Writing Condition Checks in Complete Sentences

Make Every Condition Check Say Something

In Conditions When the Large Counts Check Fails, you saw that a condition can determine whether a one-proportion \(z\)-procedure is justified. In a written response, it is not enough to list condition names or say “conditions are met.” A complete check identifies the evidence, shows the calculation when one is needed, and states what that evidence implies.

This tutorial focuses on the wording and structure of those checks. The goal is to make your reasoning easy to follow: a reader should be able to see why you judged each condition met or not met. The conditions themselves are familiar from earlier tutorials on the random condition, the 10% condition, and the Large Counts condition.

Key idea: For each condition, write three parts: name the relevant evidence, show the calculation or comparison, and state whether the condition is met. A calculation without a conclusion is incomplete; a conclusion without supporting evidence is not justified.

Use the conditions for the procedure you are actually considering. A one-proportion \(z\)-interval checks the observed success and failure counts. A one-proportion \(z\)-test checks the expected success and failure counts under the null hypothesis. As covered in Interval Versus Test Condition Checks Compared, those checks are related but not interchangeable.

A Sentence Pattern for Each Condition

The wording can vary, but the logic should remain visible. These sentence patterns are useful starting points. Replace the bracketed material with details from the situation.

Sentence pattern: “The [sampling method] is random, so the random condition is met for making inference about [target population].”

For the random condition, name how the data were collected. “The sample is random” is stronger when it says what was randomly selected and identifies the population to which that selection applies. Do not claim that a method was random if the description does not say so. Also avoid claiming that random sampling proves the sample is perfectly representative; it supports inference to the population from which the sample was drawn.

Sentence pattern: “The sample was selected without replacement from a population of \(N=[\text{size}]\). Since \(n=[\text{sample size}]\leq0.10N=[\text{calculation}]\), the 10% condition is met, so it is reasonable to treat observations as approximately independent.”

For a sample taken without replacement from a finite population, the 10% condition is \(n\leq0.10N\). State the sample size, the population size, and the comparison. Use the population the sample was actually drawn from, not a broader population that was not sampled. If the sample was taken with replacement, or the sampling process otherwise makes observations independent, explain that rather than inventing a finite \(N\).

For the Large Counts condition, state which counts belong to the procedure and show both calculations. A one-proportion \(z\)-interval uses the observed counts \(x=n\hat{p}\) and \(n-x\). A one-proportion \(z\)-test of \(H_0:p=p_0\) uses the expected counts \(np_0\) and \(n(1-p_0)\). The Large Counts condition is met only if both relevant counts are at least 10.

Sentence pattern for an interval: “There are \(x=[\text{successes}]\) observed successes and \(n-x=[\text{failures}]\) observed failures. Since \(x\geq10\) and \(n-x\geq10\), the Large Counts condition for the one-proportion \(z\)-interval is met.”

Sentence pattern for a test: “Under \(H_0:p=p_0\), the expected success count is \(np_0=[\text{calculation}]\), and the expected failure count is \(n(1-p_0)=[\text{calculation}]\). Since both are at least 10, the Large Counts condition for the one-proportion \(z\)-test is met.”

If a count is below 10, name that count and say the condition fails. If a count equals 10, it meets the stated threshold: “Since the expected success count is exactly 10 and the expected failure count is 115, both are at least 10, so the condition is met.” Do not round a count up or describe a failed check as “close enough.”

Worked Examples

Worked Example: Writing the Checks for an Interval

A fictional random sample of 180 rechargeable scooters is selected without replacement from a shipment of 2,400 scooters. Sixty-three have a particular battery fault. A technician wants to use a one-proportion \(z\)-interval to estimate the proportion of scooters in this shipment with the fault. Write complete condition checks.

Define success as “the scooter has the battery fault.” The observed number of successes is \(x=63\), and the number of failures is \(180-63=117\). The sample proportion is

\hat{p}=\frac{x}{n}=\frac{63}{180}=0.35

A complete random-condition sentence is: “The 180 scooters were selected by a random sample from the shipment, so the random condition is met for making inference about the scooters in this shipment.” This says what was randomized and identifies the target population.

For the 10% condition, show the comparison:

n=180 \qquad\text{and}\qquad 0.10N=0.10(2400)=240 \qquad\text{so}\qquad 180\leq240

A complete sentence is: “The sample was selected without replacement from a shipment of 2,400 scooters. Since \(180\leq0.10(2400)=240\), the 10% condition is met, so it is reasonable to treat the observations as approximately independent.”

For the interval’s Large Counts condition, use the observed counts:

n\hat{p}=180(0.35)=63 \qquad\text{and}\qquad n(1-\hat{p})=180(0.65)=117

A complete sentence is: “There are 63 observed successes and 117 observed failures. Since \(63\geq10\) and \(117\geq10\), the Large Counts condition for the one-proportion \(z\)-interval is met.”

Together, these sentences explain why the three conditions support using a one-proportion \(z\)-interval. They do not yet give the interval or its interpretation; those require the interval calculation and a separate conclusion.

Worked Example: A Four-Step Condition Check for a Test

A fictional random sample of 125 urban gardeners is selected without replacement from a directory of 2,500 gardeners. A researcher wants to test whether more than 8% of gardeners in that directory use a particular water-saving method. Seventeen of the sampled gardeners use it. Write a four-step response that checks the conditions.

