Conditions Check for a Proportion Confidence Interval Scenario

Why Check Conditions for a Hospital-Patient Interval?

A one-proportion \(z\)-interval estimates an unknown population proportion from sample data. Before using it, check whether the way patients were selected, the sample’s size relative to its source population, and the observed counts support the interval procedure. Here, the Large Counts check uses the sample’s observed successes and failures—not a proposed null proportion.

In Conditions Check for a Proportion Test Scenario, you applied the Random, 10%, and Large Counts conditions to a test. For an interval, the first two checks still concern how the data were collected and how the sample relates to its source population. The Large Counts check changes: as explained in Large Counts Condition for Confidence Intervals and Interval Versus Test Condition Checks Compared, use the observed counts.

Hospital settings make it especially important to name the population carefully. A random sample from one hospital’s records may represent patients in those records, but it does not automatically represent patients at other hospitals, or all patients with the same condition. Identify the source population before making a condition claim.

Conditions: Before using a one-proportion \(z\)-interval, check: Random: The data come from a random sample or an appropriate random process. 10%: If sampling without replacement from a finite population of size \(N\), verify \(n\leq0.10N\). Large Counts: Use the observed counts and verify \(n\hat{p}\geq10\) and \(n(1-\hat{p})\geq10\), equivalently \(x\geq10\) and \(n-x\geq10\).

How the Three Checks Work for an Interval

Random condition. Describe how the patients entered the sample. If the hospital randomly selected patients from a suitable list of records, that supports treating the sample as representative of the population covered by that list. Be precise about the scope: a list of patients discharged from one hospital during a particular quarter represents that frame, not every patient in the region. A group of patients who volunteered to answer a survey is not a random sample just because many people answered.

10% condition. When patients are sampled without replacement from a finite source population, compare the sample size \(n\) with 10% of that population’s size \(N\). This is a check supporting the approximate independence of observations. Use the actual population from which the sample was drawn. For example, if records were sampled from a particular hospital’s quarterly discharge list, use the size of that list—not a larger regional patient population.

n\leq0.10N

Large Counts condition. For a confidence interval, the check is based on the observed sample proportion \(\hat{p}\). If \(x\) patients have the characteristic defined as a success, then \(n\hat{p}=x\) is the observed number of successes and \(n(1-\hat{p})=n-x\) is the observed number of failures. Both counts must be at least 10 for the usual Normal-based interval check.

n\hat{p}=x\geq10 \qquad\text{and}\qquad n(1-\hat{p})=n-x\geq10

The word “success” is a label for the outcome of interest, not a judgment about whether the outcome is good. If the interval is about the proportion of patients who received a follow-up call, define receiving a call as success and not receiving one as failure. Count all sampled patients in one of those two categories.

A condition check is a brief argument. Name the sampling method or source population, show the relevant calculation, and state whether the condition is met. If any condition fails, explain that the usual one-proportion \(z\)-interval is not supported by all the stated checks. Passing the other conditions does not erase that failure.

Worked Examples

Worked Example: Follow-Up Calls After Hospital Discharge

A hospital wants to estimate the proportion of its adult patients discharged during the last quarter who received a follow-up call within seven days. The hospital randomly selects 120 records without replacement from 1,500 adult discharge records for that quarter. Of the 120 patients, 93 received a follow-up call and 27 did not. Check all three conditions for a one-proportion \(z\)-interval.

Identify the parameter and success.
Let \(p\) be the proportion of adult patients discharged from this hospital during the quarter who received a follow-up call within seven days. A success is a patient who received that call. The sample size is \(n=120\), and the sample contains \(x=93\) successes and \(120-93=27\) failures.

Check Random.
The 120 records were randomly selected from the hospital’s adult discharge records for the quarter. The Random condition is met for inference about patients represented by that discharge list. This selection does not by itself support a conclusion about patients discharged from other hospitals.

Check 10%.
The source population is the 1,500 adult discharge records for the quarter. Ten percent of that population is \(0.10(1{,}500)=150\), and \(120\leq150\). The 10% condition is met.

Check Large Counts.
The observed success count is \(x=93\), and the observed failure count is \(n-x=120-93=27\). Both are at least 10, so the Large Counts condition is met. Equivalently, \(\hat{p}=93/120=0.775\), giving \(n\hat{p}=120(0.775)=93\) and \(n(1-\hat{p})=120(0.225)=27\).

All three conditions are met, so the checks support using a one-proportion \(z\)-interval for the proportion of adult patients represented by this hospital’s discharge records who received a follow-up call within seven days. This conclusion is about the suitability of the procedure; it is not yet the interval’s numerical endpoints.

Worked Example: A Small Number of Patients With the Outcome

A clinic randomly selects 50 patient records without replacement from a list of 1,000 patients treated during a month. The clinic wants to estimate the proportion who returned for an unscheduled visit within 30 days. Six patients in the sample returned for an unscheduled visit, and 44 did not. Check the conditions.

Let \(p\) be the proportion of patients on the clinic’s monthly list who returned for an unscheduled visit within 30 days. Define a return visit as a success. The records were randomly selected, so the Random condition is met for this clinic’s monthly patient list.

For the 10% condition, \(0.10(1{,}000)=100\), and the sample size satisfies \(50\leq100\). The 10% condition is met. For Large Counts, however, the observed counts are \(x=6\) successes and \(n-x=50-6=44\) failures. Although 44 is at least 10, 6 is less than 10. Therefore, the Large Counts condition is not met.

The usual one-proportion \(z\)-interval is not supported by all three checks for these data. The fact that the sample was random and no more than 10% of the source population does not make the observed success count large enough. Do not report the standard interval as though every condition passed.

