Worked Wald Intervals For One Proportion

From an Observed Count to an Interval

When a study records whether each sampled unit has a particular characteristic, the data can be summarized by a success count and a sample size. The sample proportion estimates the population proportion, but it varies from sample to sample. A Wald confidence interval combines that estimate with an estimate of its sampling variability to give a range of plausible values for the population proportion.

For example, a hypothetical insurer might survey policyholders to estimate the proportion who would choose an optional service. Each response is classified as either a success or a failure for the characteristic being studied. The calculation does not require a special insurance model: the relevant data are the number of successes and the total number sampled.

Definition: Suppose \(X\) counts successes among \(n\) independent observations, each with the same probability \(p\) of success. Then \(X\sim\operatorname{Binomial}(n,p)\), where \(p\) is the population proportion of interest. The sample proportion is \(\hat p=X/n\), and the sample failure proportion is \(1-\hat p=(n-X)/n\).

The Wald interval is an approximate two-sided confidence interval. Its estimated standard error uses the observed sample proportion in place of the unknown population proportion. For a confidence level of \(100(1-\alpha)\%\), let \(z_{1-\alpha/2}\) be the standard normal quantile with cumulative probability \(1-\alpha/2\).

Formula: For a sufficiently large sample under the binomial model, the Wald interval for \(p\) is

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

The estimated standard error is \(\sqrt{\hat p(1-\hat p)/n}\), and the margin of error is \(z_{1-\alpha/2}\sqrt{\hat p(1-\hat p)/n}\). The interval endpoints are \(\hat p\) minus and plus that margin of error.

Common critical values are approximately 1.645 for 90% confidence, 1.960 for 95% confidence, and 2.576 for 99% confidence. Use the value corresponding to the stated confidence level; a two-sided interval allocates \(\alpha/2\) to each tail. Keep intermediate digits through the endpoint calculations and round the final results as requested.

Why the Wald Formula Has This Form

The reason for the formula is the approximate sampling distribution of a sample proportion. Let \(Y_i=1\) if observation \(i\) is a success and \(Y_i=0\) otherwise. Under the model, the \(Y_i\) are independent Bernoulli random variables with mean \(p\) and variance \(p(1-p)\). Since \(\hat p=(Y_1+\cdots+Y_n)/n\), its mean is \(p\) and its variance is \(p(1-p)/n\).

When the sample is large enough for a normal approximation, the standardized sample proportion is approximately standard normal. The population proportion in its standard error is unknown, so the Wald method estimates it with \(\hat p\). The central probability statement can then be rearranged to put \(p\) between the sample estimate minus and plus the estimated normal margin:

\frac{\hat p-p}{\sqrt{\hat p(1-\hat p)/n}} \mathrel{\approx} N(0,1) \quad\Longrightarrow\quad P\left( \hat p-z_{1-\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}} \le p \le \hat p+z_{1-\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}} \right) \mathrel{\approx} 1-\alpha.

This is an approximate result, not an exact guarantee for every sample. The independent, common-probability Bernoulli model must be reasonable, and the normal approximation must be adequate. A common screening rule checks that both \(n\hat p\) and \(n(1-\hat p)\) are at least 10; requirements can vary by convention. If the sample is too small or the observed proportion is near a boundary, the Wald approximation may perform poorly. The focus here is calculating and interpreting the Wald interval when its approximation is appropriate.

Calculating the Estimate, Standard Error, and Margin

A reliable calculation keeps three quantities distinct: \(\hat p\), the estimated standard error, and the margin of error. The proportion is the center of the interval. The standard error measures estimated sampling variability in the proportion, while the margin scales that standard error by the critical value for the chosen confidence level.

Compute the sample proportion.
Divide the number of successes \(x\) by the sample size \(n\): \(\hat p=x/n\). Also record the number of failures \(n-x\) to check the approximation condition.

Estimate the standard error.
Substitute \(\hat p\) into \(\sqrt{\hat p(1-\hat p)/n}\). The division by \(n\) is essential.