State.
Let \(p\) be the proportion of gardeners in the directory who use the water-saving method. The hypotheses are \(H_0:p=0.08\) and \(H_a:p>0.08\).

Plan.
If the conditions hold, use a one-proportion \(z\)-test. The description states that the 125 gardeners were randomly sampled, which supports the random condition for inference about gardeners in this directory. Because the sample was taken without replacement, also check the 10% condition and the Large Counts condition under \(H_0\).

Do.
The sample is random, so the random condition is met. For the 10% condition, \(0.10N=0.10(2500)=250\), and \(n=125\leq250\); therefore, the 10% condition is met and it is reasonable to treat observations as approximately independent. Under \(H_0:p=0.08\), the expected success count is \(np_0=125(0.08)=10\), and the expected failure count is \(n(1-p_0)=125(0.92)=115\). Both expected counts are at least 10, so the Large Counts condition for the test is met.

Conclude.
The random, 10%, and Large Counts conditions are met, so the conditions support using a one-proportion \(z\)-test for the stated hypotheses. These checks alone do not determine whether there is convincing evidence that more than 8% of gardeners in the directory use the method; that decision requires carrying out the test.

The expected success count is exactly 10, so it meets the condition. Notice also that the 17 observed successes are not used for the test’s Large Counts check. The null value \(p_0=0.08\) determines the expected counts.

Worked Example: Explaining Why an Interval Is Not Supported

A fictional random sample of 75 seed packets is taken without replacement from a lot of 900 packets. Eight packets contain seeds that do not germinate in a specified test. Can a one-proportion \(z\)-interval be used to estimate the proportion of packets in the lot with this issue? Write complete condition checks and a conclusion.

Define success as “the packet contains seeds that do not germinate in the specified test.” The sample contains \(x=8\) successes and \(75-8=67\) failures. The random condition is met because the 75 packets were randomly selected from the lot, supporting inference about packets in that lot.

For the 10% condition,

0.10N=0.10(900)=90 \qquad\text{and}\qquad n=75\leq90

A complete sentence is: “The sample was selected without replacement from a lot of 900 packets. Since \(75\leq0.10(900)=90\), the 10% condition is met, so it is reasonable to treat the observations as approximately independent.”

For a one-proportion \(z\)-interval, the Large Counts check uses observed counts:

x=8<10 \qquad\text{and}\qquad n-x=75-8=67\geq10

A complete sentence is: “There are 8 observed successes and 67 observed failures. Since the observed success count is \(8<10\), the Large Counts condition for the one-proportion \(z\)-interval fails.”

The conclusion should connect that failure to the method: “Although the random and 10% conditions are met, the Large Counts condition is not met. Therefore, the usual one-proportion \(z\)-interval is not supported for estimating the proportion of packets in this lot with the issue.” As discussed in Conditions When the Large Counts Check Fails, do not report the usual Normal-based interval as reliable just because a calculator can produce endpoints.

Common Mistakes and What a Complete Answer Says

Writing only “random condition met.” Name the sampling method and the population it represents. A full-credit sentence connects the stated random selection to the target population.
Giving the 10% result without the comparison. “The sample is small” does not show the condition. Write \(n\), calculate \(0.10N\), compare them, and state the conclusion.
Using the wrong population size. The \(N\) in the 10% condition is the source population for the sample. Use the shipment, directory, lot, or other actual sampling frame described in the situation.
Checking only one Large Counts value. Show both success and failure counts. If either is below 10, the condition fails.
Using observed counts for a test or expected counts for an interval. For an interval, use \(x\) and \(n-x\). For a test, use \(np_0\) and \(n(1-p_0)\). State which procedure you are checking.
Ending with an unsupported broad claim. Passing these checks means the conditions support the procedure; it does not by itself prove that the sample is unbiased or decide the inference question. Keep the conclusion limited to what the checks establish.

AP Exam Tip: Make each check independently readable: identify the condition, include the relevant evidence, show the comparison or calculation, and state “met” or “not met.” If a condition fails, say how that limits the proposed \(z\)-procedure. Do not confuse a condition-check conclusion with a test decision or an interval interpretation.

Key Takeaway

A strong condition check is a short argument, not a label. Name the evidence, show the relevant calculation or comparison, and state what it means for the proposed procedure. Keep the counts matched to the method: observed counts for a one-proportion \(z\)-interval and expected counts under \(H_0\) for a one-proportion \(z\)-test.

Key takeaway: Write the evidence, calculation, and conclusion for every condition. A reader should be able to verify why the condition is met or not met—and understand what that means for the inference procedure.

Check Your Understanding

For each situation, write or evaluate a complete condition-check sentence.

A random sample of 90 devices is selected without replacement from a warehouse of 1,200 devices. Write the 10% condition check, including the calculation and conclusion.
A one-proportion \(z\)-interval uses a sample of 140 with 18 observed successes. Find the observed success and failure counts, then write the Large Counts check.
A one-proportion \(z\)-test uses \(H_0:p=0.06\) and \(n=200\). Calculate both expected counts and write a complete Large Counts sentence.
Why is “The Large Counts condition passes because there are 12 observed successes” not enough to justify the condition for a test of \(H_0:p=0.04\) with \(n=200\)?
Write one sentence that concludes all relevant conditions are met while making clear that the condition checks alone do not establish the test’s result.