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I am referring to this article: http://www.nytimes.com/2011/01/11/science/11esp.html

Consider the following experiment. Suppose there was reason to believe that a coin was slightly weighted toward heads. In a test, the coin comes up heads 527 times out of 1,000.

Is this significant evidence that the coin is weighted?

Classical analysis says yes. With a fair coin, the chances of getting 527 or more heads in 1,000 flips is less than 1 in 20, or 5 percent, the conventional cutoff. To put it another way: the experiment finds evidence of a weighted coin “with 95 percent confidence.”

Yet many statisticians do not buy it. One in 20 is the probability of getting any number of heads above 526 in 1,000 throws. That is, it is the sum of the probability of flipping 527, the probability of flipping 528, 529 and so on.

But the experiment did not find all of the numbers in that range; it found just one — 527. It is thus more accurate, these experts say, to calculate the probability of getting that one number — 527 — if the coin is weighted, and compare it with the probability of getting the same number if the coin is fair.

Statisticians can show that this ratio cannot be higher than about 4 to 1, according to Paul Speckman, a statistician, who, with Jeff Rouder, a psychologist, provided the example.

First question: This is new to me. Has anybody a reference where I can find the exact calculation and/or can YOU help me by giving me the exact calculation yourself and/or can you point me to some material where I can find similar examples?

Bayes devised a way to update the probability for a hypothesis as new evidence comes in.

So in evaluating the strength of a given finding, Bayesian (pronounced BAYZ-ee-un) analysis incorporates known probabilities, if available, from outside the study.

It might be called the “Yeah, right” effect. If a study finds that kumquats reduce the risk of heart disease by 90 percent, that a treatment cures alcohol addiction in a week, that sensitive parents are twice as likely to give birth to a girl as to a boy, the Bayesian response matches that of the native skeptic: Yeah, right. The study findings are weighed against what is observable out in the world.

In at least one area of medicine — diagnostic screening tests — researchers already use known probabilities to evaluate new findings. For instance, a new lie-detection test may be 90 percent accurate, correctly flagging 9 out of 10 liars. But if it is given to a population of 100 people already known to include 10 liars, the test is a lot less impressive.

It correctly identifies 9 of the 10 liars and misses one; but it incorrectly identifies 9 of the other 90 as lying. Dividing the so-called true positives (9) by the total number of people the test flagged (18) gives an accuracy rate of 50 percent. The “false positives” and “false negatives” depend on the known rates in the population.

Second question: How do you exactly judge if a new finding is "real" or not with this method? And: Isn't this as arbitrary as the 5%-barrier because of the use of some preset prior probability?

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I will answer the first question in detail.

With a fair coin, the chances of getting 527 or more heads in 1,000 flips is less than 1 in 20, or 5 percent, the conventional cutoff.

For a fair coin the number of heads in 1000 trials follows the binomial distribution with number of trials $n=1000$ and probability $p=1/2$. The probability of getting more than 527 heads is then

$$P(B(1000,1/2)>=527)$$

This can be calculated with any statistical software package. R gives us

> pbinom(526,1000,1/2,lower.tail=FALSE)
   0.04684365

So the probability that with fair coin we will get more than 526 heads is approximately 0.047, which is close to 5% cuttoff mentioned in the article.

The following statement

To put it another way: the experiment finds evidence of a weighted coin “with 95 percent confidence.”

is debatable. I would be reluctant to say it, since 95% confidence can be interpreted in several ways.

Next we turn to

But the experiment did not find all of the numbers in that range; it found just one — 527. It is thus more accurate, these experts say, to calculate the probability of getting that one number — 527 — if the coin is weighted, and compare it with the probability of getting the same number if the coin is fair.

Here we compare two events $B(1000,1/2)=527$ -- fair coin, and $B(1000,p)=527$ -- weighted coin. Substituting the formulas for probabilities of these events and noting that the binomial coefficient cancels out we get

$$\frac{P(B(1000,p)=527)}{P(B(1000,1/2)=527)}=\frac{p^{527}(1-p)^{473}}{(1/2)^{1000}}.$$

This is a function of $p$, thus we cand find minima or maxima of it. From the article we may infer that we need maxima:

Statisticians can show that this ratio cannot be higher than about 4 to 1, according to Paul Speckman, a statistician, who, with Jeff Rouder, a psychologist, provided the example.

To make maximisation easier take logarithm of ratio, calculate the derivative with respect to $p$ and equate it to zero. The solution will be

$$p=\frac{527}{1000}.$$

We can check that it is really a maximum using second derivative test for example. Substituting it to the formula we get

$$\frac{(527/1000)^{527}(473/1000)^{473}}{(1/2)^{1000}}\approx 4.3$$

So the ratio is 4.3 to 1, which agrees with the article.

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  • $\begingroup$ "Now maximise this quantity with respect to p": i think you mean minimise. $\endgroup$ – Simon Byrne Jan 17 '11 at 13:52
  • $\begingroup$ @mpiktas (+1) Nice (updated) answer. $\endgroup$ – chl Jan 17 '11 at 22:01
  • $\begingroup$ I think this example shows you exactly what a confidence interval is. I find it easiest to interpret a CI as ONE observation from a Bernouli distributed random variable with a probability parameter equal to the level of confidence. It only makes sense to me to use CIs if you are doing the experiment repetitively. Another issue is that what is the alternative hypothesis? is it p=7/10, p>0.5, p=1050/2000? p=527/1000? Another issue is what do we mean by p=$\frac{1}{2}$? is it EXACTLY $\frac{1}{2}$ or is it $p \in \left(\frac{1}{2} \pm \epsilon \right)$ where $\epsilon$ is a small number. $\endgroup$ – probabilityislogic Jan 18 '11 at 11:42
  • $\begingroup$ @Simon, why is the correction to minimize? Doesn't value of P found maximize the ratio? $\endgroup$ – user3115 Feb 8 '11 at 21:28
  • $\begingroup$ @statnovice: The original version of the answer had the numerator and denominator switched. $\endgroup$ – Simon Byrne Feb 9 '11 at 16:27

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