As I read the linked article about unemployment, I was struck by the fact that the highest unemployment in the U.S., states where unemployment is above 10%, is found in states with population densities well above the national average. Conversely, the states with low unemployment are all states with population densities well below the national average. Since free trade is practiced between all fifty states, which would tend to muddle the unemployment data, I’ve tended to discount the ability of a state-by-state comparison to provide evidence in support of the theory I presented in Five Short Blasts, but perhaps it’s worth taking a look.
More than 1-in-10 workers were unemployed in four U.S. states in January, the government said Wednesday in a report that pointed to a rapid deterioration in the job market.
The unemployment rate in Michigan was 11.6%, up 4.3 percentage points from a year earlier and the highest for any state, the Labor Department said. The unemployment rates in California, Rhode Island and South Carolina rose in January to top 10%. That was the first time those states had jobless rates in the double digits during this recession, which began in December 2007.
While the overall population density of the U.S. is about 85 people per square mile, the population density of these four states is as follows:
Rhode Island = 1,025 people per square mile.
California = 231 people per square mile.
Michigan = 178 people per square mile.
South Carolina = 141 people per square mile.
Now let’s look at the states with the lowest unemployment:
There were a few states with unemployment rates significantly below the national average. The unemployment rate in January was 3.7% in Wyoming, the lowest for any U.S. state. Jobless rates were also below 5% in Iowa, Nebraska, North Dakota, South Dakota and Utah in January.
Here’s the population density of those states:
Wyoming = 5 people per square mile.
Iowa = 53 people per square mile.
Nebraska = 23 people per square mile.
North Dakota = 9 people per square mile.
South Dakota = 10 people per square mile.
Utah = 30 people per square mile.
This piqued my curiosity, so I then plotted the state-by-state data contained in the left column of the article vs. population density of the states. The result is the following scatter chart. I experimented with inserting the trend line and settled on the “power” trend line as the best fit for the data, although each of the others (linear, exponential and logarithmic) also showed a positive correlation between population density and unemployment.
In spite of the fact that unemployment data is muddled by free trade, which is nowhere practiced more vigorously than it is between the fifty U.S. states, there remains evidence of a correlation.
Once again, the evidence supports the theory that, beyond some difficult-to-define “optimum” population density, further population growth will drive unemployment higher.
President Obama: want to fix the economy and stop rising unemployment? Enact policies to stabilize our population (like cutting immigration) and fix our trade policy to eliminate the trade deficit exported by overpopulated nations.