Quick: in April 2012, how many US states have an employment rate at or above the national average? (Hint: the US has 51 states or state-equivalents. I’m counting DC.)
This actually isn’t that surprising, once you look at the data — states with large population have larger unemployment rates. (I’m not an economist; why should this be?) In the plot below we have unemployment on the y-axis and log (base 10) of population on the x-axis. The dotted line represents the average unemployment rate of 8.1 percent; you can see that most states are above it. The upward slope indicates that low-population states have lower unemployment rates than high-population states. The solid line is the least-squares regression line for predicting unemployment from log population.
R code for generating the plot, from the file “unemployment.csv” (data is in the first comment to this post; of course you should edit the first line to wherever you store the file):
unemployment = read.csv('c:/users/michael/desktop/blog/unemployment.csv')
x = log(unemployment$pop)/log(10)+3;
y = unemployment$unemp;
plot(x, y, xlab="log_10 population", ylab="unemployment", main="Population of state vs. unemployment rate, April 2012")
text(x, y, labels=unemployment$state, cex=0.5, adj=c(0,-1))
abline(8.1, 0, lty=2)
The equation of the line is . To interpret the slope, we say that if we multiply a state’s population by 10 then we expect to add 1.66 percent to the unemployment rate. It’s probably easier to think in terms of doublings; we expect a state with twice the population to have an unemployment rate percent, or 0.50 percent, larger.
So does half the population live in states with a higher-than-average unemployment rate? Pretty much.
gives the sum of the populations in those 17 states; it returns 159398, for 159.398 million. (My population data are in thousands.) Total population at that time was 308.746 million; so 51.6% of the population lived in a state with unemployment at or above average. (If you throw out Washington state, which had unemployment equal to the national average of 8.1 percent, you get 49.4%.)
Perhaps somewhat ironically given the content of this post, I’m looking for a job, in the SF Bay Area. See my linkedin profile.