# 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:

- Would it be reasonable to assume that there is a 50/50 chance for a toast to land on either side?
- Which probability should I assume?

If I believe the toast is fair, then the probability for landing on the buttered (B) or unbuttered (U) side is 50%.

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, I am not 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 por/tability 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
```