One-Sided Confidence Bounds for a Proportion

Why Use a One-Sided Bound?

A confidence interval for a population proportion often gives both a lower and an upper endpoint. But sometimes an investigation calls for only one of them. A safety assessment may ask how large a failure rate could plausibly be, while a service evaluation may ask whether the proportion of satisfied customers is at least some level. In such cases, an upper or lower confidence bound directly answers the question of interest.

A one-sided bound is not simply half of a two-sided interval. Its critical value is chosen so that the stated confidence level lies in one tail of the standard normal distribution. Choosing the right tail probability is essential to getting the intended confidence level.

Definition: For a population proportion \(p\), an upper one-sided confidence bound at confidence level \(C\) reports a range from the smallest possible value of \(p\) up to an estimated upper endpoint \(U\). A lower one-sided confidence bound reports a range from an estimated lower endpoint \(L\) up to the largest possible value of \(p\). Here \(C\) is the long-run proportion of intervals produced by the procedure that would contain \(p\), under its assumptions.

For a sample with \(x\) successes among \(n\) observations, the sample proportion is \(\hat{p}=x/n\). The normal-based one-sided bounds use the estimated standard error \(\sqrt{\hat{p}(1-\hat{p})/n}\). We use this method only when random sampling and independence are reasonable and the success-failure condition supports a normal approximation. In particular, the observed numbers of successes and failures should each be sufficiently large; this tutorial uses the common check that both are at least 10.

Critical Values and Formulas

Let \(z_C\) be the standard normal value with cumulative probability \(C\) to its left, so \(P(Z\le z_C)=C\) for \(Z\sim N(0,1)\). For example, \(z_{0.95}\) is about 1.645: 95% of the standard normal distribution lies below it. The same value is used for a 95% one-sided bound, whether the bound is upper or lower.

Formula: For confidence level \(C\), the estimated standard error is \(SE=\sqrt{\hat{p}(1-\hat{p})/n}\). The one-sided normal bounds are

Upper: \(U=\hat{p}+z_C SE\), reported as \(0\le p\le U\).

Lower: \(L=\hat{p}-z_C SE\), reported as \(L\le p\le 1\).

The plus sign gives an upper bound because it moves upward from the sample proportion; the minus sign gives a lower bound because it moves downward. The parameter \(p\) is a proportion, so its possible values are between 0 and 1. The normal approximation is an approximate method, not a guarantee of exact coverage. If its conditions are poor or a calculated endpoint falls outside the possible range, the ordinary Wald calculation should not be treated as a reliable bound merely by changing the endpoint.

Worked Examples

Worked Example: An Upper Bound on a Defect Proportion

Suppose a hypothetical quality-control team takes a random sample of 200 items from a very large production process. Of these, 126 are classified as defective. Find a 95% upper confidence bound for the process defect proportion.

Check conditions. The sample is described as random, and the process is large enough that sampling 200 items is less than 10% of the population being sampled. The observed success and failure counts are 126 defective and \(200-126=74\) nondefective, both at least 10. The normal-based procedure is therefore reasonable under these stated assumptions.

The sample proportion is \(\hat{p}=126/200=0.6300\). For a 95% one-sided bound, \(z_C=z_{0.95}=1.645\). First calculate the estimated standard error:

SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =\sqrt{\frac{0.6300(0.3700)}{200}} =\sqrt{0.0011655} \approx 0.03414.

The upper endpoint is

U=\hat{p}+z_C SE =0.6300+1.645(0.03414) \approx 0.6862.

Thus the 95% upper confidence bound is \(0\le p\le 0.6862\). In context, the method gives a 95% confidence bound of about 68.62% for the long-run defect proportion in this process. It does not mean there is a 95% probability that this fixed \(p\) is below 0.6862; the confidence level describes the long-run performance of the procedure.

Worked Example: A Lower Bound on a Satisfaction Proportion

In a hypothetical random sample of 150 customers from a large customer base, 84 say they are satisfied with a service. Find a 90% lower confidence bound for the population proportion who would say they are satisfied.

Check conditions. Assume the sample was obtained randomly and represents less than 10% of the customer base, so independence is reasonable. There are 84 satisfied customers and \(150-84=66\) not satisfied; both observed counts exceed 10.

The sample proportion is \(\hat{p}=84/150=0.5600\). For a 90% one-sided bound, \(z_{0.90}=1.282\). The estimated standard error is

SE=\sqrt{\frac{0.5600(1-0.5600)}{150}} =\sqrt{\frac{0.2464}{150}} \approx 0.04053.

For a lower bound, subtract the margin:

L=\hat{p}-z_C SE =0.5600-1.282(0.04053) \approx 0.5080.

The 90% lower confidence bound is \(0.5080\le p\le 1\). In context, the procedure gives a lower confidence bound of about 50.80% for the population satisfaction proportion. This supports a statement that the proportion is at least about 50.80% with 90% confidence, using the stated normal-approximation method.

