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vendredi 31 janvier 2014

What types of health information are consumers compelled to divulge under Affordable Care Act rules?

I just selected a new health insurance plan and they’re asking me questions about my health. I thought insurance companies can’t do this anymore?

Under health care reform laws, insurance companies can’t require you to answer health questions in order to buy insurance.

However, once you are covered by a health plan, companies are allowed to ask questions to determine whether you qualify for one of their disease management programs or for case management services. Disease management programs help consumers learn how to manage chronic health conditions such as diabetes, heart disease, or depression.

Case management programs help consumers who have very serious health conditions, such as leukemia, cut through the red tape to get the insurance company to pay for their treatment. For example, to prevent misdiagnosis, some insurance companies won’t start paying for cancer treatment until consumers get a second opinion to confirm the diagnosis. The average consumer wouldn’t know this, but a case manager will tell the consumer about this requirement and, if necessary, help the consumer schedule an appointment for a second opinion. Many case managers are also licensed nurses, so they can also suggest solutions to problems, such as side effects, that consumers experience during treatment.

Both of these services are voluntary, so you don’t have to answer the questions. Even if you do answer the questions, you don’t have to participate in the programs.

I recently applied for life insurance and they made me answer questions about my health. I thought insurance companies can’t do this anymore?

Under health care reform laws, insurance companies can’t require you answer health questions in order to buy medical insurance. However, insurance companies can still require to you answer health questions to buy other types of insurance, including:

  • Life insurance 
  • Long-term care insurance 
  • Dental insurance
  • Vision insurance 
  • Disability insurance
  • Medicare Supplement plans and Medicare Advantage plans, under certain circumstances

Read more information about health care reform.

Our consumer experts can answer your questions about any type of insurance. They are available by phone at 1-800-562-6900 or by submitting an inquiry through our website.

jeudi 30 janvier 2014

Kreidler: Nearly 14.5 percent of Washingtonians were uninsured in 2012

Today, we issued our fourth report on the number of Washingtonians who have no health insurance. At the end of 2012, some 990,000 people -- approximately 14.5 percent of the state's population -- were uninsured.

From 2010 through 2012:
  • The number of uninsured people in Washington grew by more than 44,000.
  • Four out of five people with individual insurance were underinsured, meaning they had plans that only paid for 25-40 percent of their medical costs.
  • Employer-sponsored coverage grew increasingly scarce.
  • Uncompensated care ballooned to nearly $1 billion per year.
The Affordable Care Act fully took effect on Jan. 1 and the uninsured rate is expected to drop to 6 percent by 2016. Early provisions of the Affordable Care Act prevented an estimated 100,000 people from joining the ranks of the uninsured prior to 2014.

�For many families who have struggled to get or keep health coverage, health reform couldn�t come soon enough,� said Insurance Commissioner Mike Kreidler. �Regardless of how you feel about �Obamacare,� it�s hard to argue that we�re not making progress in stopping the growth of uninsured or that the status quo was sustainable. Before health reform, we had hundreds of thousands of people living one bad diagnosis away from bankruptcy.�


Obama's Fact Check Highlights Misstatements in Speech



I've stumble upon an interesting article by Calvin Woodward AP, talking about misstatements made by Barrack Obama on his State of Union address. He gave his State of the Union address on Capitol Hill in Washington,last Tuesday January 28, 2014.

My take about the article is I don't think retiring baby boomers is that significant and it is not the same as unemployed. In reality, most of Americans wants to find full time job but can't, because of the continues exporting of jobs to other countries. 

check out the article after the jump!




WASHINGTON (AP) � It seems to be something of an occupational hazard for President Barack Obama: When he talks about his health care law, he's bound to hit a fact bump sooner or later.


So it went Tuesday night, when he declared Medicare premiums have stayed flat thanks to the law, when they've gone up. As for an even bigger theme of his State of the Union address, the president's assertion that "upward mobility has stalled" in America runs contrary to recent research, while other findings support him.

A look at some of the facts and political circumstances behind his claims, along with a glance at the Republican response to his speech:

OBAMA: "Because of this (health care) law, no American can ever again be dropped or denied coverage for a preexisting condition like asthma, back pain or cancer. No woman can ever be charged more just because she's a woman. And we did all this while adding years to Medicare's finances, keeping Medicare premiums flat, and lowering prescription costs for millions of seniors."

