Deriving The Wald Interval For One Proportion

Why a Proportion Interval Has This Form

A sample proportion is an estimate of an unknown population proportion. A confidence interval adds a measure of sampling uncertainty to that estimate: it gives a range of plausible values produced by a procedure designed to cover the fixed population proportion at a specified rate over repeated samples. The Wald interval is the direct large-sample construction based on a normal approximation.

The key steps are to find the sampling variance of the sample proportion, estimate its standard error using the observed data, and use a standard normal critical value to form a margin of error. The result is compact, but each step depends on the sampling model and on having enough successes and failures for the normal approximation to be reasonable. This derivation is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Suppose \(X_1,\ldots,X_n\) are independent Bernoulli random variables with a common success probability \(p\). The number of successes is \(X=\sum_{i=1}^n X_i\), and the sample proportion is \(\hat{p}=X/n\). The Wald confidence interval for \(p\), at confidence level \(1-\alpha\), is

\hat{p}\ \pm\ z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},

where \(z_{\alpha/2}\) is the standard normal critical value satisfying \(P(Z>z_{\alpha/2})=\alpha/2\) for \(Z\sim N(0,1)\). The two endpoints are the lower endpoint, obtained with the minus sign, and the upper endpoint, obtained with the plus sign.

From the Binomial Model to the Interval

The model assumes each observation has two possible outcomes, success or failure; the success probability \(p\) is the same for every observation; and observations are independent. Then \(X\sim\operatorname{Binomial}(n,p)\). The target \(p\) is an unknown but fixed number, while \(X\) and \(\hat p\) vary from sample to sample.

For a Bernoulli observation, \(E[X_i]=p\) and \(\operatorname{Var}(X_i)=p(1-p)\). Independence means the variance of the sum is the sum of the variances. Dividing the sum by \(n\) divides its variance by \(n^2\). Therefore, the exact mean and variance of the sample proportion are

E[\hat{p}]=p,\qquad \operatorname{Var}(\hat{p})= \operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^n X_i\right) =\frac{np(1-p)}{n^2} =\frac{p(1-p)}{n}.

When the sample is large enough for a normal approximation to be reasonable, the standardized sample proportion is approximately standard normal. In this expression, \(p\) appears in the standard deviation:

\frac{\hat{p}-p}{\sqrt{p(1-p)/n}}\ \approx\ N(0,1).

The true standard error \(\sqrt{p(1-p)/n}\) cannot be calculated directly because \(p\) is unknown. The Wald method estimates it by substituting \(\hat p\) for \(p\). The estimated standard error is therefore \(\sqrt{\hat{p}(1-\hat{p})/n}\). This substitution gives a usable estimate of sampling variability from the same sample used to estimate \(p\).

Justification: For a standard normal variable \(Z\), \(P(-z_{\alpha/2}\leq Z\leq z_{\alpha/2})=1-\alpha\). Applying this central probability to the standardized sample proportion gives, approximately,

P\left( -z_{\alpha/2}\leq \frac{\hat{p}-p}{\sqrt{p(1-p)/n}} \leq z_{\alpha/2} \right)\approx 1-\alpha.

After replacing the unknown standard error with its estimate, the Wald procedure treats the observed \(\hat p\) as the center and uses the estimated standard error to set the margin of error. This produces \(\hat p\pm z_{\alpha/2}\sqrt{\hat p(1-\hat p)/n}\). The replacement is an approximation, not an exact rearrangement that yields an exact-coverage interval.

For a two-sided 95% interval, \(z_{\alpha/2}=z_{0.025}\approx1.96\). For a 90% interval, the critical value is approximately 1.645; for a 99% interval, it is approximately 2.576. The higher the confidence level, the larger the critical value and, with the same sample proportion and sample size, the wider the interval.

Constructing and Reading the Wald Interval

Use the following sequence. Keep the estimate, estimated standard error, critical value, and endpoints distinct; that makes both the calculation and its interpretation easier to check.

Identify the data and model.
Record the sample size \(n\), the number of successes \(x\), and the stated confidence level. Confirm that the observations fit the independent, common-probability Bernoulli model.

Calculate the sample proportion.
Set \(\hat{p}=x/n\). This is the point estimate and the center of the Wald interval.

Estimate the standard error.
Calculate \(\sqrt{\hat{p}(1-\hat{p})/n}\). The square root is essential: the expression inside it is an estimated variance.

Find the margin and endpoints.
Multiply the estimated standard error by \(z_{\alpha/2}\), then subtract and add that margin to \(\hat p\).

The interval is a procedure with approximate repeated-sampling coverage. In repeated samples under the model, a 95% Wald procedure is intended to produce intervals that contain the fixed value \(p\) about 95% of the time. It does not mean there is a 95% probability that \(p\) lies in this particular calculated interval. After the data are observed, the endpoints are fixed.

Worked Example: A 95% Interval for an Audit Rate

A hypothetical audit samples 200 independent account records from a population in which each record either meets a specified review standard or does not. Assume a common probability \(p\) of meeting the standard. Of the 200 records, 84 meet it. Construct a 95% Wald interval for \(p\). Method: estimate \(p\) with the sample proportion and use the plug-in standard error. Common mistake avoided: treating the estimated variance as the standard error.

Step 1: Estimate the proportion. Here \(n=200\) and \(x=84\), so

\hat p=\frac{x}{n}=\frac{84}{200}=0.42,\qquad 1-\hat p=0.58.

Step 2: Calculate the estimated standard error.

\widehat{\operatorname{SE}}(\hat p) =\sqrt{\frac{0.42(0.58)}{200}} =\sqrt{\frac{0.2436}{200}} =\sqrt{0.001218} \approx 0.03490.

