COLUMBUS, Ohio (WCMH) – A recent study ranked three Ohio municipalities among the top 10 U.S. cities where residents are the most burdened by grocery costs.
The personal finance website WalletHub looked at 100 of the largest cities to discover where residents spend the highest percentage of their household income on common grocery items. Researchers collected data from the U.S. Census Bureau and the nonprofit Council for Community and Economic Research.
Multiple Ohio cities placed high on the list, including Cleveland, where the average cost of groceries accounts for about 3.77% of the median household income. While Cleveland experiences "reasonable" grocery prices, the city has the lowest household income of those included in the study, leading to food costs making up a larger portion of residents' paychecks, according to WalletHub.
Source: WalletHubThe other Ohio cities in the top 10 are Toledo at No. 5, where residents allocate around 3.09% of their income to groceries each month, and Cincinnati (2.90%). Columbus ranked the lowest of all Ohio cities included, but still placed relatively high on the list, at No. 26. On average, residents of Ohio's capital spend 2.28% of their monthly income on groceries, the study says.
The analysis found that Detroit residents spend the highest percentage of their incomes on groceries, with a median cost of 3.78%. While grocery prices are somewhat average, the median household income is the second lowest, WalletHub said.
The full top 10 can be found below:
Detroit, Michigan Cleveland, Ohio Birmingham, Alabama Newark, New Jersey Toledo, Ohio Hialeah, Florida Buffalo, New York Cincinnati, Ohio Milwaukee, Wisconsin Memphis, TennesseeIn contrast, California cities San Francisco (1.22%), San Jose (1.16%) and Fremont (0.96%) spend the smallest percentage of income, on average, at the grocery store.
WalletHub’s full study can be found here.
The Nexstar Media Wire contributed to this report.
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