THE FACTS: He's right that insurers can no longer turn people down because of medical problems, and they can't charge higher premiums to women because of their sex. The law also lowered costs for seniors with high prescription drug bills. But Medicare's monthly premium for outpatient care has gone up in recent years.

Although the basic premium remained the same this year at $104.90, it increased by $5 a month in 2013, up from $99.90 in 2012. Obama's health care law also raised Medicare premiums for upper-income beneficiaries, and both the president and Republicans have proposed to expand that.

Finally, the degree to which the health care law improved Medicare finances is hotly debated. On paper, the program's giant trust fund for inpatient care gained more than a decade of solvency because of cuts to service providers required under the health law. But in practice those savings cannot simultaneously be used to expand coverage for the uninsured and shore up Medicare.

___

OBAMA: "Today, after four years of economic growth, corporate profits and stock prices have rarely been higher, and those at the top have never done better. But average wages have barely budged. Inequality has deepened. Upward mobility has stalled."

THE FACTS: The most recent evidence suggests that mobility hasn't worsened. A team of economists led by Harvard's Raj Chetty released a study last week that found the United States isn't any less socially mobile than it was in the 1970s. Looking at children born between 1971 and 1993, the economists found that the odds of a child born in the poorest 20 percent of families making it into the top 20 percent hasn't changed.

"We find that children entering the labor market today have the same chances of moving up in the income distribution (relative to their parents) as children born in the 1970s," the authors said.

Still, other research has found that the United States isn't as mobile a society as most Americans would like to believe. In a study of 22 countries, economist Miles Corak of the University of Ottawa found that the United States ranked 15th in social mobility. Only Italy and Britain among wealthy countries ranked lower. By some measures, children in the United States are as likely to inherit their parents' economic status as their height.

___

OBAMA: "We'll need Congress to protect more than 3 million jobs by finishing transportation and waterways bills this summer.  But I will act on my own to slash bureaucracy and streamline the permitting process for key projects, so we can get more construction workers on the job as fast as possible."

THE FACTS: Cutting rules and regulations doesn't address what's holding up most transportation projects, which is lack of money. The federal Highway Trust Fund will run out of money in August without action. To finance infrastructure projects, Obama wants Congress to raise taxes on businesses that keep profits or jobs overseas, but that idea has been a political nonstarter.

The number of projects affected by the administration's efforts to cut red tape is relatively small, said Joshua Schank, president and CEO of the Eno Center for Transportation, a think tank. "The reason most of these projects are delayed is they don't have enough money. So it's great that you are expediting the review process, but the review process isn't the problem. The problem is we don't have enough money to invest in our infrastructure in the first place."

___

OBAMA: "More than 9 million Americans have signed up for private health insurance or Medicaid coverage."

THE FACTS: That's not to say 9 million more Americans have gained insurance under the law.

The administration says about 6 million people have been determined to be eligible for Medicaid since Oct. 1 and an additional 3 million roughly have signed up for private health insurance through the new markets created by the health care law. That's where Obama's number of 9 million comes from. But it's unclear how many in the Medicaid group were already eligible for the program or renewing existing coverage.

Likewise, it's not known how many of those who signed up for private coverage were previously insured. A large survey released last week suggests the numbers of uninsured gaining coverage may be smaller. The Gallup-Healthways Well-Being Index found that the uninsured rate for U.S. adults dropped by 1.2 percentage points in January, to 16.1 percent. That would translate to roughly 2 million to 3 million newly insured people since the law's coverage expansion started Jan. 1.

___

OBAMA: "In the coming weeks, I will issue an executive order requiring federal contractors to pay their federally funded employees a fair wage of at least $10.10 an hour, because if you cook our troops' meals or wash their dishes, you shouldn't have to live in poverty."

THE FACTS: This would be a hefty boost in the federal minimum wage, now $7.25, but not many would see it.

Most employees of federal contractors already earn more than $10.10. About 10 percent of those workers, roughly 200,000, might be covered by the higher minimum wage. But there are several wrinkles. The increase would not take effect until 2015 at the earliest and it doesn't apply to existing federal contracts, only new ones. Renewed contracts also will be exempt from Obama's order unless other terms of the agreement change, such as the type of work or number of employees needed.

