Constructing a One-Proportion z-Interval by Hand

Putting the Parts of the Interval Together

In Structure of a One-Proportion z-Interval, you saw that the interval is centered at the sample proportion and extends a margin of error in each direction. In Finding Critical Values \(z^*\) for Common Confidence Levels, you found that a 95% confidence level uses \(z^*=1.96\). Now we will combine those pieces and calculate an interval by hand for 46 successes in a sample of 176.

The sample proportion is the point estimate of the population proportion \(p\). For a one-proportion \(z\)-interval, its estimated standard error uses \(\hat{p}\), not the unknown \(p\). Multiply that standard error by \(z^*\) to get the margin of error, then subtract and add the margin to \(\hat{p}\).

Formula: For \(x\) successes in a sample of size \(n\), calculate \(\hat{p}=x/n\), then use \(\hat{p}\pm z^*\sqrt{\hat{p}(1-\hat{p})/n}\). The endpoints are \(\hat{p}-\text{margin of error}\) and \(\hat{p}+\text{margin of error}\).

A by-hand calculation is more than substituting numbers. First check whether the interval method is appropriate; then show how each number is obtained. Keeping extra digits during the calculation helps avoid endpoints that differ just because an intermediate value was rounded too soon.

Check Conditions Before Calculating

The conditions for a one-proportion \(z\)-interval were introduced in Structure of a One-Proportion z-Interval and developed in the earlier tutorials on the 10% condition and the Large Counts condition. Check them in the situation at hand. For a random sample taken without replacement, the sample should be no more than 10% of the population. For the interval’s Normal-based method, check the observed number of successes and failures.

Conditions:

Random: The data come from a random sample or a suitable random process.
10% condition: If sampling without replacement from a finite population of size \(N\), verify \(n\leq0.10N\).
Large Counts: The observed numbers of successes and failures are each at least 10: \(x\geq10\) and \(n-x\geq10\).

For this interval, the Large Counts check uses the observed counts. That differs from checking a Normal model using a specified population proportion \(p\), as in earlier sampling-distribution work. Here \(p\) is not known, so the interval estimates it using the observed data.

Worked Example: Constructing the Interval for 46 out of 176

Suppose a district team wants to estimate the proportion of households in its district that have a home compost bin. The team takes a random sample of 176 households from a district containing 4,000 households; 46 sampled households have a compost bin. Construct a 95% one-proportion \(z\)-interval by hand, using \(z^*=1.96\).

State.
Let \(p\) be the proportion of all households in this district that have a home compost bin. We will estimate \(p\) with a 95% confidence interval.

Plan and check conditions.
The problem states that the 176 households were selected randomly. Because the sample is taken without replacement, check the 10% condition: \(0.10(4000)=400\), and \(176\leq400\). There are 46 successes and \(176-46=130\) failures, both at least 10, so the Large Counts condition is met. A one-proportion \(z\)-interval is appropriate.

Do.
Calculate the sample proportion, estimated standard error, margin of error, and endpoints, as shown below.

Conclude.
Interpret the interval as an estimate of the district’s population proportion, with the confidence level referring to the long-run performance of the method.

Find the point estimate. Divide the number of successes by the sample size:

\hat{p}=\frac{x}{n}=\frac{46}{176}\approx0.26136

Thus, about 26.1% of the sampled households have a compost bin. This is the center of the interval, not the population proportion itself.

Find the estimated standard error. Substitute \(\hat{p}\) and \(n=176\) into the standard error formula:

SE_{\hat{p}} =\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =\sqrt{\frac{(0.26136)(1-0.26136)}{176}} \approx\sqrt{0.0010969} \approx0.03312

Find the margin of error. Multiply the estimated standard error by the supplied critical value \(z^*=1.96\):

\text{margin of error} =z^*SE_{\hat{p}} =(1.96)(0.033119\ldots) \approx0.06491

Calculate the endpoints. Subtract the margin of error for the lower endpoint and add it for the upper endpoint. Use the unrounded sample proportion and standard error in these calculations:

\begin{aligned} \text{Lower endpoint} &=\frac{46}{176}-1.96\sqrt{\frac{(46/176)(130/176)}{176}} \approx0.19645,\\ \text{Upper endpoint} &=\frac{46}{176}+1.96\sqrt{\frac{(46/176)(130/176)}{176}} \approx0.32628. \end{aligned}

The interval is approximately \((0.196,\,0.326)\), or about 19.6% to 32.6%. In context, we are 95% confident that between about 19.6% and 32.6% of all households in this district have a home compost bin. The interval estimates a population proportion, so its endpoints are proportions (or percentages), not counts of households.

The 95% confidence level describes the method: if we repeatedly took random samples of 176 households in the same way and constructed an interval each time, about 95% of those intervals would capture the district’s true proportion, provided the method’s conditions hold. It does not mean there is a 95% probability that this particular, already-calculated interval contains the fixed population proportion.

Keep Precision Until the End

In hand calculations, it is useful to write rounded intermediate values for readability, but use more digits—or the original fraction—when finding the final endpoints. If you replace \(\hat{p}\) with a rounded value too early, a small change can occur in the standard error, margin of error, and endpoints. That difference is usually minor, but carrying precision makes the calculation more reliable.

