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Do What Now?

Maybe I'm completely daft but, upon reading this article, one sentence made no sense to me.

From: http://www.afp.com/english/news/stories/070907143140.wzsuhlfa.html

US economy loses jobs for first time in 4 years

WASHINGTON (AFP) - The world's largest economy was hit with surprise job losses in August as the housing downturn and a credit crunch sparked increased layoffs, a government report revealed Friday.

The Labor Department said US employers unexpectedly shed 4,000 jobs in August, marking the first drop in payrolls since August of 2003.

The unexpected drop in nonfarm payrolls caught Wall Street off guard as most economists had been calling for around 110,000 new jobs to be created in August.

Economists said the report raises the odds that the Federal Reserve will soon move to slash borrowing costs.

The job report, viewed as one of the best indicators of economic momentum, suggests the US employment market has been jolted by a housing slump and rising mortgage defaults which have triggered a credit squeeze that has roiled financial markets.

The national unemployment rate held steady at 4.6 percent despite the month's job losses.

Wait, what? How can a percentage of nonworking people stay the same if the number of jobs decreased? Isn't one job defined as one person working? Let's do the math here.

300,000,000 peopletotal * 4.6% = 13,800,000 peopleunemployed before

13,800,000 peopleunemployed before + 4,000 peoplenewly unemployed = 13,804,000 peopleunemployed now

13,804,000 peopleunemployed now / 300,000,000 peopletotal = 4.6013333...%

Just because you round a number away, that doesn't mean it didn't change. Best case, that was an unnecessary sentence. Worst case it was misleading.

Ok, I admit it. I'm addicted to subscripting.


( 12 comments — Leave a comment )
Sep. 7th, 2007 05:19 pm (UTC)
Well, I remember from my economic class that if someone stops looking for work, they aren't considered part of the workforce, and therefore aren't considered unemployed. I also want to say, bsed on a vague memory from that class, that someone is assumed (by the economists) to no longer be looking for work if they are unemployed for more than 6 months. Which, to me, is clearly ridiculous.
Sep. 7th, 2007 05:30 pm (UTC)
Innnnteresting... although these 4,000 apparently all just recently lost their jobs, so that must mean that 4,000 people who were unemployed all quit looking for work. What an unusual measurement.
Sep. 7th, 2007 05:48 pm (UTC)
A quick look around wikipedia confirms that people that have stopped looking for work are no longer considered part of the labor force, and are not counted as unemployed. I'm not sure where I came up with the 6 month cut off - my teacher may have been wrong.

Anyways, here's a interesting glossary of terms used by the U.S. Bureau of Labor Statistics. http://www.bls.gov/bls/glossary.htm.

Note that 4.6% doesn't include *underemployment*. That guy who waits your table, but would rather be doing something related to his degree - underemployed. Anyone doing part-time work (even 1 hour!) but wants to be doing full-time - underemployed. There is no official survey of underemployment, and I'd love to see someone attempt those numbers.
Sep. 7th, 2007 07:55 pm (UTC)
What do they do about students that are over 18 (I'm assuming they only count citizens over 18)? Megan's not working, but I wouldn't call her unemployed or even underemployed.
Sep. 7th, 2007 08:23 pm (UTC)
She's not seeking work (I assume), so she isn't part of the labor pool. Only people that have actively work sought in the last 4 weeks and haven't found it are considered unemployed. All kind of people who aren't working aren't counted as unemployed. Self-employed people. Students. The wealthy who choose not to work. Discouraged workers who are no longer actively looking for work. Etc.
Sep. 7th, 2007 05:42 pm (UTC)
I don't know how the department of labor collects its data, but it seems to me that the difference of 0.001 may not be statistically significant, so in a well-defined way, the unemployment rate did not change. They don't write an error estimate on the unemployment rate, eg 4.6 ± 0.05, but in truth it should be there. If the calculated increase of 0.001 is smaller than the error estimate, we cannot confidently say that the unemployment rate actually rose. It may not just be a matter of rounding.