Obama also said he'll press Congress to raise the federal minimum wage overall. He tried that last year, seeking a $9 minimum, but Congress didn't act.

___

REP. CATHY McMORRIS RODGERS of Washington, in her prepared Republican response: "Last month, more Americans stopped looking for a job than found one. Too many people are falling further and further behind because, right now, the president's policies are making people's lives harder."

THE FACTS: She leaves out a significant factor in the high number of people who aren't looking for jobs: Baby boomers are retiring.

It's true that a large part of the still-high unemployment rate is due to weak job creation in a slowly recovering economy. There are roughly three people seeking every job opening. But one big reason people aren't seeking employment is that there are so many boomers � the generation born in the immediate aftermath of World War II � and therefore more than the usual number of retirements.

sources: http://news.yahoo.com/fact-check-obama-medicare-premiums-035611405--politics.html

mardi 28 janvier 2014

Binomial testing with buttered toast

Rasmus' post of last week on binomial testing made me think about p-values and testing again. In my head I was tossing coins, thinking about gender diversity and toast. The toast and tossing a buttered toast in particular was the most helpful thought experiment, as I didn't have a fixed opinion on the probabilities for a toast to land on either side. I have yet to carry out some real experiments.

Suppose I tossed 6 buttered toasts and observed that all but one toast landed on their buttered side.

Now I have two questions:
  1. Would it be reasonable to assume that there is a 50/50 chance for a toast to land on either side?
  2. Which probability should I assume?
Suppose the toast is fair, so that the probability for landing on the buttered (B) or unbuttered (U) side is 50%.

Then, the probability of observing one ore more B (right tail event) is:
(gt <- 1-1/2^6)
# [1] 0.984375
and the probability of observing one or fewer B (left tail event) is:
(lt <- 1/2^6*choose(6,1) + 1/2^6)
# [1] 0.109375
while the probability of either extreme event, one or fewer B (or U), is:
2*min(c(gt, lt))
# [1] 0.21875
In summary, if the toast has an equally probability to fall on either side, then there is 22% chance to observe one or fewer B (or U) in 6 tosses. That's not that unlikely and hence I would not dismiss the hypothesis that the toast is fair, despite the fact that the sample frequency is only 1/6.

Actually, the probabilities I calculated above are exactly the p-values I get from the classical binomal tests:
## Right tail event
binom.test(1, 6, alternative="greater")
## Left tail event
binom.test(1, 6, alternative="less")
## Double tail event
binom.test(1, 6, alternative="two.sided")
Additionally I can read from the tests that my assumption of a 50% probability is on the higher end of the 95 percent confidence interval. Thus, wouldn't it make sense to update my belief about the toast following my observations? In particular, as unlike with a coin, I am far less convinced that a 50/50 probability is a good assumption to start with. Arguably the toast is biased by the butter.

Here the concept of a conjugate prior becomes handy again. The idea is to assume that the parameter \(\theta\) of the binomial distribution is a random variable itself. Suppose I have no prior knowledge about the true probability of the toast falling on either side, then a flat prior, such as the Uniform distribution would be reasonable. However, the beta distribution with parameter \(\alpha=1\) and \(\beta=1\) has the same property and is a conjugate to the binomial distribution with parameter \(\theta\). That means there is an analytical solution, in this case the posterior distribution is beta-binomial with hyperparaemters:
\[\alpha':=\alpha + \sum_{i=1}^n x_i,\; \beta':=\beta + n - \sum_{i=1}^n x_i,\]
and the posterior predictor for one trial is given as
\[\frac{\alpha'}{\alpha' + \beta'}\]
so in my case:
alpha <- 1; beta <- 1; n <- 6; success <- 1
alpha1 <- alpha + success
beta1 <- beta + n - success
(theta <- alpha1 / ( alpha1 + beta1))
# [1] 0.25
My updated believe about the toast landing on the unbuttered side is a probability of 25%. That's lower than my prior of 50% but still higher than the sample frequency of 1/6. If I would have more toasts I could run more experiments and update my posterior predictor.