Find the margin of error.
Multiply the estimated standard error by the two-sided normal critical value for the stated confidence level.

Form and interpret the interval.
Subtract and add the margin of error to \(\hat p\). State the interval in context and, if useful, express the margin as percentage points.

Worked Example: A 95% Interval From a Service Survey

A hypothetical benefits administrator surveys 200 eligible members about whether they would select an optional online service. Of the 200 members, 124 say yes. Calculate a 95% Wald confidence interval for the population proportion who would select the service. Report the margin of error and endpoints to four decimal places.

Method and estimate. Treat each response as an independent Bernoulli observation with a common population probability \(p\). The observed number of successes is \(x=124\), the sample size is \(n=200\), and the number of failures is \(200-124=76\). Thus,

\hat p=\frac{124}{200}=0.62.

The observed success and failure counts, 124 and 76, both exceed 10. For 95% confidence, use \(z_{0.975}=1.960\). The estimated standard error is

\widehat{\operatorname{SE}}(\hat p) =\sqrt{\frac{0.62(1-0.62)}{200}} =\sqrt{\frac{0.62(0.38)}{200}} =\sqrt{\frac{0.2356}{200}} =\sqrt{0.001178} \approx 0.0343220.

The margin of error, retaining unrounded digits for the calculation, is

1.960(0.0343220)\approx 0.0672711.

Therefore, the endpoints are \(0.62-0.0672711=0.5527289\) and \(0.62+0.0672711=0.6872711\). Rounded to four decimal places, the interval is \((0.5527,0.6873)\), and the margin of error is \(0.0673\).

Interpretation and check. The sample estimate is 62%, and the margin of error is about 6.73 percentage points. The interval is centered at 0.62, and its width is approximately \(2(0.0672711)=0.1345422\), as it should be. The common mistake this calculation avoids is taking \(\sqrt{0.62(0.38)}\approx0.4854\) as the standard error; that quantity must still be divided by \(\sqrt{200}\), equivalently by dividing the product inside the square root by 200.

Worked Example: A 90% Interval for a Training Completion Rate

A hypothetical pension-administration team reviews 300 randomly selected training records. Eighty-four records show that the required module was completed. Calculate a 90% Wald confidence interval for the completion proportion, and report the margin and endpoints to four decimal places.

Method and estimate. Use the binomial model with \(x=84\), \(n=300\), and \(n-x=216\). The sample proportion is

\hat p=\frac{84}{300}=0.28.

Both observed counts are at least 10. For a two-sided 90% interval, \(z_{0.95}=1.645\). The estimated standard error is

\widehat{\operatorname{SE}}(\hat p) =\sqrt{\frac{0.28(0.72)}{300}} =\sqrt{\frac{0.2016}{300}} =\sqrt{0.000672} \approx0.02592296.

The margin of error is

1.645\sqrt{\frac{0.28(0.72)}{300}} \approx1.645(0.02592296) \approx0.0426433.

Using that unrounded margin, the lower endpoint is \(0.28-0.0426433=0.2373567\), and the upper endpoint is \(0.28+0.0426433=0.3226433\). Rounded to four decimal places, the interval is \((0.2374,0.3226)\); the margin rounds to \(0.0426\).

Interpretation and check. The estimated completion proportion is 28%, with a margin of about 4.26 percentage points. The interval is symmetric around 0.28 before rounding. The mistake to avoid is rounding the margin too early and then using that rounded number to produce endpoints; retain the unrounded calculation until the final rounding step.

Worked Example: A 99% Interval for an Election in a Hypothetical Association

A hypothetical professional association asks 1,600 randomly selected members whether they support a proposed change to its continuing-education rules. A total of 1,360 support it. Calculate a 99% Wald confidence interval for the support proportion. Report the standard error and margin to six decimal places and the endpoints to four decimal places.

Method and estimate. Here \(x=1360\), \(n=1600\), and the failure count is \(1600-1360=240\). The estimated proportion is

\hat p=\frac{1360}{1600}=0.85.

