Pennsylvania had nearly as many underinsured residents as it did uninsured in 2012, according to a new report from the Commonwealth Fund.
About 10 percent, or 1,114,294 people, were underinsured in Pennsylvania in 2011-12, compared to about 13 percent, or 1,426,872, who were uninsured, the report said. The national totals were higher, at 12 and 18 percent respectively, together totaling 78,729,718 people.
The report defined the underinsured as those whose households spent a high share of annual income on medical care -- 10 percent or more of income on medical care, excluding premiums, or 5 percent or more if income was under 200 percent of the federal poverty level. It did not include insured people who needed care but went without it because of the out-of-pocket costs they would incur or the insured who stayed healthy during the year but whose health insurance would have exposed them to high medical costs had they needed and sought care.
“The state level estimates provide a baseline to assess changes in premiums affordability relative to income over time,” the report said. Obamacare has the potential to reduce high medical care cost burdens while also covering the uninsured, it said, but “the extent of improvement will critically depend on state decisions and the plans people select.”
The report said 12 percent of insured Pennsylvanians had premiums in 2011-2012 that exceeded the Obamacare threshold for maximum premium contribution as a share of income in the Marketplace or Medicaid. It also showed a rise in the average health insurance premium as percent of median household income; in Pennsylvania, it increased from 14 percent in 2003 to 19.6 percent in 2012, with the national change at 14.9 percent in 2003 to 21.6 percent in 2012.
The lower their income, the more likely Pennsylvanians were to be underinsured -- 34 percent of those with incomes less than 100 percent of FPL, 19 percent of those at 100 to 199 percent of FPL, 3 percent of 200 to 399 percent of FPL, and 1 percent of those at 400 percent of FPL or more.