I get the same answer from Rasmus' bayes.binom.test function:
> library(BayesianFirstAid)
> bayes.binom.test(1, 6)

Bayesian first aid binomial test

data: 1 and 6
number of successes = 1, number of trials = 6
Estimated relative frequency of success:
0.25
95% credible interval:
0.014 0.527
The relative frequency of success is more than 0.5 by a probability of 0.061
and less than 0.5 by a probability of 0.939
Of course I could change my view on the prior and come to a different conclusion. I could follow the Wikipedia article on buttered toast and believe that the chance of the toast landing on the buttered side is 62%. I further have to express my uncertainty, say a standard deviation of 10%, that is a variance of 1%. With that information I can update my belief of the toast landing on the unbuttered side following my observations (and transforming the variables):
x <- 0.38
v <- 0.01
alpha <- x*(x*(1-x)/v-1)
beta <- (1-x)*(x*(1-x)/v-1)
alpha1 <- alpha + success
beta1 <- beta + n - success
(theta <- alpha1 / ( alpha1 + beta1))
# [1] 0.3351821
I would conclude that for my toasts / tossing technique the portability is 34% to land on the unbuttered side.

In summary, although their is no sufficient evidence to reject the hypothesis that the 'true' probability is not 50% (at the typical 5% level), I would work with 34% until I have more data. Toast and butter please!

Session Info

R version 3.0.1 (2013-05-16)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets methods
[7] base

other attached packages:
[1] BayesianFirstAid_0.1 rjags_3-12 coda_0.16-1
[4] lattice_0.20-24

loaded via a namespace (and not attached):
[1] grid_3.0.1 tools_3.0.1

Binomial testing with buttered toast

Rasmus' post of last week on binomial testing made me think about p-values and testing again. In my head I was tossing coins, thinking about gender diversity and toast. The toast and tossing a buttered toast in particular was the most helpful thought experiment, as I didn't have a fixed opinion on the probabilities for a toast to land on either side. I have yet to carry out some real experiments.

Suppose I tossed 6 buttered toasts and observed that all but one toast landed on their buttered side.

Now I have two questions:
  1. Would it be reasonable to assume that there is a 50/50 chance for a toast to land on either side?
  2. Which probability should I assume?
Suppose the toast is fair, so that the probability for landing on the buttered (B) or unbuttered (U) side is 50%.

Then, the probability of observing one ore more B (right tail event) is:
(gt <- 1-1/2^6)
# [1] 0.984375
and the probability of observing one or fewer B (left tail event) is:
(lt <- 1/2^6*choose(6,1) + 1/2^6)
# [1] 0.109375
while the probability of either extreme event, one or fewer B (or U), is:
2*min(c(gt, lt))
# [1] 0.21875
In summary, if the toast has an equally probability to fall on either side, then there is 22% chance to observe one or fewer B (or U) in 6 tosses. That's not that unlikely and hence I would not dismiss the hypothesis that the toast is fair, despite the fact that the sample frequency is only 1/6.

Actually, the probabilities I calculated above are exactly the p-values I get from the classical binomal tests:
## Right tail event
binom.test(1, 6, alternative="greater")
## Left tail event
binom.test(1, 6, alternative="less")
## Double tail event
binom.test(1, 6, alternative="two.sided")
Additionally I can read from the tests that my assumption of a 50% probability is on the higher end of the 95 percent confidence interval. Thus, wouldn't it make sense to update my belief about the toast following my observations? In particular, as unlike with a coin, I am far less convinced that a 50/50 probability is a good assumption to start with. Arguably the toast is biased by the butter.

Here the concept of a conjugate prior becomes handy again. The idea is to assume that the parameter \(\theta\) of the binomial distribution is a random variable itself. Suppose I have no prior knowledge about the true probability of the toast falling on either side, then a flat prior, such as the Uniform distribution would be reasonable. However, the beta distribution with parameter \(\alpha=1\) and \(\beta=1\) has the same property and is a conjugate to the binomial distribution with parameter \(\theta\). That means there is an analytical solution, in this case the posterior distribution is beta-binomial with hyperparaemters:
\[\alpha':=\alpha + \sum_{i=1}^n x_i,\; \beta':=\beta + n - \sum_{i=1}^n x_i,\]
and the posterior predictor for one trial is given as
\[\frac{\alpha'}{\alpha' + \beta'}\]
so in my case:
alpha <- 1; beta <- 1; n <- 6; success <- 1
alpha1 <- alpha + success
beta1 <- beta + n - success
(theta <- alpha1 / ( alpha1 + beta1))
# [1] 0.25
My updated believe about the toast landing on the unbuttered side is a probability of 25%. That's lower than my prior of 50% but still higher than the sample frequency of 1/6. If I would have more toasts I could run more experiments and update my posterior predictor.

