The Binomial test evaluate the probability P_bin of seeing k or more extreme number of positive cases in a sample of n cases, if the reference proportion is P_ref.

Please note : Binomial probability assumes that the observation is one off. In most quality assurance programs repeated observations are made, and the probability estimates must take into account variations from repeated measurements, and different statistics should be used.

Three parameters are involved :

• Reference proportion (P_ref) : also known as expected proportion or probability, is the expected, theoretical, or reference proportion of cases that are positive in the indexed outcome (deaths, cures, winners, losers, complications, etc). This is usually expressed as a number between 0 and 1, so that 0.25 represents 25%. The abbreviation P_ref is used in these programs.
• Sample size (n) : is the number of cases observed in the current study, and this is abbreviated to n.
• Numbers positives (k) : is the number of cases found to be positive in the sample observed in the current study, and this is abbreviated to k.

The test result is the Binomial Probability, abbreviated to P_bin. Two such values are calculated, although usually one is required in a particular study.

• P_bin<=k is the probability of observing k or less number of positive cases amongst a sample of n cases, if the true proportion is P_ref
• P_bin>=k is the probability of observing k or more number of positive cases amongst a sample of n cases, if the true proportion is P_ref

Calculations use the Binomial Coefficient, which required calculation of Factorial numbers. The duration of calculations therefore increases exponentially with large numbers, and sample size in excess of 1000 may exceed the time limits allowed by the server and crash the program. Should this happens, an alternative algorithm using the approximate Normal transformation of the proportion, using the following algorithm.

1. mean (μ) = P_ref * n
2. Standard Deviation (sd) = sqrt(n * P_ref * (1 - P_ref))
3. y = k
   • if k<μ then y = k + 0.5
   • if k>μ then y = k - 0.5
4. z = (y-μ)/sd
5. P_bin = probability of z, calculated in the Probability of z Program Page

Reference : Nonparametric Statistics for the Behavioral Sciences. (2nd. Ed. 1998) S. Siegal & N. J. Castellan Jr. McGraw-Hill New York. ISBN 0-07-057357-3 p. 38

Example 1

A gambler arriving at a casino requests that he be allowed to use his own dice, as it gives him confidence. He insisted that the dice was properly made, and contains no bias. The casino authority agreed to this, providing they could satisfy themselves that this was an unbiased dice after 1000 throws of the dice.

As the dice has 6 sides, the chance of having any single value is 1/6=0.17 (P_ref = 0.17 or 17%) if the dice is unbiased. When the dice was thrown 1000 times (n=1000), the value 6 was seen 200 times (k=200), in other words the number 6 was obtained in 20% of the time. Binomial test shows that the probability of obtaining k=200 or more times for any number (number 6 in this case) in n=1000 throws, when the expected proportion P_ref=0.17, is P_bin = 0.03. Although this result is marginal, it was considered statistically significant using the standard of p<0.05. The casino therefore concluded that the dice was possibly biased, and disallowed its use.

Example 2

A dangerous operation on the heart is expected to have a mortality rate of 14% (P_ref=0.14). A new surgeon was appointed, and amongst his first 8 (n=8) operations 5 patients died (k=5). In other words, his death rate in the first 8 operations was 62.5%.

When the issue was raised by the anaesthetists, the surgeon replied that this should not be unexpected as the operation was known to be dangerous and associated with a high mortality rate, and the number of cases was too few for any conclusions to be drawn in any case.

The governance committee was not satisfied with this answer and sought statistical clarification. The Binomial test produce a probability of P_bin = 0.0021 in observing k>=5 deaths in n=8 cases, when the expected deaths rate was 14% (P_ref=0.14). A conclusion that the mortality rate was significantly higher than that expected could therefore be drawn, and further investigations warranted.

0.6 100 55 0.6 1001 550 0.6 100 70 0.6 1001 700

Calculate Binomial Probability (Pbin) Using Reference P{robability (Pref), Sample Size (n), and number of positives (k) Data Entry    The data is a 3 column table of numbers, each row a separate test       Column 1 is the Reference Probability or proportion (Pref) from 0 (0%) to 1 (100%)       Column 2 is the sample size of the data collected (n)       Column 3 is the number of positives (k) in that sample

The R code for the Binomial Test is copied from the reference web page https://www.rdocumentation.org/packages/stats/versions/3.6.2/topics/binom.test, the basic formular is

binom.test(x, n, p=0.5, alternative = c("two.sided", "less", "greater"), conf.level = 0.95)

Please note that

• x is the number of positives observed
• n is the sample size
• p is the reference or expected proportion of positives. If not stated, the default value is 0.5
• Alternatives are to detect the probability of x in n not being p. If not stated the default is two.sided
• confidence interval is the confidence interval of calculated probability x/n. If not stated, the default is 95% confidence interval (0.95)

Acepting the default 95% confidence interval, the three ways the formular can be used are as follows.

The data for demonstration is the first row in the Javascript program where the regerence proportion is 60% (0.6), and 55 positives found in 100 observations (x = 55, n = 100)

p = 0.6
n = 100
x = 55

The first option is to test the null hypothesis probability that 55 in 100 is significantly less than 60%

binom.test(x, n, p,alternative = "less")

The results are as follows

number of successes = 55, number of trials = 100, p-value = 0.1789
alternative hypothesis: true probability of success is less than 0.6
95 percent confidence interval: 0.0000000 0.6348377
sample estimates: probability of success 0.55 

In the firrst line p-value = 0.1789 means the probability of the null hypothesis that 55 in 100 is not less than 60% (0.6) is 0.1789

The second and third line indicates that 55 positives in 100 reprsents a 95% confidence of being less than 60& to be 0 to 0.6348

The last line merely reiterates 55/100 = 0.55

The second option is to test the null hypothesis probability that 55 in 100 is significantly greater than 60%

binom.test(x, n, p,alternative = "greater")

The results are as follows

number of successes = 55, number of trials = 100, p-value = 0.8689
alternative hypothesis: true probability of success is greater than 0.6
95 percent confidence interval: 0.4628896 1.0000000
sample estimates: probability of success 0.55 

In the firrst line p-value = 0.8689 means the probability of the null hypothesis that 55 in 100 is not greater than 60% (0.6) is 0.8689

The second and third line indicates that 55 positives in 100 reprsents a 95% confidence of being greater than 60% to be 0.4629 to 1

The last line merely reiterates 55/100 = 0.55

The third option is to test the null hypothesis probability that 55 in 100 is significantly not 60%

binom.test(x, n, p,alternative = "two.sided")

The results are as follows

number of successes = 55, number of trials = 100, p-value = 0.3092
alternative hypothesis: true probability of success is not equal to 0.6
95 percent confidence interval: 0.4472802 0.6496798
sample estimates: probability of success 0.55 

In the firrst line p-value = 0.3096 means the probability of the null hypothesis that 55 in 100 is not 60% (0.6) is 0.3092

The second and third line indicates that 55 positives in 100 reprsents a 95% confidence of the true proportion is 0.4472 to 0.6497

The last line merely reiterates 55/100 = 0.55