# College Math Teaching

## April 5, 2019

Let’s start with an example from sports: basketball free throws. At a certain times in a game, a player is awarded a free throw, where the player stands 15 feet away from the basket and is allowed to shoot to make a basket, which is worth 1 point. In the NBA, a player will take 2 or 3 shots; the rules are slightly different for college basketball.

Each player will have a “free throw percentage” which is the number of made shots divided by the number of attempts. For NBA players, the league average is .672 with a variance of .0074.

Now suppose you want to determine how well a player will do, given, say, a sample of the player’s data? Under classical (aka “frequentist” ) statistics, one looks at how well the player has done, calculates the percentage ( $p$) and then determines a confidence interval for said $p$: using the normal approximation to the binomial distribution, this works out to $\hat{p} \pm z_{\frac{\alpha}{2}} \sqrt{n}\sqrt{p(1-p)}$

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Yes, I know..for someone who has played a long time, one has career statistics ..so imagine one is trying to extrapolate for a new player with limited data.

That seems straightforward enough. But what if one samples the player’s shooting during an unusually good or unusually bad streak? Example: former NBA star Larry Bird once made 71 straight free throws…if that were the sample, $\hat{p} = 1$ with variance zero! Needless to say that trend is highly unlikely to continue.

Classical frequentist statistics doesn’t offer a way out but Bayesian Statistics does.

This is a good introduction:

But here is a simple, “rough and ready” introduction. Bayesian statistics uses not only the observed sample, but a proposed distribution for the parameter of interest (in this case, p, the probability of making a free throw). The proposed distribution is called a prior distribution or just prior. That is often labeled $g(p)$

Since we are dealing with what amounts to 71 Bernoulli trials where p = .672 so the distribution of each random variable describing the outcome of each individual shot has probability mass fuction $p^{y_i}(1-p)^{1-y_i}$ where $y_i = 1$ for a make and $y_i = 0$ for a miss.

Our goal is to calculate what is known as a posterior distribution (or just posterior) which describes $g$ after updating with the data; we’ll call that $g^*(p)$.

How we go about it: use the principles of joint distributions, likelihood functions and marginal distributions to calculate $g^*(p|y_1, y_2...,y_n) = \frac{L(y_1, y_2, ..y_n|p)g(p)}{\int^{\infty}_{-\infty}L(y_1, y_2, ..y_n|p)g(p)dp}$

The denominator “integrates out” p to turn that into a marginal; remember that the $y_i$ are set to the observed values. In our case, all are 1 with $n = 71$.

What works well is to use the beta distribution for the prior. Note: the pdf is $\frac{\Gamma (a+b)}{\Gamma(a) \Gamma(b)} x^{a-1}(1-x)^{b-1}$ and if one uses $p = x$, this works very well. Now because the mean will be $\mu = \frac{a}{a+b}$ and $\sigma^2 = \frac{ab}{(a+b)^2(a+b+1)}$ given the required mean and variance, one can work out $a, b$ algebraically.

Now look at the numerator which consists of the product of a likelihood function and a density function: up to constant $k$, if we set $\sum^n_{i=1} y_i = y$ we get $k p^{y+a-1}(1-p)^{n-y+b-1}$
The denominator: same thing, but $p$ gets integrated out and the constant $k$ cancels; basically the denominator is what makes the fraction into a density function.

So, in effect, we have $kp^{y+a-1}(1-p)^{n-y+b-1}$ which is just a beta distribution with new $a^* =y+a, b^* =n-y + b$.

So, I will spare you the calculation except to say that that the NBA prior with $\mu = .672, \sigma^2 =.0074$ leads to $a = 19.355, b= 9.447$

Now the update: $a^* = 71+19.355 = 90.355, b^* = 9.447$.

What does this look like? (I used this calculator) That is the prior. Now for the posterior: Yes, shifted to the right..very narrow as well. The information has changed..but we avoid the absurd contention that $p = 1$ with a confidence interval of zero width.

We can now calculate a “credible interval” of, say, 90 percent, to see where $p$ most likely lies: use the cumulative density function to find this out: And note that $P(p < .85) = .042, P(p < .95) = .958 \rightarrow P(.85 < p < .95) = .916$. In fact, Bird’s lifetime free throw shooting percentage is .882, which is well within this 91.6 percent credible interval, based on sampling from this one freakish streak.

## October 12, 2016

### P-values and precision of language

Filed under: media, popular mathematics — Tags: , — collegemathteaching @ 2:00 am

I read yet another paper proclaiming that it is “now time to do away with p-values.” And yes, I can recommend reading the article.

From my point of view, one of the troubles with p-values is that there is a misunderstanding as to what they actually mean.

So here goes: the p-value is the probability that, given the null hypothesis is true, one obtains an observation as extreme (or greater) than the given observation. That is, if $Y$ is a random variable with a probability distribution as given by the null hypothesis, and $Y^*$ is the observation, $P(Y \geq Y^*) = p$.

