Start by Defining the Outcome
Before checking counts for one-proportion inference, decide what the word success means in the question. In this setting, success does not necessarily mean something beneficial. It means the particular outcome being counted to define the population proportion \(p\). The other outcome is a failure, meaning the outcome does not have that characteristic.
For example, if \(p\) is the proportion of devices with a cracked screen, a device with a cracked screen is a success for the analysis—even though a cracked screen is undesirable. If \(p\) is the proportion of residents who support a proposal, support is the success. Naming the outcome carefully keeps the sample count, the sample proportion, and the population proportion connected to the same characteristic.
This is a two-outcome description: each observation either has the characteristic or does not. Do not choose “success” based on which outcome is larger, more favorable, or more interesting. Choose it based on how the population proportion in the question is defined.
In Large Counts Condition for Confidence Intervals and Checking the Success-Failure Condition for Tests, you learned why the success and failure counts matter. Interval Versus Test Condition Checks Compared emphasized that an interval uses observed counts, while a test uses counts expected under the null hypothesis. Here, the central skill is preparing those counts correctly in the first place.
Translate Between Percentages and Counts
If a sample has \(n\) observations and a proportion \(\hat{p}\) of them are successes, multiply the proportion by the sample size to find the success count. Then subtract that count from \(n\) to find the failure count. A percentage must first be written as a proportion by dividing by 100.
For instance, 14% becomes \(0.14\), not 14. If the sample size is 450, the success count is \(450(0.14)=63\). The failure count is the rest of the sample, \(450-63=387\). As a check, the complementary percentage is \(100\%-14\%=86\%\), and \(450(0.86)=387\).
The two counts should add back to the full sample size. The success percentage and failure percentage should add to 100%. These are simple but useful checks for misplaced decimal points, subtraction errors, or counting the wrong outcome.
Observed Counts and Expected Counts Are Different
The phrase “success count” can refer to different quantities depending on the question. For a one-proportion \(z\)-interval, the Large Counts check uses the successes and failures actually observed in the sample: \(x\) and \(n-x\). For a one-proportion \(z\)-test of \(H_0:p=p_0\), the check uses the counts expected under the null proportion: \(np_0\) and \(n(1-p_0)\). The earlier tutorial Interval Versus Test Condition Checks Compared develops this distinction.
The outcome designated as success stays the same for both procedures. What changes is the source of the counts: the interval uses the sample, and the test uses the null model. Expected counts need not be whole numbers. They describe what the null model predicts on average, not a tally of actual people or objects.
When a problem gives a percentage and a total, first determine whether the percentage describes what happened in the sample or is the null proportion in a test. Do not use a null percentage to claim how many successes were actually observed. Likewise, do not substitute the observed sample percentage for \(p_0\) when finding expected counts under \(H_0\).
Worked Examples
Worked Example: Find Counts from a Sample Percentage
A fictional transit survey asks 450 riders whether they commute by bicycle at least once a week. Suppose exactly 14% answer yes. Define success as “the rider commutes by bicycle at least once a week.” Find the number of successes and failures.
The total sample size is \(n=450\). The success percentage is 14%, which as a proportion is \(0.14\). Therefore,
There are 63 observed successes. The number of failures is
Check by using the complementary percentage: \(100\%-14\%=86\%\), or \(0.86\). Then \(450(0.86)=387\), agreeing with the subtraction. The counts add to \(63+387=450\), the original sample size.
The word “success” here is just the label for the yes outcome being counted. It does not suggest that commuting by bicycle is better or worse than another way of commuting.
Worked Example: Count a Less Common Outcome
A fictional equipment team inspects 320 portable sensors. Exactly 7.5% have a loose connector. For a study of the proportion of sensors with loose connectors, define success as “the sensor has a loose connector.” How many successes and failures are in the inspected sample?
The sample size is \(n=320\). Convert 7.5% to a proportion:
Multiply this proportion by the total to find the observed success count:
So 24 sensors are successes for this question. The remaining sensors do not have loose connectors and are failures:
As a check, the complementary percentage is \(100\%-7.5\%=92.5\%\), and \(320(0.925)=296\). Also, \(24+296=320\). Notice that “failure” in the count terminology means “does not have the characteristic being studied”; it does not mean that the sensor itself failed.
