is through the standard normal distribution, in which a z-test is performed of the test statistic , given by where is the number of successes observed in a sample of size and is the probability of success according to the null hypothesis. An improvement on this approximation is possible by introducing a continuity correction: For very large , this continuity correction will be unimportant, but for intermediate values, where the exact binomial test doesn't work, it will yield a substantially more accurate result. In notation in terms of a measured sample proportion , null hypothesis for the proportion , and sample size , where and , one may rearrange and write the z-test above as by dividing by in both numerator and denominator, which is a form that may be more familiar to some readers. Example Suppose we have a board game that depends on the roll of one die and attaches special importance to rolling a 6. In a particular game, the die is rolled 235 times, and 6 comes up 51 times. If the die is fair, we would expect 6 to come up times. We have now observed that the number of 6s is higher than what we would expect on average by pure chance had the die been a fair one. But, is the number significantly high enough for us to conclude anything about the fairness of the die? This question can be answered by the binomial test. Our null hypothesis would be that the die is fair (probability of each number coming up on the die is 1/6). To find an answer to this question using the binomial test, we use the binomial distribution with pmf . As we have observed a value greater than the expected value, we could consider the probability of observing 51 6s or higher under the null, which would constitute a one-tailed test (here we are basically testing whether this die is biased towards generating more 6s than expected). In order to calculate the probability of 51 or more 6s in a sample of 235 under the null hypothesis we add up the probabilities of getting exactly 51 6s, exactly 52 6s, and so on up to probability of getting exactly 235 6s: If we have a significance level of 5%, then this result (0.02654 < 5%) indicates that we have evidence that is significant enough to reject the null