Step 3: Calculate the margin of error. For 95% confidence, use \(z_{0.025}=1.96\):

1.96(0.03490)\approx0.06840.

Step 4: Calculate both endpoints.

0.42-0.06840=0.35160,\qquad 0.42+0.06840=0.48840.

Thus, the 95% Wald interval is approximately \((0.3516,\,0.4884)\), or 35.16% to 48.84%. As a check, the midpoint is 0.42, and each endpoint is 0.06840 from the midpoint. The interval estimates the population proportion; it does not describe a range containing 95% of individual records.

Worked Example: A 90% Interval for a Service Completion Proportion

In a hypothetical service review, 104 of 160 independently selected requests are completed within a stated target time. Assume each request has the same probability \(p\) of meeting the target. Construct a 90% Wald interval for that probability. Method: use the 90% standard normal critical value and the estimated standard error. Common mistake avoided: using the 95% critical value when the requested confidence level is 90%.

Step 1: Calculate the observed proportion and its complement.

\hat p=\frac{104}{160}=0.65,\qquad 1-\hat p=0.35.

Step 2: Find the estimated standard error.

\widehat{\operatorname{SE}}(\hat p) =\sqrt{\frac{0.65(0.35)}{160}} =\sqrt{\frac{0.2275}{160}} =\sqrt{0.001421875} \approx0.03771.

Step 3: Use the 90% critical value. With \(z_{0.05}\approx1.645\), the margin is

1.645(0.03771)\approx0.06203.

Step 4: Form the interval.

0.65-0.06203=0.58797,\qquad 0.65+0.06203=0.71203.

The 90% Wald interval is \((0.5880,\,0.7120)\) to four decimal places. Its midpoint is 0.6500, consistent with the observed proportion. Choosing 1.96 instead would produce a wider 95% interval, not the requested 90% interval.

Worked Example: A 99% Interval for a Participation Proportion

A hypothetical research team samples 400 eligible participants independently from a defined population. In the sample, 112 participate in a voluntary program. Assume a common participation probability \(p\), and construct a 99% Wald interval. Method: use the 99% critical value with the plug-in standard error. Common mistake avoided: keeping the 95% multiplier even though a more demanding confidence level is requested.

Step 1: Calculate the sample proportion.

\hat p=\frac{112}{400}=0.28,\qquad 1-\hat p=0.72.

Step 2: Calculate the estimated standard error.

\widehat{\operatorname{SE}}(\hat p) =\sqrt{\frac{0.28(0.72)}{400}} =\sqrt{\frac{0.2016}{400}} =\sqrt{0.000504} \approx0.02245.

Step 3: Find the 99% margin of error. Use \(z_{0.005}\approx2.576\):

2.576(0.02245)\approx0.05783.

Step 4: Calculate the endpoints.

0.28-0.05783=0.22217,\qquad 0.28+0.05783=0.33783.

The 99% Wald interval is approximately \((0.2222,\,0.3378)\). The midpoint is 0.2800, and the larger 99% critical value gives a wider interval than a 95% interval would, with the same data and estimated standard error.

Assumptions, Checks, and Common Traps

The derivation depends on the binomial sampling model and an adequate normal approximation. A commonly used diagnostic is to check whether the expected counts of successes and failures, \(n\hat p\) and \(n(1-\hat p)\), are sufficiently large; a course or problem may state a specific rule of thumb. This check supports the approximation but does not turn Wald coverage into an exact result. If observations are dependent or their success probabilities differ, the stated binomial model and this standard error need not apply.

The most frequent computational trap is omitting the square root. For the first worked example, \(\hat p(1-\hat p)/n=0.001218\) is the estimated variance, not the estimated standard error. Using 0.001218 as the standard error would produce a false margin of \(1.96(0.001218)\approx0.002387\), rather than the correct margin of about 0.06840. The variance is in squared proportion units; taking its square root returns the scale of the proportion.

Wrong tail value: a two-sided 95% interval uses 1.96, not the one-sided 95% value 1.645.
Wrong denominator: the sample proportion is \(x/n\), where \(n\) is the total number observed, not just the number of failures.
Wrong interpretation: the confidence level describes the long-run performance of the method, not a probability assigned to the fixed parameter after the interval is calculated.
Ignoring model conditions: a normal-based interval is not justified solely because a formula can be evaluated. The data and sample size must support the approximation.

The Wald interval is useful because its center, standard error, and margin are easy to identify and derive. Its simplicity should not be confused with exact coverage: it is a large-sample approximation, and the estimated standard error depends on the observed proportion. For this tutorial, the central skill is to derive and apply that construction correctly while stating the model and recognizing its limits.

Check Your Understanding

For each problem, use the Wald formula, select the critical value for the stated confidence level, and round interval endpoints to four decimal places.

A hypothetical inspection finds that 75 of 250 independently sampled components pass a specified test. Assuming a common pass probability, construct a 95% Wald interval. (Answer: \((0.2432,\,0.3568)\))
Of 180 independently sampled applications in a hypothetical program, 117 are complete. Assuming the same completion probability for each application, find a 90% Wald interval. (Answer: \((0.5915,\,0.7085)\))
A hypothetical poll records 215 affirmative responses among 500 independent respondents. Under a common response probability, calculate a 95% Wald interval. (Answer: \((0.3866,\,0.4734)\))
In a hypothetical sample of 1,200 independently selected cases, 36 have a specified outcome. Assuming a common probability, calculate a 95% Wald interval. (Answer: \((0.0203,\,0.0397)\))
A hypothetical study records participation for 320 independent eligible individuals; 128 participate. Calculate a 99% Wald interval using \(z_{0.005}=2.576\). (Answer: \((0.3295,\,0.4705)\))

What You'll Learn