Example: suppose you assume that a coin is fair (the null hypothesis), and you toss it 100 times and observe 65 heads. It can be shown that $P(Y \geq 65) = 0.00175882086148504$. So that is the p-value of that particular experiment. That is, IF the coin really were fair, you’d expect to 65 or more heads .1716 percent of the time.

That seems clear enough, statistically speaking.

But when one gets down to the science, one wants to determine whether there is evidence enough to believe one thing or another thing. So, is this coin biased or did this result happen “just by chance”? And strictly speaking, we don’t really know. For example, it could be that we did a precision scientific measurement on the coin and found it to be fair before doing the above experiment. Or it could be that this was just some coin we came across, or it could be that we were asked to examine this coin because of previous suspicious results. This information matters.

And think of it this way: suppose the above experiment was repeated, say, 100,000 times with a coin known to be fair. Then we’d expect to see the above result about 176 times and ALL of those “positives” would be “due to chance”.

Upshot: when it comes to scientific experiments, we still need replication.

## July 12, 2013

### Hypothesis Testing: Frequentist and Bayesian

Filed under: science, statistics — Tags: , — collegemathteaching @ 4:24 pm

I was working through Nate Silver’s book The Signal and the Noise and got to his chapter about hypothesis testing. It is interesting reading and I thought I would expand on that by posing a couple of problems.

Problem one: suppose you knew that someone attempted some basketball free throws.
If they made 1 of 4 shots, what would the probability be that they were really, say, a 75 percent free throw shooter?
Or, what if they made 5 of 20 shots?

Problem two: Suppose a woman aged 40-49 got a digital mammagram and got a “positive” reading. What is the probability that she indeed has breast cancer, given that the test catches 80 percent of the breast cancers (note: 20 percent is one estimate of the “false negative” rate; and yes, the false positive rate is 7.8 percent. The actual answer, derived from data, might surprise you: it is : 16.3 percent.

I’ll talk about problem two first, as this will limber the mind for problem one.

So, you are a woman between 40-49 years of age and go into the doctor and get a mammogram. The result: positive.

So, what is the probability that you, in fact, have cancer?

Think of it this way: out of 10,000 women in that age bracket, about 143 have breast cancer and 9857 do not.
So, the number of false positives is 9857*.078 = 768.846; we’ll keep the decimal for the sake of calculation;
The number of true positives is: 143*.8 = 114.4.
The total number of positives is therefore 883.246.

The proportion of true positives is $\frac{114.4}{883.246} = .1628$ So the false positive rate is 83.72 percent.

It turns out that, data has shown the 80-90 percent of positives in women in this age bracket are “false positives”, and our calculation is in line with that.

I want to point out that this example is designed to warm the reader up to Bayesian thinking; the “real life” science/medicine issues are a bit more complicated than this. That is why the recommendations for screening include criteria as to age, symptoms vs. asymptomatic, family histories, etc. All of these factors affect the calculations.

For example: using digital mammograms with this population of 10,000 women in this age bracket adds 2 more “true” detections and adds 170 more false positives. So now our calculation would be $\frac{116.4}{1055.25} = .1103$ , so while the true detections go up, the false positives also goes up!

Our calculation, while specific to this case, generalizes. The formula comes from Bayes Theorem which states: $P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|not(A))P(not(A))}$. Here: $P(A|B)$ is the probability of event A occurring given that B occurs and P(A) is the probability of event A occurring. So in our case, we were answering the question: given a positive mammogram, what is the probability of actually having breast cancer? This is denoted by P(A|B) . We knew: P(B|A) which is the probability of having a positive reading given that one has breast cancer and P(B|not(A)) is the probability of getting a positive reading given that one does NOT have cancer. So for us: $P(B|A) = .8, P(B|not(A)) = .078$ and $P(A) = .0143, P(not(A)) = .9857$ .

The bottom line: If you are testing for a condition that is known to be rare, even a reasonably accurate test will deliver a LOT of false positives.

Here is a warm up (hypothetical) example. Suppose a drug test is 99 percent accurate in that it will detect that a certain drug is there 99 percent of the time (if it is really there) and only yield a false positive 1 percent of the time (gives a positive result even if the person being tested is free of this drug). Suppose the drug use in this population is “known” to be, say 5 percent.

Given a positive test, what is the probability that the person is actually a user of this drug?

Answer: $\frac{.99*.05}{.99*.05+.01*.95} = .839$ . So, in this population, about 16.1 percent of the positives will be “false positives”, even though the test is 99 percent accurate!

Now that you are warmed up, let’s proceed to the basketball question:

Question: suppose someone (that you don’t actually see) shoots free throws.

Case a) the player makes 1 of 4 shots.
Case b) the player makes 2 of 8 shots.
Case c) the player makes 5 of 20 shots.

Now you’d like to know: what is the probability that the player in question is really a 75 percent free throw shooter? (I picked 75 percent as the NBA average for last season is 75.3 percent).