I get the same answer from Rasmus' bayes.binom.test function:
> library(BayesianFirstAid)
> bayes.binom.test(1, 6)

Bayesian first aid binomial test

data: 1 and 6
number of successes = 1, number of trials = 6
Estimated relative frequency of success:
0.25
95% credible interval:
0.014 0.527
The relative frequency of success is more than 0.5 by a probability of 0.061
and less than 0.5 by a probability of 0.939
Of course I could change my view on the prior and come to a different conclusion. I could follow the Wikipedia article on buttered toast and believe that the chance of the toast landing on the buttered side is 62%. I further have to express my uncertainty, say a standard deviation of 10%, that is a variance of 1%. With that information I can update my belief of the toast landing on the unbuttered side following my observations (and transforming the variables):
x <- 0.38
v <- 0.01
alpha <- x*(x*(1-x)/v-1)
beta <- (1-x)*(x*(1-x)/v-1)
alpha1 <- alpha + success
beta1 <- beta + n - success
(theta <- alpha1 / ( alpha1 + beta1))
# [1] 0.3351821
I would conclude that for my toasts / tossing technique the portability is 34% to land on the unbuttered side.

In summary, although their is no sufficient evidence to reject the hypothesis that the 'true' probability is not 50% (at the typical 5% level), I would work with 34% until I have more data. Toast and butter please!

Session Info

R version 3.0.1 (2013-05-16)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets methods
[7] base

other attached packages:
[1] BayesianFirstAid_0.1 rjags_3-12 coda_0.16-1
[4] lattice_0.20-24

loaded via a namespace (and not attached):
[1] grid_3.0.1 tools_3.0.1

mardi 21 janvier 2014

Fun with the Raspberry Pi

Since Christmas I have been playing around with a Raspberry Pi. It is certainly not the fastest computer, but what a great little toy! Here are a few experiences and online resources that I found helpful.

Setup

Initially I connected the Raspberry Pi via HDMI to a TV; together with keyboard, mouse and an old USB Wifi adapter. Everything worked out of the box and I could install Raspbian and set up the network.

HDMI to VGA

Using an old VGA computer monitor via an adapter required changes to the file /boot/config.txt. You can find the parameters that match your monitor on Raspberry Pi StackExchange. In my case I had to set:
hdmi_group=2
hdmi_mode=35 # 1280x1024 @ 60Hz

Remote Access

But who needs a monitor when you can access the Pi remotely anyway? The command ifconfig tells me the local IP address of the Raspberry Pi.

With XQuartz on my Mac running I can connect to the Raspberry Pi via ssh with the X session forwarded:
ssh -X pi@your.ip.address.here
However, the performance is a bit sluggish and online comments suggest to use VNC instead. Nothing easier than that, install the VNC server on the Pi and use Screen Sharing on your Mac to access the Pi. Mitch Malone has a great post on this subject. Following the VNC setup on the Raspberry Pi I can type:
vnc://pi@your.ip.address.here:5901
into Safari and that will bring up the Screen Sharing App; see screen shot below.

AirPrint and AirPlay

Ok, let's give the Pi something to do: Rohan Kapoor explains how to set up the Raspberry Pi as a print server with AirPrint.

How about AirPlay as well? Follow Thorin Klosowski's steps on Lifehacker and you can stream music from your iOS devices to the Raspberry Pi's audio out.

Mathematica and R



Just before Christmas Stephen Wolfram announced that Mathematica would be made freely available for the Raspberry Pi. I had used Mathematica a little at university and was curious to see it on the Pi. Alex Newman posted the installation instructions on the Wolfram community site. It is as simple as:
sudo apt-get update && sudo apt-get install wolfram-engine
As a little toy example I run Paul Nylander's Mathematica code for a Feigenbaum diagram:

Surprisingly, it took about 3.5 minutes to run. Curious to find out if my old R routine would run faster I installed R on my Pi:
sudo apt-get install r-base
and adapted the code to match the parameters of the Mathematica routine (well, so I think):

The R code finished after about 10 seconds. Surely, there must be ways to speed up the Mathematica code that I have to investigate.

