4 ms·

Estimate: Top US Firms Have Over 700,000 Job Openings

5 points by gne1963 18y ago

nostrademons 18y ago
The "100 companies is a representative sample" is a big assumption. The linked article says those 100 companies were all Fortune 500: that implies that there's a very large likelihood that they are bigger than the other companies in the Fortune 1000. Bigger companies naturally have more job openings. Even if this is true, it doesn't necessarily mean the economy's doing better. It's pretty likely that the job openings require skills that the newly unemployed workers do not have. Otherwise, they'd be filled. You can't hire a former real estate agent at Google and expect him to be immediately productive. (Hell, you usually can't hire a software engineer at Google and expect him to be immediately productive, but at least you can hire him.) The issue with recessions is misallocation of capital, and it can't be fixed with a magic wand. You generally have to wait for people to retrain in new, productive areas of the economy, and for capital to be liquidated and entrepreneurs to find new uses for it. That's why it takes ~4 years instead of ~4 months.
- gne1963 18y ago
  The samples were taken from a random selection of 100 of the top 1000 firms. The real question is how to make that selection of 100 firms so that statistically the sample size is a representative sampling. The goal is not to determine whether or not the jobs are available to those folks who may have just been laid off. The goal is to determine just how many job opening do exist in the Fortune 1000 without needing to hit every single job board of every one of those 1000 companies... surely there is a statistician who can solve that one? ;-)
  - nostrademons 18y ago
    From the link on your page: "Yesterday we embarked on a research project to find current job openings. We started with 21 firms (with 18,839 open positions) and now we're up to data from 100 of them. (All Fortune 500)" The chance that a random sample of 100 firms are all in the top half of the sample universe is 2^100 = about a decillion. Doesn't really look random to me.
    - gne1963 18y ago
      Thanks, nice catch and that post needs to be changed... momentarily... The universe was truely from the Fortune 1000 from the list publish online... not the top 500 only...
      - gne1963 18y ago
        You can see my confusion is somewhat born out by hitting this link... it is labeled 500 when actually it lists the top 1000... http://money.cnn.com/magazines/fortune/fortune500/2008/full_list/ http://money.cnn.com/magazines/fortune/fortune500/2008/full_...
        gne1963 18y ago
        So here is the methodology: 1. Hit the site above. 2. Randomly pick a company from the list... for instance: 3. Pick Paychex Systems... company 921 on Fortunes list... 4. On their site they list 137 job openings... 5. Add that to the total... 6. Repeat until you have a representative sample size Question: Using an appropriate random number generation between 1 and 1000, what is the sample size required to then estimate the total job openings on the 1000 sites?
gne1963 18y ago
So here is the methodology: 1. Hit http://money.cnn.com/magazines/fortune/fortune500/2008/full_list/ http://money.cnn.com/magazines/fortune/fortune500/2008/full_... 2. Randomly pick a company from the list... for instance: Pick Paychex Systems... company 921 on Fortunes list...On their career site they list 137 current job openings... 3. Add that to your total... 4. Repeat until you have a representative sample. 5. Extrapolate to 1000... Question: Using an appropriate random number generation between 1 and 1000, what is the sample size required to then estimate the total job openings on the 1000 sites?
- NyxWulf 18y ago
  Well here are a couple of things: you show 99 records on your excel spreadsheet, with an average of 709 and a standard deviation of 1073. With that large of a standard deviation and that average you can't assume a normal distribution so you need to use the T distribution. Using the T distribution at a 95% confidence level I come up with an estimated average of [495, 923] which results in a number of jobs in the top 1000 from 495,000 to 923,000 with a 95% certainty. Just for kicks I wanted to see what would happen if you sampled 500 times but had that same standard deviation. At that point your spread would be about +/- 100 for the average value of X. If you are looking for a singular number you'll need to check all 1,000 of them, otherwise it will be an estimated range.
  - gne1963 18y ago
    Thanks! So what if I did another 100 random sample resulting a "better" (lower) std dev? At what level of std dev could one assume a normal distribution out to 1000? gne
anamax 18y ago
Job opening on a company's "career site" are a very poor predictor of the number of people that company would hire if could find suitable candidates. In some cases, the listing is to satisfy an HR rule and the person to be hired has already been selected. In other cases, it's "maybe someone wonderful will wander in". And then there are the hires that happen without a listing. I'd guess that the actual uncertainty in the total is more a function of the uncertainty in the numbers than it is in the sampling.
- gne1963 18y ago
  >I'd guess that the actual uncertainty >in the total is more a function of the >uncertainty in the numbers than it is in >the sampling. totally agree. As with any indicator there is quite a bit of uncertainty. And as you suggest "there are the hires that happen without a listing..." Most career counselors would agree that there are actually more hidden jobs than what the boards suggest... So... your argument might just boost the number if we add some multiplier into the model based on the jobs actually posted. thanks! gne1963

