Deriving the One-Proportion Z Interval

From a Sample Proportion to an Interval

A sample proportion is an estimate of a population proportion, but a single estimate does not show how much it might vary from sample to sample. A confidence interval combines the estimate with a measure of that sampling variability. The one-proportion z interval is obtained by starting with the approximate normal distribution of a sample proportion, standardizing it, and solving for the unknown population proportion.

The derivation has an important wrinkle: the standard error of the sample proportion depends on the unknown population proportion $p$. To make an interval from observed data, the usual one-proportion z procedure estimates that standard error by substituting the observed sample proportion $\hat{p}$ for $p$. The resulting interval is useful under suitable conditions, but this substitution makes its coverage approximate rather than exact.

The Model and Its Sampling Variability

Suppose each sampled individual has a binary outcome: the outcome either has the characteristic of interest or it does not. Let $X_i=1$ when the $i$th sampled individual has the characteristic and $X_i=0$ otherwise. The sample proportion is the fraction of the $n$ observations with $X_i=1$, so $\hat{p}$ estimates the population proportion $p$.

Model: For the standard one-proportion interval, the observations are modeled as independent Bernoulli trials with a common probability $p$ of success. Under this model, $\hat{p}$ has mean $p$ and standard error $\sqrt{p(1-p)/n}$. When a normal approximation to its sampling distribution is adequate, $\hat{p}$ is approximately normal with that mean and standard error.

The standard error in this model describes the typical sample-to-sample variation in $\hat{p}$. It depends on $p$, which is the quantity the interval is meant to estimate. The earlier material on sampling distributions establishes why a standardized sample proportion can be approximated by a standard normal random variable when the approximation is suitable.

The model presumes a meaningful binary outcome, a common population probability for the observations, and sampling that supports independence. The normal approximation is an additional requirement: it is not guaranteed merely because a sample proportion has been calculated. The formal success-failure check is treated separately; here, worked examples state when the approximation is being assumed adequate.

Standardize, Then Use a Critical Value

If $p$ were known, the model-based standard error would be $\sqrt{p(1-p)/n}$. Standardizing $\hat{p}$ by subtracting its mean and dividing by this standard error gives a quantity that is approximately standard normal:

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

For a chosen confidence level $1-\alpha$, let $z^*$ be the positive standard normal critical value that leaves probability $\alpha/2$ in each tail. Thus, for a standard normal variable $Z$, $P(-z^*\leq Z\leq z^*)=1-\alpha$. The earlier material on critical values supplies the value of $z^*$ for a chosen confidence level.

Applying this central probability statement to the standardized sample proportion gives an approximate statement about repeated samples:

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

This expression is the starting point, not yet a usable interval calculated from the data: its standard error still contains the unknown $p$. First, it helps to see how the probability bounds lead to limits for $p$ when the standard error is treated as a fixed positive quantity.

Rearranging the Bounds and Estimating the Standard Error

Write $SE=\sqrt{p(1-p)/n}$ temporarily. For any fixed $p$ with $0<p<1$, this standard error is positive, so multiplying all parts of the inequality by it does not reverse either inequality:

-z^*SE\leq\hat{p}-p\leq z^*SE.

The left-hand inequality gives an upper bound for $p$, and the right-hand inequality gives a lower bound. In detail, $-z^*SE\leq\hat{p}-p$ implies $p\leq\hat{p}+z^*SE$; and $\hat{p}-p\leq z^*SE$ implies $p\geq\hat{p}-z^*SE$. Together they give:

\hat{p}-z^*SE\leq p\leq\hat{p}+z^*SE.

This is the algebraic shape of the interval: an estimate plus or minus a critical value times a standard error. But the expression still cannot be evaluated directly because $SE$ contains $p$. The usual z interval estimates the standard error by substituting $\hat{p}$ for $p$:

\widehat{SE}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.

Using this estimated standard error produces the one-proportion z interval, also called the Wald interval:

Definition: For a sample proportion $\hat{p}$, sample size $n$, and critical value $z^*$, the one-proportion z interval for the population proportion $p$ is $$ \left[ \hat{p}-z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \ \hat{p}+z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right]. $$ The endpoints are included. This interval uses an estimated standard error and a normal approximation, so its confidence level is approximate.

The substitution is a practical plug-in step, not a claim that the unknown $p$ equals $\hat{p}$. It makes the standard error calculable from the sample. Because it is an estimate, the resulting interval does not have exactly the stated coverage for every possible $p$ and $n$.

Worked Examples

Worked Example: A 95% Interval Under the Model

Suppose a hypothetical simple random sample of 400 people from a large population finds that 232 have a specified characteristic. Assume the binary observations are independent with a common population proportion, and that the normal approximation is adequate. Find the 95% one-proportion z interval.