Another possibility, as Lis mentioned, is that the total number of people seeking jobs changed.
Sep. 7th, 2007 06:47 pm (UTC)
Nah, see, the layoffs were in the construction industry, so they were all Mexicans, and Mexicans don't count because they don't qualify for unemployment.

Besides, they all went to work in their uncle's landscaping business or their aunt's restaurant. (Funny that the hardest-working people in our economy are the ones that the government is so bound and determined to get rid of, but that's another subject).

I may be wrong, but I believe that unemployment statistics are based at least in part from data on unemployment compensation, which doesn't go on indefinitely. That's why the cutoff. And it makes sense, really, because there are a lot of people who do drift into and out of the employment pool -- kids and students, women who decide to stay home with the kids unless a really great job comes along, seasonal workers, retirement-age people.

When I was a physics TA, back in the dinosaur days, hand calculators had just reached the price point where the more well-to-do students could afford one. Every problem got answered to eight digits. I had a hard time convincing them that you can't have four digit accuracy unless all the numbers going into your calculation are known to four digits. That's pretty obvious if you use a slide rule, but it's not at all apparent to calculator calculators. Nobody really knows exactly how many people are working at any time, and how many are seeking work, so that fourth digit doesn't mean anything. 4.601 or 4.604 or 4.598, they're all the same: 4.6 .

Sep. 7th, 2007 07:57 pm (UTC)
You know, I never did understand significant digits. They taught it to us in Chemistry, but it never stuck with me. Like if you multiply 4.4 * 4.3, you get absolutely 18.92, but you can only keep 18.9 because you didn't start with 4.40 and 4.30?
Sep. 7th, 2007 09:03 pm (UTC)
The idea is that you cannot make gains in precision. Let's say that I am given a distance x traversed in a time t and am asked to compute the average velocity, v=x/t. The precision with which I can compute v is limited by the precision to which I know x and t. Assume that I somehow know x = 10 cm exactly. If I measure t and find t = 3.2 s, I might be inclined to say v = 3.125 cm/s. The problem with this is that I am claiming to know the velocity to 4 digits of precision when I know the time only to two digits. If a more careful measurement found t = 3.23 s, I would calculate v = 3.099 cm/s. If a still more careful measurement found t = 3.231 s, I would get v = 3.095 cm/s. So let's look at the results:

With sig figs Without sig figs
t = 3.231 sv = 3.095 cm/sv = 3.095 cm/s
t = 3.23 sv = 3.10 cm/sv = 3.099 cm/s
t = 3.3 sv = 3.1 cm/sv = 3.125 cm/s

With sig figs, as I decrease the precision with which I know the time, I likewise decrease the precision to which I calculate v. The progression in v should make sense: it's basically rounding. WIthout sig figs, though, I acquire artificial digits. I know from my more accurate measurement that v ≠ 3.125 cm/s. The 2 and 5 are artifacts of the fact that I am attempting to make a calculation to 4 digits when I only know my input number to 2 digits.
Sep. 7th, 2007 09:10 pm (UTC)
I understand that logic. But how do ever know x = 10cm exactly? And if you don't, which original value's number of significant digits matter? Is it that the calculation can have no more precision than the least precise original measurement?
Sep. 7th, 2007 09:15 pm (UTC)
I understand that logic. But how do ever know x = 10cm exactly?

In reality you wouldn't. For illustrative purposes, I chose to set x to an exact value so that I would only have to worry about the accuracy in t. If I were actually trying to compute something's average velocity, I would have to worry about the accuracy with which both x and t were measured.

And if you don't, which original value's number of significant digits matter? Is it that the calculation can have no more precision than the least precise original measurement?

Yes, exactly.
Sep. 7th, 2007 10:03 pm (UTC)
what Hans is trying to say is, yes.
( 12 comments — Leave a comment )

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