Now suppose you knew NOTHING else about this situation; you know only that someone attempted free throws and you got the following data.

The traditional “hypothesis test” uses the “frequentist” model: you would say: if the hypothesis that the person really is a 75 percent free throw shooter is true, what is the probability that we’d see this data?

So one would use the formula for the binomial distribution and use n = 4 for case A, n = 8 for case B and n = 20 for case C and use p = .75 for all cases.

In case A, we’d calculate the probability that the number of “successes” (made free throws) is less than or equal to 1; 2 for case B and 5 for case C.

For you experts: the null hypothesis would be, say for the various cases would be $P(Y \le 1 | p = .75), P(Y \le 2 | p = .75), P(Y \le 5 | = .75)$ respectively, where the probability mass function is adjusted for the different values of n .

We could do the calculations by hand, or rely on this handy calculator.

Case A: .0508
Case B: .0042
Case C: .0000 ( $3.81 \times 10^{-6}$)

By traditional standards: Case A: we would be on the verge of “rejecting the null hypothesis that p = .75 and we’d easily reject the null hypothesis in cases B and C. The usual standard (for life science and political science) is p = .05).

(for a refresher, go here)

So that is that, right?

Well, what if I told you more of the story?

Suppose now, that in each case, the shooter was me? I am not a good athlete and I played one season in junior high, and rarely, some pickup basketball. I am a terrible player. Most anyone would happily reject the null hypothesis without a second thought.

But now: suppose I tell you that I took these performances from NBA box scores? (the first one was taken from one of the Spurs-Heat finals games; the other two are made up for demonstration).

Now, you might not be so quick to reject the null hypothesis. You might reason: “well, he is an NBA player and were he always as bad as the cases show, he wouldn’t be an NBA player. This is probably just a bad game.” In other words, you’d be more open to the possibility that this is a false positive.

Now you don’t know this for sure; this could be an exceptionally bad free throw shooter (Ben Wallace shot 41.5 percent, Shaquille O’Neal shot 52.7 percent) but unless you knew that, you’d be at least reasonably sure that this person, being an NBA player, is probably a 70-75 shooter, at worst.

So “how” sure might you be? You might look at NBA statistics and surmise that, say (I am just making this up), 68 percent of NBA players shoot between 72-78 percent from the line. So, you might say that, prior to this guy shooting at all, the probability of the hypothesis being true is about 70 percent (say). Yes, this is a prior judgement but it is a reasonable one. Now you’d use Bayes law: $P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|not(A))P(not(A))}$

Here: A represents the “75 percent shooter” being actually true, and B is the is the probability that we actually get the data. Note the difference in outlook: in the first case (the “frequentist” method), we wondered “if the hypothesis is true, how likely is it that we’d see data like this”. In this case, called the Bayesian method, we are wondering: “if we have this data, what is the probability that the null hypothesis is true”. It is a reverse statement, of sorts.

Of course, we have $P(A) = .7, P(not(A)) = .3$ and we’ve already calculated P(B|A) for the various cases. We need to make a SECOND assumption: what does event not(A) mean? Given what I’ve said, one might say not(A) is someone who shoots, say, 40 percent (to make him among the worst possible in the NBA). Then for the various cases, we calculate $P(B|not(A)) = .4752, .3154, .1256$ respectively.

So, we now calculate using the Bayesian method:

Case A, the shooter made 1 of 4: .1996. The frequentist p-value was .0508
Case B, the shooter made 2 of 8: .0301. The frequentist p-value was .0042
Case C, the shooter made 5 of 20: 7.08 x 10^-5 The frequentist p-value was 3.81 x 10^-6

We see the following:
1. The Bayesian method is less likely to produce a “false positive”.
2. As n, the number of data points, grows, the Bayesian conclusion and the frequentist conclusions tend toward “the truth”; that is, if the shooter shoots enough foul shots and continues to make 25 percent of them, then the shooter really becomes a 25 percent free throw shooter.

So to sum it up:
1. The frequentist approach relies on fewer prior assumptions and is computationally simpler. But it doesn’t include extra information that might make it easier to distinguish false positives from genuine positives.
2. The Bayesian approach takes in more available information. But it is a bit more prone to the user’s preconceived notions and is harder to calculate.

How does this apply to science?
Well, suppose you wanted to do an experiment that tried to find out which human gene alleles correspond so a certain human ailment. So a brute force experiment in which every human gene is examined and is statistically tested for correlation with the given ailment with null hypothesis of “no correlation” would be a LOT of statistical tests; tens of thousands, at least. And at a p-value threshold of .05 (we are willing to risk a false positive rate of 5 percent), we will get a LOT of false positives. On the other hand, if we applied bit of science prior to the experiment and were able to assign higher prior probabilities (called “posterior probability”) to the genes “more likely” to be influential and lower posterior probability to those unlikely to have much influence, our false positive rates will go down.

Of course, none of this eliminates the need for replication, but Bayesian techniques might cut down the number of experiments we need to replicate.