Fun with the Raspberry Pi

Since Christmas I have been playing around with a Raspberry Pi. It is certainly not the fastest computer, but what a great little toy! Here are a few experiences and online resources that I found helpful.

Setup

Initially I connected the Raspberry Pi via HDMI to a TV; together with keyboard, mouse and an old USB Wifi adapter. Everything worked out of the box and I could install Raspbian and set up the network.

HDMI to VGA

Using an old VGA computer monitor via an adapter required changes to the file /boot/config.txt. You can find the parameters that match your monitor on Raspberry Pi StackExchange. In my case I had to set:
hdmi_group=2
hdmi_mode=35 # 1280x1024 @ 60Hz

Remote Access

But who needs a monitor when you can access the Pi remotely anyway? The command ifconfig tells me the local IP address of the Raspberry Pi.

With XQuartz on my Mac running I can connect to the Raspberry Pi via ssh with the X session forwarded:
ssh -X pi@your.ip.address.here
However, the performance is a bit sluggish and online comments suggest to use VNC instead. Nothing easier than that, install the VNC server on the Pi and use Screen Sharing on your Mac to access the Pi. Mitch Malone has a great post on this subject. Following the VNC setup on the Raspberry Pi I can type:
vnc://pi@your.ip.address.here:5901
into Safari and that will bring up the Screen Sharing App; see screen shot below.

AirPrint and AirPlay

Ok, let's give the Pi something to do: Rohan Kapoor explains how to set up the Raspberry Pi as a print server with AirPrint.

How about AirPlay as well? Follow Thorin Klosowski's steps on Lifehacker and you can stream music from your iOS devices to the Raspberry Pi's audio out.

Mathematica and R



Just before Christmas Stephen Wolfram announced that Mathematica would be made freely available for the Raspberry Pi. I had used Mathematica a little at university and was curious to see it on the Pi. Alex Newman posted the installation instructions on the Wolfram community site. It is as simple as:
sudo apt-get update && sudo apt-get install wolfram-engine
As a little toy example I run Paul Nylander's Mathematica code for a Feigenbaum diagram:

Surprisingly, it took about 3.5 minutes to run. Curious to find out if my old R routine would run faster I installed R on my Pi:
sudo apt-get install r-base
and adapted the code to match the parameters of the Mathematica routine (well, so I think):

The R code finished after about 10 seconds. Surely, there must be ways to speed up the Mathematica code that I have to investigate.

samedi 18 janvier 2014

Jimmy Kimmel Take on ObamaCare (Affordable Care Act)



Jimmy Kimmel live opening monologue is about ObamaCare, he ridiculed the young people who are supporting the Law since they will be the one to be disadvantaged. Check out the video clip above.

lundi 13 janvier 2014

How many more R-bloggers posts can I expect?

I noticed that the monthly number of posts on R-bloggers stopped increasing over the last year. Indeed, the last couple of months saw a decline in posts compared to the previous year. Thus, has most been said and written about R already?


Who knows? Well, I took a stab at looking into the future. However, I can tell you already that I am not convinced by my predictions. But maybe someone else will be inspired to take this work forward.

First, I have to get the data - that's easy, I can scrape the monthly post counts from the R-bloggers homepage.


Looking at the incremental and cumulative plots, and believing that eventually the number of R posts will decrease, I thought that a logistic growth function would provide a nice fit to the data and also give an asymptotic view of the total number of posts on R-bloggers.

Although the fit, see below, looks reasonable at first glance, I don't believe it provides a sensible prediction of the future. The model would forecast only another 1,269 post by the end of 2016 with not much more to expect after that. Indeed the asymptotic total number of posts K is only 14,396. I don't believe this can be right, not even as a proxy, when the current count of monthly posts is well above 100.



I played around with data and the logistic growth function a little further, using annual instead of monthly data, changing the time horizon and fixing K, yet without much success.

Eventually I recalled a talk by Rob Hyndman's about his forecast package. After all, I have a time series here. So, applying the forecast function to the incremental data provides a somewhat more realistic prediction of 2,695 posts for the next 12 months, but with an increasing trend in monthly posts for 2014, which I find hard to believe given the observations over the last year.



