Estimate a population proportion

Try

An original synthetic random sample records 120 successes in 300 independent binary trials. Before calculating uncertainty, define success and the target process. The sample proportion is 120/300=0.4. A different random sample could yield a different proportion even if the process probability stayed unchanged.

Predict

Predict why estimated uncertainty becomes smaller with more independent observations. Also predict what goes wrong with only a few successes or failures. A normal curve is symmetric and continuous, whereas a sparse binary count is discrete, bounded and often strongly skewed. Numerical computability does not establish approximation quality.

Experiment

For the sample, the estimated standard error is √[0.4(0.6)/300]=√0.0008≈0.028284. Using the supplied 95% normal critical value z*=1.96 gives margin about 0.055437 and interval [0.344563,0.455437]. Round reported endpoints sensibly, for example [0.345,0.455], while retaining extra digits during calculation.

Both observed counts, 120 successes and 180 failures, exceed the common introductory threshold of 10. This checks a distributional approximation, not random selection or independence. Add a design note explaining how the trials were obtained before interpreting the interval for a population.

Understand

The sample proportion p̂ is the observed success count divided by n. For independent Bernoulli trials with common probability p, its standard deviation is √[p(1−p)/n]. Since p is unknown, the basic normal or Wald interval substitutes p̂ to estimate that spread, then uses p̂ ± z√[p̂(1−p̂)/n]. Here z is a standard-normal quantile chosen for the desired central coverage.

This plug-in interval is an introductory approximation, not a universally reliable proportion method. It can have poor coverage near zero or one, especially with small samples. With no successes it estimates zero variability, an unreasonable assertion of certainty. Boundaries outside [0,1] are another warning. Simply clipping an invalid endpoint to zero or one does not repair the method’s coverage properties.

A common classroom condition requires at least 10 observed successes and 10 observed failures. Treat it as a practical approximation check, not a theorem guaranteeing exact coverage. Score-based or suitable exact methods can be preferable, particularly for sparse data, but their construction and interpretation must be taught rather than replacing an endpoint without explanation.

Sampling without replacement from a finite population produces dependence. When a simple random sample is no more than about 10% of the population, the independent-trial variance can be a useful approximation; otherwise consider the design and finite-population adjustment. Repeated measurements within people or clusters are another source of dependence that a large total n does not eliminate.

The interval estimates the target proportion only insofar as the sample represents that target. A random sample supports population generalization under the design; a randomized experiment supports its own treatment comparison. Nonresponse, inconsistent outcome definitions and selective exclusions can invalidate a polished numerical report.

Apply

Create a four-part solution: define p, verify design and counts, calculate the interval, and interpret it in context. Then contrast the well-populated example with zero successes out of eight. Explain why the second needs a different justified method rather than an announcement that the true rate is zero.

Continue

Next plan a sample size before collecting outcomes. A planning bound trades precision against effort, but still depends on a credible collection process.