The sample proportion is $\hat{p}=232/400=0.5800$. For a 95% central confidence level, use $z^*=1.96$. The estimated standard error is $$ \widehat{SE}=\sqrt{\frac{0.5800(1-0.5800)}{400}} =\sqrt{\frac{0.2436}{400}} =\sqrt{0.000609} \approx 0.02468. $$

The margin in this derivation is $z^*\widehat{SE}$, which is $$ 1.96(0.02468)\approx 0.04837. $$

Therefore, the interval is $$ [0.5800-0.04837,\ 0.5800+0.04837] =[0.5316,\ 0.6284]. $$

On the assumptions stated, the 95% procedure gives an approximate interval from 0.5316 to 0.6284 for the population proportion, or about 53.16% to 62.84%. It is the procedure’s repeated-sampling coverage that is described by the confidence level, not a probability assigned to this fixed $p$ after the interval is observed.

Worked Example: A 90% Interval With a Different Critical Value

In a hypothetical independent sample of 250 eligible items, 40 meet a specified binary criterion. Assume the common-proportion model applies and the normal approximation is adequate. Construct a 90% one-proportion z interval.

Here $\hat{p}=40/250=0.1600$. For a 90% central confidence level, $z^*\approx1.645$. The plug-in standard error is $$ \widehat{SE}=\sqrt{\frac{0.1600(1-0.1600)}{250}} =\sqrt{\frac{0.1344}{250}} =\sqrt{0.0005376} \approx0.02319. $$

The margin is $$ 1.645(0.02319)\approx0.03814. $$

The interval is $$ [0.1600-0.03814,\ 0.1600+0.03814] =[0.1219,\ 0.1981]. $$

Thus, under the model and approximation, the 90% interval is approximately 0.1219 to 0.1981, or 12.19% to 19.81%, for the population proportion. Using a different confidence level changes $z^*$, and therefore changes the width even though the sample estimate and its estimated standard error remain the same.

Worked Example: A 99% Interval and Its Width

A hypothetical random sample of 500 independently observed units has 360 units with a specified characteristic. Suppose the observations share a common population probability and the normal approximation is adequate. Derive the 99% interval.

The observed proportion is $\hat{p}=360/500=0.7200$. For a 99% central confidence level, $z^*\approx2.576$. The estimated standard error is $$ \widehat{SE}=\sqrt{\frac{0.7200(1-0.7200)}{500}} =\sqrt{\frac{0.2016}{500}} =\sqrt{0.0004032} \approx0.02008. $$

The margin is $$ 2.576(0.02008)\approx0.05173. $$

Consequently, the interval is $$ [0.7200-0.05173,\ 0.7200+0.05173] =[0.6683,\ 0.7717]. $$

The estimated population proportion is therefore bounded by about 0.6683 and 0.7717 in this 99% interval. Its margin is larger than it would be at a lower confidence level with the same data because the larger critical value multiplies the same estimated standard error.

Why the Derivation Matters

The derivation explains each part of the interval rather than treating the formula as a recipe. The center, $\hat{p}$, is the observed estimate. The estimated standard error describes its sampling variability under the binary-outcome model. The critical value sets how many estimated standard errors extend on either side of the estimate. This structure also shows why an interval can change when the confidence level changes even if the sample itself does not.

The underlying standardized expression uses $\sqrt{p(1-p)/n}$, not $\sqrt{\hat{p}(1-\hat{p})/n}$. The latter appears only when the unknown standard error is estimated for practical use. Thus the resulting interval is not an exact inversion of a known standard normal distribution. It is a normal-approximation interval with a plug-in estimate of variability.

Common Mistake

A frequent error is to solve the probability inequality for $p$ while reversing which side gives the lower and upper bounds. For example, from $-a\leq\hat{p}-p$, one might incorrectly conclude $p\geq\hat{p}-a$. Rearranging carefully instead gives $p\leq\hat{p}+a$. The other side, $\hat{p}-p\leq a$, gives $p\geq\hat{p}-a$. The complete result is therefore $\hat{p}-a\leq p\leq\hat{p}+a$, not the reversed bounds.

Another mistake is to treat the plug-in standard error as though it made the interval exact. It does not: both the normal approximation and the estimate of the standard error contribute to the approximation. The one-proportion z interval should be used only when the sampling model and approximation are appropriate; checking the success-failure condition is a separate step.

Key takeaway: Standardization leads to a central probability statement; rearranging it places $p$ between the estimate minus and plus a critical-value multiple of the standard error. Replacing the unknown standard error with its estimate based on $\hat{p}$ gives the one-proportion z interval, with approximate coverage under suitable conditions.

Check Your Understanding

Use the model and derivation above to answer these questions.

What is the model-based standard error of $\hat{p}$ before the unknown population proportion is replaced by an estimate?
Why must the standard error be positive when multiplying through the standardized inequality?
Starting from $-z^*SE\leq\hat{p}-p\leq z^*SE$, which inequality gives the upper bound for $p$, and which gives the lower bound?
What substitution makes the one-proportion z interval calculable from sample data?
Why is the coverage of the resulting interval approximate rather than exact?