Both observed counts exceed 10. For 99% confidence, use \(z_{0.995}=2.576\). The estimated standard error is

\widehat{\operatorname{SE}}(\hat p) =\sqrt{\frac{0.85(0.15)}{1600}} =\sqrt{\frac{0.1275}{1600}} =\sqrt{0.0000796875} \approx0.00892677.

The margin of error, calculated from the unrounded standard error, is

2.576\sqrt{\frac{0.1275}{1600}} \approx 2.576(0.00892677) \approx 0.0229954.

The lower endpoint is \(0.85-0.0229954=0.8270046\), and the upper endpoint is \(0.85+0.0229954=0.8729954\). To four decimal places, the interval is \((0.8270,0.8730)\). To six decimal places, the standard error is 0.008927 and the margin is 0.022995.

Interpretation and check. The sample estimate is 85%, and the margin is about 2.30 percentage points. The endpoints average to 0.85 before rounding. The common mistake here is rounding the standard error before multiplying by 2.576; using the unrounded value gives a margin of approximately 0.0229954, which rounds to 0.0230 at four decimal places.

Interpretation and Common Traps

A 95% confidence procedure does not mean there is a 95% probability that the fixed population proportion lies in this particular interval. In the repeated-sampling interpretation, if the same method were applied to many samples under the model, approximately 95% of the resulting intervals would contain the population proportion. For the single observed sample, the interval is a range of plausible values produced by that procedure.

The margin of error is measured in proportion units. For example, a margin of 0.0426 is 4.26 percentage points, not a 4.26% relative increase from the sample estimate. When reporting a result in context, distinguish the estimated proportion from the uncertainty around it.

Omitting the sample-size division. The standard error is \(\sqrt{\hat p(1-\hat p)/n}\), not merely \(\sqrt{\hat p(1-\hat p)}\).
Using the wrong critical value. For a two-sided interval, use \(z_{1-\alpha/2}\), not the one-sided quantile \(z_{1-\alpha}\).
Confusing the estimate and the margin. The interval is centered at \(\hat p\); the margin is the distance from that center to either endpoint.
Rounding too soon. Keep full calculator precision through the margin and endpoint calculations, then round the reported values.
Calling the result exact. The Wald interval relies on a normal approximation and its quality depends on the data and model conditions.

Key takeaway: For a one-proportion Wald interval, calculate \(\hat p=x/n\), estimate its standard error with \(\sqrt{\hat p(1-\hat p)/n}\), multiply by the appropriate two-sided normal critical value, and add and subtract the margin from \(\hat p\). State the confidence level, give the result in context, and preserve unrounded values until the final step.

Check Your Understanding

For each problem, calculate the Wald interval using the stated confidence level. Unless specified otherwise, report the margin of error and interval endpoints to four decimal places.

A hypothetical call center checks 120 randomly selected claims, and 48 were resolved during the first contact. Find the 90% Wald interval for the proportion resolved at first contact. (Answer: margin of error \(0.0736\); interval \((0.3264,0.4736)\).)
A random sample of 250 hypothetical retirement-plan participants includes 75 who use automatic contribution increases. Find the 95% Wald interval for the population proportion. (Answer: margin of error \(0.0568\); interval \((0.2432,0.3568)\).)
In a hypothetical audit of 240 benefit statements, 204 have no detected data-entry error. Find the 95% Wald interval for the proportion with no detected error. (Answer: margin of error \(0.0452\); interval \((0.8048,0.8952)\).)
A hypothetical association surveys 500 members, and 310 favor an annual meeting format change. Find the 98% Wald interval, using \(z_{0.99}=2.326\). (Answer: margin of error \(0.0505\); interval \((0.5695,0.6705)\).)
A sample of 180 hypothetical policyholders includes 27 who elect a new paperless option. Find the 90% Wald interval and state the estimated standard error to four decimal places. (Answer: estimated standard error \(0.0266\); margin of error \(0.0438\); interval \((0.1062,0.1938)\).)

What You'll Learn