One way to check the estimated standard error is to write it directly in terms of the success and failure counts. Since \(\hat{p}=x/n\) and \(1-\hat{p}=(n-x)/n\), the estimated standard error can also be calculated as \(\sqrt{x(n-x)/n^3}\). This is the same formula expressed using counts.

Worked Example: Checking the Arithmetic with Counts

Use the compost-bin sample—46 successes in 176 households—to verify the estimated standard error and interval without rounding \(\hat{p}\) at the start.

There are \(176-46=130\) failures. Using the count form of the standard error gives:

SE_{\hat{p}} =\sqrt{\frac{x(n-x)}{n^3}} =\sqrt{\frac{(46)(130)}{176^3}} =\sqrt{\frac{5980}{5{,}451{,}776}} \approx\sqrt{0.00109689} \approx0.033119

This matches the standard error found from \(\hat{p}\). The margin of error is \((1.96)(0.033119)\approx0.06491\). Keeping the fraction for the center, the endpoints are:

\frac{46}{176}-0.064913\ldots\approx0.19645, \qquad \frac{46}{176}+0.064913\ldots\approx0.32628

Both methods give the same interval, approximately \((0.196,0.326)\). This is a useful arithmetic check: calculate the standard error once from \(\hat{p}\) and once from the success and failure counts. The two results should agree apart from rounding.

Worked Example: Constructing and Interpreting Another Interval

Imagine a community program wants to estimate the proportion of its 3,000 members who use a refillable water bottle during a typical week. In a random sample of 200 members, 71 report doing so. Construct a 95% one-proportion \(z\)-interval using \(z^*=1.96\).

Check conditions. The sample is stated to be random. For sampling without replacement, \(0.10(3000)=300\), and \(200\leq300\), so the 10% condition is met. There are 71 successes and \(200-71=129\) failures, each at least 10; the Large Counts condition is also met.

Calculate the estimate and standard error.

\hat{p}=\frac{71}{200}=0.355, \qquad SE_{\hat{p}}=\sqrt{\frac{(0.355)(1-0.355)}{200}} =\sqrt{0.001144875} \approx0.033836

Calculate the margin of error and endpoints.

\text{margin of error}=(1.96)(0.033836)\approx0.06632

0.355-0.06632\approx0.28868, \qquad 0.355+0.06632\approx0.42132

The 95% confidence interval is approximately \((0.289,0.421)\). We are 95% confident that about 28.9% to 42.1% of all members of this community program use a refillable water bottle during a typical week. As in the earlier example, the estimate is \(\hat{p}\), the margin of error is \(z^*\) times the estimated standard error, and the endpoints are formed by subtracting and adding that margin.

Common Mistakes and AP Exam Tips

Using the wrong standard error. For a confidence interval, use \(\sqrt{\hat{p}(1-\hat{p})/n}\). Do not substitute a claimed population proportion \(p\), which belongs in calculations for a sampling distribution under a specified model.
Using the wrong failure count. The failure count is \(n-x\). In the 46-out-of-176 example it is \(176-46=130\), not \(46\), and not \(176\).
Adding the margin to the lower endpoint. The lower endpoint is \(\hat{p}-\text{margin of error}\); the upper endpoint is \(\hat{p}+\text{margin of error}\).
Skipping conditions. A numerical interval alone does not show why the method is suitable. State the random-sample evidence, check the 10% condition when needed, and show that both observed counts are at least 10.
Rounding too early. Keep extra digits in \(\hat{p}\), the standard error, and the margin of error, then round the endpoints for reporting. If intermediate values are rounded, small endpoint differences may result.
Giving an incomplete interpretation. Identify the population proportion and the characteristic, give the interval in context, and explain confidence as a long-run property of the method—not as a probability that a fixed interval contains \(p\).

AP Exam Tip: Make the calculation easy to follow: show \(\hat{p}=x/n\), the estimated standard error, the margin of error, and both endpoints. Then state the interval and interpret it in context. A correct calculator-free result still needs the conditions and a complete conclusion.

Key Takeaway

To construct a one-proportion \(z\)-interval by hand, check the conditions, calculate \(\hat{p}\), find its estimated standard error, multiply by \(z^*\) for the margin of error, and subtract and add that margin. For 46 successes out of 176 with \(z^*=1.96\), the 95% interval is approximately \((0.196,0.326)\).

Key takeaway: Keep the estimate, estimated standard error, and margin of error distinct: \(\hat{p}=x/n\), \(SE_{\hat{p}}=\sqrt{\hat{p}(1-\hat{p})/n}\), and margin of error \(=z^*SE_{\hat{p}}\). Use \(\hat{p}\pm\text{margin of error}\), and interpret the resulting interval for the population proportion in context.

Check Your Understanding

Use the by-hand interval process for each question. Show your calculations and keep unrounded values until the endpoints.

For 46 successes in a sample of 176, calculate \(\hat{p}\) and the number of failures.
For that same sample, calculate the estimated standard error using \(\hat{p}=46/176\).
Using \(z^*=1.96\), calculate the margin of error and both endpoints for the 46-out-of-176 interval.
A random sample of 120 people is drawn without replacement from a population of 900. Does it meet the 10% condition? Explain.
In your own words, explain what 95% confidence means for the method used to construct an interval.