Well, I presented two models here: One predicts a rapid decline in monthly posts on R-bloggers, while the other forecasts an increase. Neither feels right to me. Of course time will tell, but have you got any ideas or views?

Session Info

R version 3.0.2 (2013-09-25)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets methods base

other attached packages:
[1] forecast_4.8 xts_0.9-7 zoo_1.7-10 XML_3.95-0.2

loaded via a namespace (and not attached):
[1] colorspace_1.2-4 fracdiff_1.4-2
[3] grid_3.0.2 lattice_0.20-23
[5] nnet_7.3-7 parallel_3.0.2
[7] quadprog_1.5-5 Rcpp_0.10.6
[9] RcppArmadillo_0.3.920.1 tools_3.0.2
[11] tseries_0.10-32

How many more R-bloggers posts can I expect?

I noticed that the monthly number of posts on R-bloggers stopped increasing over the last year. Indeed, the last couple of months saw a decline in posts compared to the previous year. Thus, has most been said and written about R already?


Who knows? Well, I took a stab at looking into the future. However, I can tell you already that I am not convinced by my predictions. But maybe someone else will be inspired to take this work forward.

First, I have to get the data - that's easy, I can scrape the monthly post counts from the R-bloggers homepage.


Looking at the incremental and cumulative plots, and believing that eventually the number of R posts will decrease, I thought that a logistic growth function would provide a nice fit to the data and also give an asymptotic view of the total number of posts on R-bloggers.

Although the fit, see below, looks reasonable at first glance, I don't believe it provides a sensible prediction of the future. The model would forecast only another 1,269 post by the end of 2016 with not much more to expect after that. Indeed the asymptotic total number of posts K is only 14,396. I don't believe this can be right, not even as a proxy, when the current count of monthly posts is well above 100.



I played around with data and the logistic growth function a little further, using annual instead of monthly data, changing the time horizon and fixing K, yet without much success.

Eventually I recalled a talk by Rob Hyndman's about his forecast package. After all, I have a time series here. So, applying the forecast function to the incremental data provides a somewhat more realistic prediction of 2,695 posts for the next 12 months, but with an increasing trend in monthly posts for 2014, which I find hard to believe given the observations over the last year.



Well, I presented two models here: One predicts a rapid decline in monthly posts on R-bloggers, while the other forecasts an increase. Neither feels right to me. Of course time will tell, but have you got any ideas or views?

Session Info

R version 3.0.2 (2013-09-25)
Platform: x86_64-apple-darwin10.8.0 (64-bit)

locale:
[1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8

attached base packages:
[1] stats graphics grDevices utils datasets methods base

other attached packages:
[1] forecast_4.8 xts_0.9-7 zoo_1.7-10 XML_3.95-0.2

loaded via a namespace (and not attached):
[1] colorspace_1.2-4 fracdiff_1.4-2
[3] grid_3.0.2 lattice_0.20-23
[5] nnet_7.3-7 parallel_3.0.2
[7] quadprog_1.5-5 Rcpp_0.10.6
[9] RcppArmadillo_0.3.920.1 tools_3.0.2
[11] tseries_0.10-32

samedi 11 janvier 2014

Accenture is now the lead contractor for Obamacare website HealthCare.gov

Accenture, Obamacare website HealthCare.gov, Obamacare, Affordable Care Act, Medicare

Reuters recently reported that the Obama government awarded Accenture as the lead contractor for the Obamacare enrollment website. They will be replacing CGI Federal, which encountered  a lot of failures and bugs when they launched the website HealthCare.gov in October for millions of Americans shopping for health insurance. This was announced by the U.S. Centers for Medicare and Medicaid Services on Saturday.

When CGI build the website HealthCare.gov it was riddled with errors, slow speeds, and glitches that gave President Barack Obama a lot of problems and gave Republican another reason to repeal Obamacare.

CGI Federal announced last Friday that its contract that was awarded in 2011, will end on February 28, and would not be renewed.

Accenture said the contract was worth $45 million for the initial phase of the project, and the final value of the one-year contract would be about $90 million.

March 31 is the deadline for signing up for a 2014 health insurance under the Affordable Care Act, Accenture will need to be ready and take on the surge of last-minute sign-ups.

mardi 7 janvier 2014

Whale charts - Visualising customer profitability

The Christmas and New Year's break is over, yet there is still time to return unwanted presents. Return to Santa was the title of an article in the Economist that highlighted the impact on online retailers, as return rates can be alarmingly high.

