### More Measles: Vaccination Rates and School Funding

I took a look at California's personal belief exemption rate (PBE) for kindergarten vaccinations in Part I. California also provides poverty information for public schools through the Free or Reduced Price Meals data sets, both of which conveniently include California's school codes. Cleaned versions of these data sets and my R code are in my vaccination GitHub.

We can use the school code as a key to join these two data sets. But remember, the FRPM data set only includes data about public schools, so we'll have to retain the private school data for PBEs by doing what's called a left outer join. This still performs a join on the school code key, but if any school codes included in the left data don't have corresponding entries in the right data set we still retain them. The missing values for the right data set in this case are set to NULL.

We can perform a left outer join in R by using "merge" with the option "all.x=TRUE". I'll start by looking at how the PBE rate varies between charter, non-charter public and private schools, so we'll need to replace those missing values for funding source after our join. If the funding source is missing, it's a private school. The FRPM data also denotes non-charter public schools with funding type "", so I'll replace those with "aPublic" for convenience. For factors, R will by default set the reference level to be the level that comes first alphabetically.

Here's a subset of the output. The addition of the funding source is an improvement over the model that doesn't include it, and the estimates for the odds ratios for funding source is the highest for directly funded charter schools, followed by locally funded charter schools and private schools. Remember, public schools are the reference level, so for the public level $$\log(\text{odds ratio}) = 0$$. Everything else being equal, our odds ratio estimates based on funding source would be: \begin{align*}
\mathrm{OR}_{\text{public}} &= e^{-3.820}\times e^{0} &= 0.022\\
\mathrm{OR}_{\text{private}} &= e^{-3.820}\times e^{0.752} &= 0.047\\
\mathrm{OR}_{\text{charter-local}} &= e^{-3.820}\times e^{1.049} &= 0.063\\
\mathrm{OR}_{\text{charter-direct}} &= e^{-3.820}\times e^{1.348} &= 0.085
\end{align*}
Converting to estimated PBE rates, we get: \begin{align*}
\mathrm{PBE}_{\text{public}} &= \frac{0.022}{1+0.022} &= 0.022\\
\mathrm{PBE}_{\text{private}} &= \frac{0.047}{1+0.047} &= 0.045\\
\mathrm{PBE}_{\text{charter-local}} &= \frac{0.063}{1+0.063} &= 0.059\\
\mathrm{PBE}_{\text{charter-direct}} &= \frac{0.085}{1+0.085} &= 0.078
\end{align*}

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows $\frac{\partial W}{\partial P}(P=1) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …