Checking Null Expected Successes and Failures

Why Check Both Expected Counts?

A one-proportion test often uses a normal approximation to describe how far a sample result is from the proportion stated by the null hypothesis. Before using that approximation, check whether the null model expects enough successes and enough failures. A sample could have many observations overall but still have very few expected outcomes in one category when the null proportion is close to zero or one.

This check concerns the distribution of results that would occur if the null hypothesis were true. It is not a check of whether the observed data happen to contain enough successes and failures. Keeping those two ideas separate is essential: the expected counts come from the null proportion and the sample size, while the observed counts come from the data.

The Null Model Behind the Check

Suppose each individual or trial has a binary outcome, classified as either a success or a failure. Under the null hypothesis, the model assumes that each outcome has the same success probability \(p_0\), and that outcomes are independent. If \(X\) is the total number of successes in the sample, then under this model:

Here, “success” is simply the label for the outcome being counted; it does not have to mean a desirable result. The number of failures is \(n-X\). Under the null model, the expected number of successes is \(np_0\), and the expected number of failures is \(n(1-p_0)\). These are averages across repeated samples imagined under the null, not predictions that a particular sample must match exactly.

The independence and common-probability assumptions give a clear model for the count. In a random sample drawn without replacement from a finite population, observations are not literally independent; a common practical check is that the sample is no more than 10% of that population. The expected-count check does not replace these design and model considerations. Nor does it determine whether the sample represents the population of interest.

A Short Justification for the Expected Counts

The formulas follow from adding the indicators for the individual outcomes. Let \(I_i=1\) if observation \(i\) is a success and \(I_i=0\) otherwise. Under the null, \(P(I_i=1)=p_0\), so \(E(I_i)=p_0\). The total number of successes is \(X=I_1+\cdots+I_n\). By linearity of expectation:

For failures, define \(J_i=1-I_i\). Then \(E(J_i)=1-p_0\), and the total number of failures is \(n-X=J_1+\cdots+J_n\). Therefore:

The normal approximation is most useful when the binomial distribution is not too lopsided. If the null probability is small, successes are relatively rare; if it is large, failures are relatively rare. Requiring enough expected outcomes in both categories helps guard against approximating a strongly skewed count distribution with a symmetric curve. The guideline is a diagnostic, not a proof that the binomial distribution is exactly normal.

How to Apply the Check

Use the null value \(p_0\), not the observed sample proportion \(\hat{p}\). First calculate \(np_0\), then calculate \(n(1-p_0)\). Compare each result with the stated guideline. If either count is below the cutoff, do not declare the check passed just because the other count is large. The direction of the alternative hypothesis does not change these expected counts.

Worked Checks

Why the Null Counts Matter

The test asks how unusual the observed result would be under the null hypothesis. Therefore, the relevant distribution is the one generated by assuming the null proportion is true. The observed proportion \(\hat{p}\) can be very different from \(p_0\); substituting \(\hat{p}\) into the expected-count check would assess a different model from the null model used by the test.

For example, a sample might contain many observed successes even though the null model expects few. That discrepancy could be evidence against the null, but it does not make the null expected-count calculation change. The check is performed using \(p_0\) precisely because it assesses whether the reference distribution under \(H_0\) is suitable for a normal approximation.

Likewise, the alternative hypothesis might state that the proportion is greater than, less than, or different from \(p_0\). The expected counts remain \(np_0\) and \(n(1-p_0)\) in all three cases. The alternative determines which departures are relevant to the test, but it does not alter the null model used for this condition.

Common Mistakes

• Checking only one category. Seeing that \(np_0\) is large is not enough. Check \(n(1-p_0)\) as well; if either count is below the chosen guideline, the condition has not passed.
• Using the observed proportion. Calculating \(n\hat{p}\) and \(n(1-\hat{p})\) checks observed counts, not expected counts under the null. Use \(p_0\) for this test condition.
• Treating the cutoff as an exact law. The “at least 10” guideline is a practical rule. Passing it supports a normal approximation but does not guarantee exact accuracy; failing it signals caution rather than proving that every normal approximation would be unusable.
• Confusing expected with actual counts. An expected count of 6.4 is an average under repeated samples, not a claim that a particular sample must contain a fractional number of successes.
• Assuming this is the only assumption. Adequate expected counts do not establish that outcomes are independent, that the probability is constant, or that the data were sampled appropriately.