Worked Example: Separate Sample Counts from Null Expected Counts
A fictional survey of 240 residents finds that exactly 62.5% support a proposed neighborhood garden. A one-proportion \(z\)-test is being considered for \(H_0:p=0.60\), where \(p\) is the proportion of all residents in the target population who support the proposal. Define success as “a resident supports the proposal.” Find the observed sample counts and the expected counts under the null hypothesis.
First use the observed sample percentage to find the sample success count:
There are \(240-150=90\) observed failures—residents in the sample who do not support the proposal. These are the sample’s actual counts: 150 successes and 90 failures.
For the test check, use the null proportion \(p_0=0.60\), not the observed proportion \(0.625\). The expected success count under the null is
The expected failure count under the null is
The two pairs answer different questions. The sample contains 150 supporters and 90 nonsupporters. If the null hypothesis were true, the model would predict 144 supporters and 96 nonsupporters on average in samples of this size. Do not report 144 and 96 as what the survey actually found.
Percentages That Have Been Rounded
A percentage printed in a report may be rounded. If the percentage is explicitly exact, multiplying it by \(n\) can give the exact count, as in the examples above. If it is rounded, the multiplication may produce a non-integer value. An observed count of people or objects must be a whole number, so a non-integer result is a signal to investigate the rounding rather than to report a fractional person or object.
For example, suppose a report gives a whole-number percentage but does not say it is exact. Multiplying that displayed percentage by the sample size gives a useful estimate of the count, but the true count may differ because the percentage was rounded for display. If exact observed counts are needed for a one-proportion procedure, use the raw count when available. If only the rounded percentage is provided, say that the count calculated from it is approximate and do not claim more precision than the report supports.
Before calculating, also check what the percentage describes. It might be the sample percentage, a claimed population percentage, or a null proportion stated in a hypothesis. Only the sample percentage can be directly converted into an observed sample count. A claimed or null population proportion can be used to find expected counts for a test, together with \(n\).
Common Mistakes and AP Exam Tip
- Choosing success to mean “good.” Success is the outcome named by the population proportion, not necessarily a desirable result. State the outcome in words before calculating.
- Forgetting to convert a percentage to a proportion. For example, 7.5% is \(0.075\), not \(7.5\). Multiply the decimal proportion by \(n\).
- Subtracting the percentage instead of the count. First calculate \(x=n\hat{p}\), then find failures with \(n-x\). You can also use the complementary proportion \(1-\hat{p}\).
- Reversing the categories midway through a solution. If success means “supports the proposal,” failures must mean “does not support the proposal.” Keep the same definition through every calculation.
- Calling null expected counts observed counts. In a test, \(np_0\) and \(n(1-p_0)\) are expected under the null hypothesis; they are not a tally of the sample.
- Treating a rounded percentage as exact. If multiplying a displayed percentage by \(n\) gives a non-integer, identify the result as approximate unless the question establishes that the percentage is exact.
For a one-proportion \(z\)-interval, the observed success and failure counts are used in the Large Counts condition, as covered in Large Counts Condition for Confidence Intervals. For a one-proportion \(z\)-test, the expected counts based on \(p_0\) are used, as covered in Checking the Success-Failure Condition for Tests. In either case, the relevant counts must both be at least 10 for the Large Counts check to pass. Correctly identifying the outcome comes before applying that check.
Key Takeaway
Counting correctly begins with the question, not the arithmetic. Define success as the characteristic whose population proportion is being studied, convert percentages to proportions before multiplying by the total, and use the complement to find failures. Then identify whether the question asks for actual sample counts or null expected counts.
Check Your Understanding
For each situation, define success, calculate the requested counts, and show a check that the two counts correspond to the stated total.
- A fictional sample contains 180 shoppers, and exactly 35% used a reusable bag. If success means “used a reusable bag,” how many successes and failures are there?
- A fictional inspection finds that exactly 12% of 250 seedlings show new growth. Define success as “shows new growth” and find both sample counts.
- A survey of 160 residents reports exactly 45% support for a park renovation. If support is the success, find the observed success and failure counts.
- For the sample in question 3, a test uses \(H_0:p=0.50\). Find the null expected success and failure counts. Explain why they differ from the observed sample counts.
- A report says that 18% of a sample of 75 people chose an option, but the report does not say whether the percentage is exact or rounded. What should you check before reporting an exact success count?