Worked Example: A Large Volunteer Patient Survey

A hospital sends a survey invitation to 2,400 patients and asks whether they felt they understood their discharge instructions. Patients choose whether to respond. Of the 200 who respond, 154 say they understood the instructions and 46 say they did not. The hospital proposes a one-proportion \(z\)-interval about all 2,400 patients. Assess the conditions.

Let \(p\) be the proportion of the 2,400 invited patients who understood their discharge instructions, and define “understood” as success. The respondents were not randomly selected from the invited patients; they chose whether to answer. The Random condition is not met for inference about all 2,400 patients. Patients with stronger opinions or more time to respond might be more likely to complete the survey.

The numerical comparison \(200\leq0.10(2{,}400)=240\) does not turn voluntary responses into a random sample. The 10% check applies to a random sample taken without replacement from a finite population; here, response depends on patients’ choices, so the hospital cannot use this calculation to justify a random-sampling model for the respondents.

The observed counts among respondents are 154 successes and 46 failures, both at least 10. Thus the Large Counts check passes for the respondent data. But that does not repair the failed Random condition. The hospital can describe the responses: \(154/200=0.77\), or 77% of the people who answered said they understood the instructions. It should not treat this voluntary-response result as an interval estimate for all invited patients based on the usual one-proportion \(z\)-interval.

Worked Example: A Random Sample That Exceeds 10% of the Source Population

A small hospital has 600 patients in a particular follow-up program. It randomly selects 90 patients without replacement to estimate the proportion who completed a required follow-up visit. Of those selected, 63 completed the visit and 27 did not. Check the conditions for a one-proportion \(z\)-interval.

Let \(p\) be the proportion of the 600 patients in this program who completed the required visit. A patient who completed the visit is a success. Because the 90 patients were randomly selected, the Random condition is met.

For the 10% condition, calculate \(0.10(600)=60\). The sample size is 90, and \(90\not\leq60\); the sample is 15% of the source population. The 10% condition is not met. For Large Counts, the observed successes are \(x=63\) and the observed failures are \(n-x=90-63=27\). Both are at least 10, so Large Counts is met.

The usual one-proportion \(z\)-interval is not supported by all three checks as described. A random sample and adequate observed counts do not compensate for a sample that exceeds 10% of the finite source population. The relevant comparison is with the 600 patients in the program—not with a larger group of patients served by the hospital.

Writing Condition Checks for Full Credit

A concise response can still show complete evidence. For the follow-up-call example, a strong condition check would say: “The 120 records were randomly selected from the hospital’s adult discharge list for the quarter. Since \(120\leq0.10(1{,}500)=150\), the 10% condition is met. There are 93 observed successes and \(120-93=27\) observed failures, both at least 10, so the Large Counts condition is met.”

Notice what that response does not do: it does not use a hypothetical population proportion for Large Counts, and it does not claim that the interval represents patients beyond the sampling frame. Those choices connect the arithmetic to the study that produced the data.

Common Mistakes and AP Exam Tips

Using expected counts from a claimed proportion. For a one-proportion confidence interval, use observed counts \(x\) and \(n-x\). Expected counts based on \(p_0\) belong to the Large Counts check for a test.
Calling a survey random because invitations were sent randomly. If patients decide whether to respond, the actual respondents may be self-selected. Describe who supplied data, not just who received an invitation.
Comparing \(n\) with the wrong population size. Use the source population from which the sample was actually drawn, such as the hospital’s program list or quarterly discharge list.
Treating a numerical 10% comparison as a cure for selection bias. A ratio such as \(200/2{,}400\) does not make voluntary responses a random sample. Each condition addresses a different aspect of the data.
Checking only the larger observed count. Both the success count and the failure count must be at least 10. In the clinic example, 44 failures do not compensate for only 6 successes.
Writing only “the conditions are met.” A full-credit response states how the sample was chosen, shows the 10% comparison when relevant, and gives both observed counts for Large Counts.

AP Exam Tip: For a one-proportion interval, define \(p\) and success in context, then check Random, 10%, and observed Large Counts separately. Show the evidence for each condition and limit the inference to the population represented by the sampling frame.

Key Takeaway

The three conditions answer different questions. Random selection supports inference to a represented population, the 10% condition supports approximate independence when sampling without replacement, and Large Counts supports the Normal approximation using observed successes and failures. Keep the checks tied to the hospital’s actual records and patients.

Key takeaway: For a one-proportion \(z\)-interval, describe the random process, verify \(n\leq0.10N\) when sampling without replacement from a finite population, and check \(x\geq10\) and \(n-x\geq10\). If any condition fails, say which one and why the usual interval is not supported by all the checks.

Check Your Understanding

For each situation, identify the population proportion and assess the Random, 10%, and Large Counts conditions for a one-proportion \(z\)-interval.

A hospital randomly selects 160 records from 2,000 patients discharged during a month. In the sample, 118 patients attended a follow-up appointment and 42 did not. Which conditions are met?
A clinic randomly samples 45 patients from a list of 900. Eight patients had an outcome of interest and 37 did not. Which condition fails, if any?
A hospital invites 500 patients to an optional online survey. Of the 180 respondents, 135 report understanding their care instructions. What does the response method imply for the Random condition?
A program randomly samples 75 patients without replacement from a finite list of 700. What comparison should be used for the 10% condition, and does it pass?
In a sample of 100 patients, 91 have the characteristic of interest and 9 do not. Does Large Counts pass? Explain why checking only the success count would be incomplete.