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A Binomial Test in R is a statistical method used to determine whether the observed success rate of a binary outcome (such as success or failure) is significantly different from a hypothesized success rate. This test can be performed by first setting the desired significance level and then using the binom.test() function in R, which takes in the observed number of successes, total number of trials, and the hypothesized success rate as input. The output of this test includes the p-value, indicating the likelihood of obtaining the observed result if the null hypothesis (that the success rate is equal to the hypothesized rate) is true. A p-value lower than the chosen significance level indicates that the observed success rate is significantly different from the hypothesized rate. This test is commonly used in various fields such as economics, psychology, and biology to evaluate the effectiveness of a treatment or intervention.
Perform a Binomial Test in R
A binomial test compares a sample proportion to a hypothesized proportion. The test has the following null and alternative hypotheses:
H0: π = p (the population proportion π is equal to some value p)
HA: π ≠ p (the population proportion π is not equal to some value p)
The test can also be performed with a one-tailed alternative that the true population proportion is greater than or less than some value p.
To perform a binomial test in R, you can use the following function:
binom.test(x, n, p)
where:
- x: number of successes
- n: number of trials
- p: probability of success on a given trial
The following examples illustrate how to use this function in R to perform binomial tests.
Example 1: Two-tailed Binomial Test
You want to determine whether or not a die lands on the number “3” during 1/6 of the rolls so you roll the die 24 times and it lands on “3” a total of 9 times. Perform a Binomial test to determine if the die actually lands on “3” during 1/6 of rolls.
#perform two-tailed Binomial test binom.test(9, 24, 1/6) #output Exact binomial test data: 9 and 24 number of successes = 9, number of trials = 24, p-value = 0.01176 alternative hypothesis: true probability of success is not equal to 0.1666667 95 percent confidence interval: 0.1879929 0.5940636 sample estimates: probability of success 0.375
The p-value of the test is 0.01176. Since this is less than 0.05, we can reject the null hypothesis and conclude that there is evidence to say the die does not land on the number “3” during 1/6 of the rolls.
Example 2: Left-tailed Binomial Test
You want to determine whether or not a coin is less likely to land on heads compared to tails so you flip the coin 30 times and find that it lands on heads just 11 times. Perform a Binomial test to determine if the coin is actually less likely to land on heads compared to tails.
#perform left-tailed Binomial test binom.test(11, 30, 0.5, alternative="less") #output Exact binomial test data: 11 and 30 number of successes = 11, number of trials = 30, p-value = 0.1002 alternative hypothesis: true probability of success is less than 0.5 95 percent confidence interval: 0.0000000 0.5330863 sample estimates: probability of success 0.3666667
The p-value of the test is 0.1002. Since this is not less than 0.05, we fail to reject the null hypothesis. We do not have sufficient evidence to say that the coin is less likely to land on heads compared to tails.
A shop makes widgets with 80% effectiveness. They implement a new system that they hope will improve the rate of effectiveness. They randomly select 50 widgets from a recent production run and find that 46 of them are effective. Perform a binomial test to determine if the new system leads to higher effectiveness.
#perform right-tailed Binomial test binom.test(46, 50, 0.8, alternative="greater") #output Exact binomial test data: 46 and 50 number of successes = 46, number of trials = 50, p-value = 0.0185 alternative hypothesis: true probability of success is greater than 0.8 95 percent confidence interval: 0.8262088 1.0000000 sample estimates: probability of success 0.92
The p-value of the test is 0.0185. Since this is less than 0.05, we reject the null hypothesis. We have sufficient evidence to say that the new system produces effective widgets at a higher rate than 80%.