5 points by gne1963 18y ago

The "100 companies is a representative sample" is a big assumption. The linked article says those 100 companies were all Fortune 500: that implies that there's a very large likelihood that they are bigger than the other companies in the Fortune 1000. Bigger companies naturally have more job openings. Even if this is true, it doesn't necessarily mean the economy's doing better. It's pretty likely that the job openings require skills that the newly unemployed workers do not have. Otherwise, they'd be filled. You can't hire a former real estate agent at Google and expect him to be immediately productive. (Hell, you usually can't hire a software engineer at Google and expect him to be immediately productive, but at least you can hire him.) The issue with recessions is misallocation of capital, and it can't be fixed with a magic wand. You generally have to wait for people to retrain in new, productive areas of the economy, and for capital to be liquidated and entrepreneurs to find new uses for it. That's why it takes ~4 years instead of ~4 months.

gne1963 18y ago
The samples were taken from a random selection of 100 of the top 1000 firms. The real question is how to make that selection of 100 firms so that statistically the sample size is a representative sampling. The goal is not to determine whether or not the jobs are available to those folks who may have just been laid off. The goal is to determine just how many job opening do exist in the Fortune 1000 without needing to hit every single job board of every one of those 1000 companies... surely there is a statistician who can solve that one? ;-)
- nostrademons 18y ago
  From the link on your page: "Yesterday we embarked on a research project to find current job openings. We started with 21 firms (with 18,839 open positions) and now we're up to data from 100 of them. (All Fortune 500)" The chance that a random sample of 100 firms are all in the top half of the sample universe is 2^100 = about a decillion. Doesn't really look random to me.
  - gne1963 18y ago
    Thanks, nice catch and that post needs to be changed... momentarily... The universe was truely from the Fortune 1000 from the list publish online... not the top 500 only...
    - gne1963 18y ago
      You can see my confusion is somewhat born out by hitting this link... it is labeled 500 when actually it lists the top 1000... http://money.cnn.com/magazines/fortune/fortune500/2008/full_list/ http://money.cnn.com/magazines/fortune/fortune500/2008/full_...
      - gne1963 18y ago
        So here is the methodology: 1. Hit the site above. 2. Randomly pick a company from the list... for instance: 3. Pick Paychex Systems... company 921 on Fortunes list... 4. On their site they list 137 job openings... 5. Add that to the total... 6. Repeat until you have a representative sample size Question: Using an appropriate random number generation between 1 and 1000, what is the sample size required to then estimate the total job openings on the 1000 sites?

gne1963 18y ago

So here is the methodology: 1. Hit http://money.cnn.com/magazines/fortune/fortune500/2008/full_list/ http://money.cnn.com/magazines/fortune/fortune500/2008/full_... 2. Randomly pick a company from the list... for instance: Pick Paychex Systems... company 921 on Fortunes list...On their career site they list 137 current job openings... 3. Add that to your total... 4. Repeat until you have a representative sample. 5. Extrapolate to 1000... Question: Using an appropriate random number generation between 1 and 1000, what is the sample size required to then estimate the total job openings on the 1000 sites?

NyxWulf 18y ago
Well here are a couple of things: you show 99 records on your excel spreadsheet, with an average of 709 and a standard deviation of 1073. With that large of a standard deviation and that average you can't assume a normal distribution so you need to use the T distribution. Using the T distribution at a 95% confidence level I come up with an estimated average of [495, 923] which results in a number of jobs in the top 1000 from 495,000 to 923,000 with a 95% certainty. Just for kicks I wanted to see what would happen if you sampled 500 times but had that same standard deviation. At that point your spread would be about +/- 100 for the average value of X. If you are looking for a singular number you'll need to check all 1,000 of them, otherwise it will be an estimated range.
- gne1963 18y ago
  Thanks! So what if I did another 100 random sample resulting a "better" (lower) std dev? At what level of std dev could one assume a normal distribution out to 1000? gne

anamax 18y ago

Job opening on a company's "career site" are a very poor predictor of the number of people that company would hire if could find suitable candidates. In some cases, the listing is to satisfy an HR rule and the person to be hired has already been selected. In other cases, it's "maybe someone wonderful will wander in". And then there are the hires that happen without a listing. I'd guess that the actual uncertainty in the total is more a function of the uncertainty in the numbers than it is in the sampling.

gne1963 18y ago
>I'd guess that the actual uncertainty >in the total is more a function of the >uncertainty in the numbers than it is in >the sampling. totally agree. As with any indicator there is quite a bit of uncertainty. And as you suggest "there are the hires that happen without a listing..." Most career counselors would agree that there are actually more hidden jobs than what the boards suggest... So... your argument might just boost the number if we add some multiplier into the model based on the jobs actually posted. thanks! gne1963