A Condition Check Is an Evidence-Based Argument
In What Happens to Results When Conditions Are Violated, you saw why a failed condition can undermine the usual guarantees of a one-proportion \(z\)-procedure. On an exam, it is not enough to list condition names or say that a sample is “large.” You need to identify the evidence in the study description, show the relevant calculation, and explain what that evidence means for the proposed procedure.
This tutorial brings together the checks practiced in earlier lessons. The goal is to make an exam response easy to follow: identify the procedure, connect each condition to the appropriate evidence, and give a clear verdict. A condition may be met, violated, or not established from the information provided. Those are different conclusions, and your wording should make the distinction clear.
The conditions for a one-proportion \(z\)-interval and a one-proportion \(z\)-test overlap, but their Large Counts checks differ. As explained in Interval Versus Test Condition Checks Compared, an interval uses the observed sample counts. A test uses counts expected under the null hypothesis.
| Condition | For a one-proportion \(z\)-interval | For a one-proportion \(z\)-test of \(H_0:p=p_0\) |
|---|---|---|
| Random | Describe the random sample or suitable random process. | Describe the random sample or appropriate randomized process. |
| 10% condition | For sampling without replacement, compare \(n\) with \(0.10N\) for the source population. | For sampling without replacement, compare \(n\) with \(0.10N\) for the source population. |
| Large Counts | Check observed successes \(x\) and failures \(n-x\). | Check expected successes \(np_0\) and failures \(n(1-p_0)\). |
The 10% condition is relevant when observations are sampled without replacement from a finite population. Do not use a target population count if the sample was actually drawn from a different source population. If the study description does not provide enough information to verify a condition, say it is not established rather than inventing evidence.
A Reliable Exam Response Structure
A useful technique is to treat the condition check as a short evidence audit. First name the proposed procedure and parameter. Then address the conditions one by one, using the study description for the Random condition, the actual source population for the 10% condition, and the correct counts for Large Counts. Finish by saying whether the conditions support using the procedure.
State whether the question proposes a one-proportion \(z\)-interval or \(z\)-test, and identify the population proportion \(p\) being estimated or tested.
State who or what was selected or assigned, and whether chance guided that process. Avoid treating a large sample or a random-sounding phrase as proof of randomness.
For sampling without replacement from a finite source population of size \(N\), verify \(n\leq0.10N\). Use the population from which the sample was actually selected.
For an interval, calculate observed successes \(x\) and failures \(n-x\). For a test, calculate expected successes \(np_0\) and failures \(n(1-p_0)\).
Say whether the evidence supports each condition and whether it supports using the proposed \(z\)-procedure. If a condition is not established, identify the missing information.
You do not need to repeat every formula from earlier tutorials. You do need to show enough work that a reader can verify your reasoning. For example, “Large Counts holds because the expected counts are adequate” is incomplete. A stronger response gives both expected counts and compares each with 10.
Worked Examples
Worked Example: Checking a One-Proportion Interval
A community garden organization wants to estimate the proportion of registered garden plots that have a compost bin. It selects a simple random sample of 120 plots without replacement from 2,100 registered plots. In the sample, 78 plots have a compost bin. Is a one-proportion \(z\)-interval supported by the conditions?
State: Let \(p\) be the true proportion of all 2,100 registered garden plots that have a compost bin. The proposed procedure is a one-proportion \(z\)-interval for \(p\).
Plan: Check the Random condition, the 10% condition for sampling without replacement, and the interval’s Large Counts condition using the observed successes and failures.
Do: The description says the 120 plots were selected by a simple random sample, so chance guided selection from the population of registered plots. The Random condition is met for estimating the proportion among those registered plots.
For the 10% condition, compare the sample size with 10% of the source population:
The sample is no more than 10% of the 2,100 plots, so the 10% condition is met. The observed number of failures is the number without a compost bin:
Both observed counts are at least 10: \(78\geq10\) and \(42\geq10\). Therefore, the interval’s Large Counts condition is met. As a check on the sample proportion, \(\hat{p}=78/120=0.65\), and the observed counts agree with \(120(0.65)=78\) and \(120(1-0.65)=42\).
Conclude: The simple random sample supports the Random condition, the sample is no more than 10% of the source population, and both observed counts exceed 10. The conditions support using a one-proportion \(z\)-interval to estimate the proportion of registered garden plots with a compost bin.
Worked Example: Checking a One-Proportion Test
A transit office is investigating whether more than 25% of residents in a particular service area use a mobile fare pass at least once a week. Staff take a simple random sample of 80 residents without replacement from the area’s 900 residents. Fourteen sampled residents report weekly use. Before carrying out a one-proportion \(z\)-test, check its conditions.
State: Let \(p\) be the true proportion of residents in this service area who use a mobile fare pass at least once a week. The proposed test is a one-proportion \(z\)-test of \(H_0:p=0.25\) against \(H_a:p>0.25\).
Plan: Check whether the sample was random, verify the 10% condition using the 900 residents from whom it was drawn, and check Large Counts using the null value \(p_0=0.25\), not the observed sample proportion.
Do: The sample is described as a simple random sample of residents, so the Random condition is met for the stated service-area population. Since sampling was without replacement, check:
The sample is at most 10% of the source population, so the 10% condition is met. For the test’s Large Counts condition, use the null proportion:
Both expected counts are at least 10. The test’s Large Counts condition is met. The observed sample proportion is \(14/80=0.175\), but that value is not used for this test condition check; the check asks whether the null model predicts enough expected successes and failures.
