Reading Proportion Interval Software Output

Reading an Interval as a Set of Decisions

Software can display a confidence interval in one compact line, but that line combines several choices: which data were analyzed, which confidence level was requested, and which interval method was used. Reading the endpoints alone does not reveal all of this. A careful check links the displayed estimate and interval back to the sample count, sample size, and method.

This tutorial focuses on intervals for one population proportion. The sample proportion is the observed fraction of sampled individuals with the outcome of interest. For a standard one-proportion $z$ interval, software uses that sample proportion in both the estimate and the estimated standard error. Other methods, such as the Wilson score method, use a different calculation, so a correct check must first identify the method.

Definition: For $x$ observed successes among $n$ sampled individuals, the sample proportion is $\hat{p}=x/n$. A software interval output should be read alongside its estimate, confidence level, lower and upper endpoints, method, and the data or summary counts used to calculate it.

A common output layout might include fields labelled “Successes,” “Trials,” “Sample proportion,” “Confidence level,” “Lower,” “Upper,” and “Method.” “Successes” means the count $x$ of observations with the specified outcome; “Trials” or “Observations” usually means the sample size $n$. The method label matters: the Wald interval and Wilson score interval can have different endpoints even when they use the same $x$, $n$, and confidence level.

The Wald Interval and Its Output Fields

The one-proportion Wald interval is the sample proportion plus or minus a standard-normal critical value times its estimated standard error. The method uses the sample proportion to estimate the standard error. Before using it, the data should support the usual normal approximation: the observations are from a random sample or an appropriate randomized design, observations are independent (or the sampling fraction is sufficiently small when sampling without replacement), and the sample has enough successes and failures for the approximation. A common check is that $n\hat{p}$ and $n(1-\hat{p})$ are each at least 10.

Formula: For a two-sided Wald interval with confidence level $C$, let $z^*$ be the standard-normal critical value with central probability $C$. Then $$ \hat{p}\ \pm\ z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. $$ The reported endpoints are $\hat{p}-z^*SE$ and $\hat{p}+z^*SE$, where $SE=\sqrt{\hat{p}(1-\hat{p})/n}$.

For instance, the usual critical values are approximately 1.645 for a two-sided 90% interval and 1.960 for a two-sided 95% interval. The stated confidence level should agree with the critical value used. Also check the scale: software may show proportions between 0 and 1 or percentages between 0% and 100%. Converting between the two is a multiplication or division by 100, not a change in the underlying interval.

Worked Example: Checking a 95% Wald Interval

Suppose hypothetical software reports 84 successes in 200 observations, an estimate of 0.420, and a 95% Wald interval from 0.352 to 0.488. Check the estimate and endpoints.

Check the estimate. The sample proportion from the counts is $$ \hat{p}=\frac{x}{n}=\frac{84}{200}=0.420. $$ This agrees with the displayed estimate.

Check the conditions and standard error. Assume, for this hypothetical example, the observations came from an appropriate random sample and are independent. The success and failure counts are $84$ and $200-84=116$, both at least 10. Thus the usual success-failure check is met. The estimated standard error is $$ SE=\sqrt{\frac{0.420(1-0.420)}{200}} =\sqrt{\frac{0.2436}{200}} =\sqrt{0.001218} \approx 0.03490. $$

Check the interval. For 95% confidence, use $z^*=1.960$. The margin from the estimate is $$ 1.960(0.03490)\approx 0.06840. $$ Therefore the endpoints are $0.420-0.06840=0.3516$ and $0.420+0.06840=0.4884$. Rounded to three decimal places, these are 0.352 and 0.488. The reported interval is consistent with the Wald calculation, allowing for rounding.

A Short Derivation for Verifying the Endpoints

The endpoint check follows directly from the form of the Wald interval. Starting with its center $\hat{p}$ and half-width $z^*SE$, subtracting the half-width gives the lower endpoint and adding it gives the upper endpoint:

L=\hat{p}-z^*SE,\qquad U=\hat{p}+z^*SE.

Adding the two equations cancels the equal-and-opposite margins, while subtracting them doubles the margin:

\frac{L+U}{2}=\hat{p},\qquad \frac{U-L}{2}=z^*SE.

These identities give two quick checks. The midpoint of a Wald interval should match the sample proportion, up to rounding. Its half-width should match the critical value times the estimated standard error. If either check fails by more than rounding can explain, inspect the method, confidence level, counts, and scale before concluding that the software or calculation is wrong. These identities are specific to the symmetric Wald interval; they should not be imposed on intervals from methods that are not centered at $\hat{p}$.

Worked Example: Checking the Confidence Level and Scale

Hypothetical software displays 132 successes out of 240 observations, an estimate of 55.0%, and a 90% Wald interval from 49.7% to 60.3%. Verify the output in percentage units.

The estimate is $132/240=0.550$, or 55.0%. The sample contains 132 successes and $240-132=108$ failures. Assume the observations satisfy random sampling and independence; both counts exceed 10, so the usual approximation check is met.

