9.3 Beta Model - Inference About Proportions

Example 1 The Sunrise Problem

The sunrise problem, as first stated by Laplace, is “What is the probability that the sun will rise tomorrow?” We'll start with the assumption that initially one has never seen a sunrise, and then observe a year of sunrises each morning with no morning without one. Thus we have the form of the data as $h$ successes (days with a sunrise) in $N$ total days. Our model of the data is specified as before with a binomial distribution, resulting in the posterior Beta, as described in Section 6.7.

After a only 10 years of watching sunrises, and no failures of a sunrise, the best estimate for the probability of a sunrise is

$\begin{matrix} {\hat{θ}}_{m e d i a n} \approx \frac{h + 2}{N + 4} \\ = \frac{3650 + 2}{3650 + 4} = 0.9995 \end{matrix}$

making it virtually certain for a sunrise.

Example 2 Cancer Rates

This example is from Donald Berry's Statistics textbook¹:

pp 192: A study (Murphy and Abbey, Cancer in Families, 1959) addressed the question of whether cancer runs in families. The investigator identified 200 women with breast cancer and another 200 women without breast cancer and asked them whether their mothers had had breast cancer. Of the 400 women in the two groups combined, 10 of the mothers had had breast cancer. If there is no genetic connection, then about half of these 10 would come from each group.

The data is that 7 of the daughters had cancer and 3 did not. Is there strong evidence of a connection?

The proper way, assuming total initial ignorance, is to use the Beta distribution:

$\begin{matrix} P (θ_{c a n c e r} | data) = B e t a (h = 7, N = 10) \end{matrix}$

which has a median of $θ hat sub c a n c e r equals 0.68$ , but a 95% credible interval of $θ hat sub c a n c e r equals 0.39$ up to $θ hat sub c a n c e r equals 0.89$ . This means there is not strong evidence of an effect.

Example 3 Cancer Rates - Normal Approximation

We can estimate the the Beta distribution median and credible intervals with a Normal distribution, by using the “assuming 2 successes and 2 failures” method.

$\begin{matrix} {\hat{θ}}_{c a n c e r} = \frac{h + 2}{N + 4} \\ = \frac{7 + 2}{10 + 4} = 0.643 \end{matrix}$

and

$\begin{matrix} σ = \sqrt{{\hat{θ}}_{c a n c e r} (1 - {\hat{θ}}_{c a n c e r}) / (N + 4)} \\ = \sqrt{0.643 (1 - 0.643) / (10 + 4)} \\ = 0.128 \end{matrix}$

So the approximate 95% credible interval is

$θ hat sub c a n c e r ± 2 σ$

which is between 0.387 and 0.899, again with the same conclusion of no strong evidence of an effect.

Example 4 Will it rain on the 4 $^{t h}$ of July?

Show solution

In the United States, the 4 $^{t h}$ of July is Independence Day, and is known for parades. The oldest continuously running parade is in Bristol, RI, and it runs rain or shine. Is it likely to rain on the parade? Climate data from nearby Providence is here from wunderground.com:

We can estimate the the Beta distribution median and credible intervals with a Normal distribution, by using the “assuming 2 successes and 2 failures” method.

$\begin{matrix} {\hat{θ}}_{r a i n} = \frac{h + 2}{N + 4} \\ = \frac{19 + 2}{48 + 4} = 0.404 \end{matrix}$

around 40%, less than an even chance (50%) of rain, but

$\begin{matrix} σ = \sqrt{{\hat{θ}}_{r a i n} (1 - {\hat{θ}}_{r a i n}) / (N + 4)} \\ = \sqrt{0.404 (1 - 0.404) / (48 + 4)} \\ = 0.068 \end{matrix}$

So the approximate 95% credible interval is

$θ hat sub r a i n ± 2 σ$

which is between 0.268 and 0.540. This is not strong evidence against a purely fair and random “coin flip” for rain on the 4 $^{t h}$ of July.

Example 5 Hot Hand Reexamined

In Tversky and Gilovich² we have the following data for Larry Bird free throws in basketball:

Given each of 53 missed shots, Larry Bird successfully shot 48 of the next attempt.
Given each of 285 successful shots, Larry Bird successfully shot 251 of the next attempt.

This data alone almost suggests an anti-hot-hand (where you're less likely to make a successful attempt following a successful shot). However, we can demonstrate that these numbers are not in fact statistically different. Given the relatively large number of attempts (greater than 30) we can use the Normal approximation to estimate the two probabilities of success:

$\begin{matrix} θ_{after a miss} = \frac{48 + 2}{53 + 4} = 0.877 \\ θ_{after a success} = \frac{251 + 2}{285 + 4} = 0.875 \end{matrix}$

and the uncertainty,

$\begin{matrix} σ_{after a miss} = \sqrt{0.877 (1 - 0.877) / (53 + 4)} = 0.044 \\ σ_{after a success} = \sqrt{0.875 (1 - 0.875) / (285 + 4) = 0.019} \end{matrix}$

making the 95% credible intervals for probability of a Larry Bird successful attempt

$\begin{matrix} 95 % {C I}_{after a miss} = 0.877 \pm 2 \cdot 0.044 = [0.789, 0.965] \\ 95 % {C I}_{after a success} = 0.875 \pm 2 \cdot 0.019 = [0.837, 0.913] \end{matrix}$

Notice that the intervals overlap, so there is no significant evidence for a difference in Larry Bird's success following another success or following a miss. Thus, there is no significant evidence for a hot hand, or an anti-hot hand.

Try it in XYZ Crunch

Both of Bird's intervals, and the difference the book leaves implicit

After a miss the estimate is Normal(0.877, 0.044), giving [0.7908, 0.9632]; after a success it is Normal(0.875, 0.019), giving [0.8378, 0.9122] - the book's two intervals, and the second sits inside the first. Going one step past the book: the difference is 0.002 plus or minus sqrt(0.044^2 + 0.019^2) = 0.0479, and the panel puts 0.4834 of it below zero. A coin flip. That is as close to no evidence as this kind of comparison gets.

Adapted from Statistical Inference for Everyone, by Brian Blais (Bryant University), licensed under CC BY-SA 4.0 (dual-licensed under the GNU FDL 1.2 or later; this adaptation uses the CC BY-SA grant). Changes were made; this adaptation is distributed under the same license. License: CC-BY-SA-4.0.