The article quotes a study by Christian Schulze of the Frankfurt School of Finance and Management, which analyses the return habits of customers who bought at least five items over a five year period from a large European online retailer. Although only a few figures are cited I will attempt to create a little model that replicates the customer behaviour and visualises the impact on overall profitability.

The study found that 5% of customers sent back more than 80% of the items they had bought; and that 1% of customers sent back at least 90% of their purchases. Or in other words 95% of customers send back less than 80% and 99% of customers send back less than 90%. To model this behaviour an S-shape curve seems appropriate, such as the logistic curve, as no-one can return more than they bought or less than nothing. With location and scale parameters m and s the logistic function can be fitted to the data, see the R code below.


The return rates do look quite high. However, if the products were shoes rather than books then I find them believable.

Additionally the article cites studies that suggest handling each returned item costs online sellers between $6 and $18, not to mention losses from items that are returned in unsaleable condition. Furthermore, without the cost of returns, online retailer's profits would be almost 50% higher.

Thus, to spin my toy model further, I assume 100 customers with revenues following an exponential distribution (?=1/250), the cost ratio of sold goods to be lognormal (?=-0.1, ?=0.1) and the cost of returns to follow a normal distribution with mean of $12 and standard deviation of $6.

In my simulation I could have made a profit of $1,979 instead of $1,441. Clearly the customers who return many items cause a real dent to my bottom line.

This situation is best visualised in what is often called a Whale Chart. Here I plot the cumulative profit against customers, with the most profitable customer on the left and the least profitable customer on the right. This chart shows me how much profit the first x number of customers generated. Often this graphs looks like a whale coming out of the water - hence its name.


In my little toy simulation I note that the first 20 most profitable customers would have generated more profit than the revenue of all customers. Indeed, profitability could have been 37% higher if it wasn't for loss making customers.

So, what shall I do? Manage my customers, know who I should reward and keep and whose loss wouldn't hurt at all. More customers are not the answer. I need more customers who return less.

R Code

Read more �

Whale charts - Visualising customer profitability

The Christmas and New Year's break is over, yet there is still time to return unwanted presents. Return to Santa was the title of an article in the Economist that highlighted the impact on online retailers, as return rates can be alarmingly high.

The article quotes a study by Christian Schulze of the Frankfurt School of Finance and Management, which analyses the return habits of customers who bought at least five items over a five year period from a large European online retailer. Although only a few figures are cited I will attempt to create a little model that replicates the customer behaviour and visualises the impact on overall profitability.

The study found that 5% of customers sent back more than 80% of the items they had bought; and that 1% of customers sent back at least 90% of their purchases. Or in other words 95% of customers send back less than 80% and 99% of customers send back less than 90%. To model this behaviour an S-shape curve seems appropriate, such as the logistic curve, as no-one can return more than they bought or less than nothing. With location and scale parameters m and s the logistic function can be fitted to the data, see the R code below.


The return rates do look quite high. However, if the products were shoes rather than books then I find them believable.

Additionally the article cites studies that suggest handling each returned item costs online sellers between $6 and $18, not to mention losses from items that are returned in unsaleable condition. Furthermore, without the cost of returns, online retailer's profits would be almost 50% higher.

Thus, to spin my toy model further, I assume 100 customers with revenues following an exponential distribution (?=1/250), the cost ratio of sold goods to be lognormal (?=-0.1, ?=0.1) and the cost of returns to follow a normal distribution with mean of $12 and standard deviation of $6.

In my simulation I could have made a profit of $1,979 instead of $1,441. Clearly the customers who return many items cause a real dent to my bottom line.

This situation is best visualised in what is often called a Whale Chart. Here I plot the cumulative profit against customers, with the most profitable customer on the left and the least profitable customer on the right. This chart shows me how much profit the first x number of customers generated. Often this graphs looks like a whale coming out of the water - hence its name.


In my little toy simulation I note that the first 20 most profitable customers would have generated more profit than the revenue of all customers. Indeed, profitability could have been 37% higher if it wasn't for loss making customers.

So, what shall I do? Manage my customers, know who I should reward and keep and whose loss wouldn't hurt at all. More customers are not the answer. I need more customers who return less.

R Code

Read more �