For 90% confidence, the two-sided critical value is approximately 1.645, not 1.960. The standard error and margin of error are $$ SE=\sqrt{\frac{0.550(0.450)}{240}} =\sqrt{0.00103125} \approx 0.03211, $$ $$ ME=1.645(0.03211)\approx 0.05283. $$

In proportion units the endpoints are $0.550-0.05283=0.4972$ and $0.550+0.05283=0.6028$. Multiplying by 100 gives approximately 49.72% and 60.28%, which round to 49.7% and 60.3%. The estimate, confidence level, method, and endpoints are consistent. Using the 95% critical value here would produce a wider interval and would not verify this reported 90% interval.

When the Method Is Not Wald

A software label such as “score,” “Wilson,” or “Wilson score” signals a different interval calculation from the Wald formula above. For the Wilson score interval, the lower and upper endpoints are not generally equally far from $\hat{p}$. Consequently, the midpoint check for a Wald interval does not apply. The reported method is part of the output, not an optional detail.

For a quick verification of a Wilson interval, one can substitute the displayed $\hat{p}$, $n$, and $z^*$ in its endpoint formula. For clarity, the center and half-width forms are shown below; their sum and difference give the upper and lower endpoints.

Wilson score calculation: Define $D=1+(z^*)^2/n$. The interval center is $C_W=[\hat{p}+(z^*)^2/(2n)]/D$, and its half-width is $H_W=z^*\sqrt{\hat{p}(1-\hat{p})/n+(z^*)^2/(4n^2)}/D$. The endpoints are $C_W-H_W$ and $C_W+H_W$.

Worked Example: Verifying a Wilson Score Interval

Suppose hypothetical output reports 8 successes in 40 observations, an estimate of 0.200, a 95% confidence level, and a Wilson score interval from 0.105 to 0.348. Check the estimate and interval.

The estimate is $8/40=0.200$. Assume the observations came from an appropriate random sample and are independent. The sample has 8 successes and 32 failures, so both counts are at least 5, though the stricter check of at least 10 successes is not met. This cautions against treating a Wald approximation as automatically suitable; here the reported method is Wilson, so we verify that method rather than applying the Wald endpoint formula.

Using $z^*=1.960$, first calculate $$ D=1+\frac{1.960^2}{40}=1.09604. $$ The Wilson center is $$ C_W=\frac{0.200+1.960^2/(2\cdot40)}{1.09604} =\frac{0.24802}{1.09604} \approx 0.2263. $$

The half-width is $$ H_W=\frac{1.960\sqrt{0.200(0.800)/40+1.960^2/(4\cdot40^2)}}{1.09604} =\frac{1.960\sqrt{0.004000+0.0006003}}{1.09604} \approx 0.1213. $$ Thus the endpoints are $0.2263-0.1213=0.1050$ and $0.2263+0.1213=0.3476$. Rounded to three decimal places, the interval is 0.105 to 0.348, consistent with the software output. Its midpoint is not 0.200, as expected for this Wilson interval.

A Practical Reading Sequence

A consistent sequence helps prevent a plausible-looking interval from being paired with the wrong method or inputs. Read the output from the data outward: first identify the count and sample size, then confirm the estimate, confidence level and method, and finally reproduce the endpoints. If the output omits counts, calculate the estimate from the displayed proportion and use the displayed sample size where possible; note that rounded proportions can cause small discrepancies.

Identify the data.
Locate the success count $x$, sample size $n$, and outcome defining a success. Check that the estimate agrees with $x/n$.

Read the settings.
Find the confidence level, interval method, and units. Convert percentages to proportions or proportions to percentages as needed.

Check the calculation.
Use the formula for the stated method and the critical value for the stated confidence level. Compare your calculated endpoints with those displayed, allowing for rounding.

Common Mistakes and Why This Check Matters

Using the wrong critical value. A calculation using 1.960 does not verify a 90% two-sided interval; the 90% critical value is about 1.645.
Assuming every interval is Wald. A Wilson interval need not be centered at the sample proportion. Apply the formula named by the software.
Mixing proportions and percentages. An estimate of 0.42 is the same fraction as 42%, but endpoints must be compared on the same scale.
Treating a small rounding difference as a method error. Software may round displayed estimates or endpoints. Recalculate with full precision when available and compare at the displayed precision.
Ignoring the sample counts. The same displayed estimate can come from different sample sizes and therefore produce different standard errors and interval widths.

The point of verification is not to replace software with hand calculation in every analysis. It is to ensure that the output corresponds to the intended sample, level, and method. This makes it easier to catch settings errors, transcription mistakes, and mismatches between a reported method and the calculation used to interpret the output.

Key takeaway: A proportion interval is verifiable only when its estimate, confidence level, endpoints, and method are read together. Recalculate using the stated method, check the scale, and treat small differences in rounded output appropriately.

Check Your Understanding

Use the formulas and checks in this tutorial to answer each question.

Software reports 45 successes in 100 observations, an estimate of 0.45, and a 95% Wald interval. What estimate should the output show, and which critical value should be used?
A reported 90% Wald interval is checked using $z^*=1.960$. What setting should be checked, and why?
Why is it not a valid check to require the midpoint of a Wilson score interval to equal $\hat{p}$?
Software displays endpoints of 0.28 and 0.44. Express them as percentages and state what must be checked before comparing them with output in percent units.
What two quantities can be checked using the midpoint and half-width of a Wald interval?