Worked Example: Using the One-Sided Critical Value

Suppose a hypothetical random sample of 400 eligible voters includes 220 who support a proposal. Assume the sample is less than 10% of the voter population. Find a 95% upper bound and compare the appropriate critical value with the one used for a two-sided 95% interval.

The sample proportion is \(\hat{p}=220/400=0.5500\). There are 220 supporters and 180 nonsupporters, so both observed counts meet the success-failure check. Under the stated random-sampling and independence assumptions, calculate

SE=\sqrt{\frac{0.5500(0.4500)}{400}} =\sqrt{0.00061875} \approx 0.02487.

For a 95% upper bound, use \(z_{0.95}=1.645\), not 1.96:

U=0.5500+1.645(0.02487) \approx 0.5909.

The 95% upper bound is \(0\le p\le 0.5909\). By contrast, 1.96 is the critical value with 97.5% to its left. Using it for this upper endpoint would produce approximately \(0.5500+1.96(0.02487)=0.5988\), which corresponds to about 97.5% one-sided coverage under the normal approximation, not 95%. A two-sided 95% interval uses 1.96 because it allocates 2.5% to each tail; a one-sided 95% bound allocates 5% to its single tail.

Why the One-Sided Formula Works

The justification comes from the approximate sampling distribution of \(\hat{p}\). When the sample is random, observations are independent, and the success-failure condition holds, the standardized sample proportion is approximately standard normal. The standard error in the calculation replaces the unknown \(p\) in the theoretical standard deviation with \(\hat{p}\), so the resulting coverage is approximate.

Justification: Write \(Z\approx(\hat{p}-p)/SE\), with \(Z\) approximately standard normal. For the upper endpoint, the event that the bound covers \(p\) is \(p\le\hat{p}+z_CSE\). Rearranging gives \((\hat{p}-p)/SE\ge -z_C\), whose approximate probability is \(P(Z\ge -z_C)=C\). For the lower endpoint, coverage means \(p\ge\hat{p}-z_CSE\), equivalently \((\hat{p}-p)/SE\le z_C\), whose approximate probability is \(P(Z\le z_C)=C\). Thus each bound has approximate one-sided coverage \(C\).

This argument also explains why the critical value is \(z_C\) for either direction: standard normal symmetry gives \(P(Z\ge -z_C)=P(Z\le z_C)=C\). The choice between adding and subtracting is determined by whether the question asks for an upper or lower bound.

Choosing and Reporting a Bound

Choose the direction based on the question being asked, rather than looking at the sample first and then selecting whichever direction gives a more favorable result. A bound chosen after inspecting the data may not have the stated long-run coverage. In a report, identify the population and outcome, state the direction and confidence level, and give the numerical bound with units or as a percentage.

A one-sided bound provides less information than a two-sided interval: it limits the parameter on one side while leaving the other side at its logical extreme, 0 or 1. This is useful when only one limit matters, but it should not be described as a two-sided interval or as evidence that the parameter equals the sample proportion.

Common Mistakes

Using the two-sided critical value. A learner may use 1.96 for a 95% upper bound because that value is familiar from two-sided intervals. For a 95% one-sided bound, use 1.645; 1.96 places 97.5% below the endpoint.
Choosing the wrong sign. Adding the margin gives an upper bound; subtracting it gives a lower bound. Check that the endpoint matches the question.
Giving a probability statement about the fixed parameter. “There is a 95% probability that \(p\le U\)” is not the frequentist interpretation. Instead, the procedure has 95% coverage in repeated sampling, under its assumptions.
Ignoring the conditions. A calculation alone does not establish that the normal approximation is suitable. Check random sampling, independence, and the observed counts of successes and failures before reporting the bound.
Choosing a direction after seeing the result. Selecting whichever bound is more convenient after examining the sample does not preserve the advertised confidence level in general. Decide which limit answers the question in advance.

Key takeaway: For a one-sided confidence bound for a proportion, calculate \(SE=\sqrt{\hat{p}(1-\hat{p})/n}\), use \(z_C\), then add the margin for an upper bound or subtract it for a lower bound. State the result in context and interpret confidence as a property of the procedure, not as a probability assigned to the fixed parameter.

Check Your Understanding

Try these questions before reviewing the formulas.

For a 90% upper confidence bound, which standard normal critical value is appropriate: \(z_{0.90}\) or \(z_{0.95}\)? Explain.
A random sample has 72 successes and 48 failures. What two observed counts should be checked for the success-failure condition?
For an upper bound, should the margin of error be added to or subtracted from \(\hat{p}\)?
In repeated sampling, what does 95% confidence mean for a one-sided bound?
Why does using 1.96 for a one-sided upper bound give a different confidence level from using 1.645?