake committed by consumers during the Christmas shopping season and one of the top sources of holiday stress.
Typical Christmas activities include shopping, dining out, and attending holiday events. Many people also are likely to attend church services.
WalletHub considered these factors to determine cities where you can enjoy a great Christmas. It compared the 100 largest U.S. cities based on 31 key indicators of a festive and affordable Christmas, such as traditions, shopping, and costs.
The top 10 cities are:
1. Orlando, Florida
2. Las Vegas
3. New York
4. Atlanta, Georgia
5. Pittsburgh
6. San Francisco
7. Los Angeles
8. Chicago
9. Seattle
10. Dallas
The bottom 10 cities are:
91. St. Petersburg, Florida
92. Garland, Texas
93. Stockton, California
94. Toledo, Ohio
95. Glendale, Arizona
96. Chula Vista, California
97. Laredo, Texa
98. North Las Vegas, Nevada
99. San Bernardino, California
100. Hialeah, Florida
So merry Christmas, where ever you are or where ever you're traveling, and happy New Year to all, too.
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