Conclude: The random sample, the 10% comparison, and the null-based expected counts support using the proposed one-proportion \(z\)-test. This conclusion only addresses whether the conditions are met; it does not determine the test’s result.
Worked Example: Large Counts Cannot Repair a Biased Selection Process
A school district wants to estimate the proportion of all district families who support adding a late bus. It posts a QR-code poll on its public website and invites any family to respond. Of the 380 responses, 246 support the late bus. A student says that a one-proportion \(z\)-interval is appropriate because there are many responses. Evaluate that claim.
State: The target is the proportion of all district families who support adding a late bus. The proposed procedure is a one-proportion \(z\)-interval.
Plan: Examine how respondents entered the poll, then check the observed counts. A large sample and adequate observed counts do not establish that the sample represents the target population.
Do: Families chose for themselves whether to visit the website and respond. They were not selected through a random sample, and the poll description gives no random process that would make the respondents representative of all district families. Therefore, the Random condition is not met for generalizing to all district families.
The observed successes and failures are:
Both observed counts are at least 10, so the Large Counts condition for the interval is met. The sample proportion is \(246/380\approx0.6474\), or about 0.647 rounded to three decimal places. Neither the large counts nor the calculated sample proportion fixes the self-selection problem.
The district-wide population count is not given, and the information does not establish that respondents were randomly sampled without replacement from that population. A student should not invent a population size to claim the 10% condition holds. More importantly, even a numerical comparison with 10% would not make this voluntary-response poll random.
Conclude: The interval’s observed counts are large enough, but the Random condition is not met for inference to all district families, and the 10% condition is not established from the description. The proposed interval is not supported for the district-wide target. The poll can describe the families who chose to respond, but it does not provide a sound basis for generalizing to every district family.
Worked Example: A Random Sample That Is Too Large for the 10% Condition
A warehouse quality manager wants to estimate the proportion of the 500 packages prepared during one shift that contain a labeling error. The manager takes a simple random sample of 75 packages without replacement and finds 18 labeling errors. Are the conditions for a one-proportion \(z\)-interval met?
State: Let \(p\) be the true proportion of the 500 packages prepared during this shift that contain a labeling error. The proposed procedure is a one-proportion \(z\)-interval.
Plan: Check the random selection, compare the sample size with 10% of the 500-package source population, and use the observed counts for Large Counts.
Do: The manager selected a simple random sample from the packages prepared during the shift, so the Random condition is met for that shift’s packages. The 10% comparison is:
The sample is more than 10% of the source population, so the 10% condition is not met. The observed number of packages without a labeling error is \(75-18=57\). Thus, the Large Counts check is:
Both observed counts meet the Large Counts threshold. Also, \(\hat{p}=18/75=0.24\), which is consistent with \(75(0.24)=18\) observed successes and \(75(1-0.24)=57\) observed failures.
Conclude: The Random and interval Large Counts conditions are met, but the 10% condition is not. Because the sample is a substantial fraction of the 500 packages, the usual independence justification from the 10% condition is unavailable. The conditions do not support proceeding with the usual one-proportion \(z\)-interval as though all requirements had been met.
Common Mistakes and What Full Credit Requires
- Listing condition names without evidence. “Random, 10%, and Large Counts are satisfied” does not show why. Full-credit writing identifies the random selection, compares \(n\) with \(0.10N\), and gives both relevant counts.
- Using the wrong Large Counts check. For an interval, use observed successes \(x\) and failures \(n-x\). For a test, use expected counts \(np_0\) and \(n(1-p_0)\). State which procedure you are checking so the reader can tell why those counts apply.
- Using the wrong population in the 10% comparison. The relevant \(N\) is the population from which the sample was actually selected. A study may discuss a broad target population but draw its sample from a narrower list or group.
- Treating a large sample as proof of randomness. A large voluntary-response sample can still be biased. Randomness depends on how the observations were obtained, not on the sample size or the word “survey.”
- Calling missing information a violation. If the description does not say whether selection was random, the Random condition may be not established rather than demonstrably violated. Use precise language: “The description does not establish…” when that is what the evidence supports.
- Letting one met condition cancel another that fails. Large Counts does not repair self-selection, and random selection does not make a sample meet the 10% condition automatically. Each condition addresses a different concern.
Key Takeaway
A strong exam response is specific to the study and the procedure. Trace how the sample was obtained, compare the sample with its actual source population when the 10% condition applies, and choose the correct Large Counts check. Then state plainly which conditions are met, violated, or not established.
Check Your Understanding
For each situation, identify the relevant evidence and give a condition verdict.
- A simple random sample of 90 students is drawn without replacement from a school of 1,500 students. For an interval, 52 report biking to school and 38 do not. Check the 10% and Large Counts conditions.
- A test uses \(H_0:p=0.40\) with \(n=50\). Calculate the two expected counts for the test’s Large Counts condition and decide whether it is met.
- A poll is posted on a neighborhood social-media page, and users choose whether to answer. Explain whether a large number of responses would establish the Random condition for estimating a proportion among all neighborhood residents.
- A random sample of 40 items is taken without replacement from a shipment of 350 items. Calculate \(0.10N\) and decide whether the 10% condition is met.
- In one sentence, explain why an interval uses observed counts for Large Counts while a test uses expected counts under the null.