AP Statistics Inference for Categorical Data: Proportions — Worked Answer Explanations

Unit 6 · 12 questions explained

Below is a complete answer key for our AP Statistics Inference for Categorical Data: Proportions practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Inference for Categorical Data: Proportions practice test and come back here to review, or head back to the Inference for Categorical Data: Proportions unit overview.

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Question 1 · Easy
A researcher wants to estimate the proportion of adults who exercise regularly. She samples 400 adults and finds 160 exercise regularly. What is the 95% confidence interval for the true proportion?
- A
  $(0.35, 0.45)$
  Why not A: Using $z^{*} = 1$ instead of $z^{*} \approx 1.96$ for 95% confidence.
- B
  $(0.352, 0.448)$ Correct
- C
  $(0.36, 0.44)$
  Why not C: Rounding the standard error to 0.02 instead of computing $(0.4) (0.6) /400 \approx 0.0245$ .
- D
  $(0.40, 0.40)$
  Why not D: Reporting only $\overset{p}{^}$ without a margin of error.
Explanation
$\overset{p}{^} = 160/400 = 0.40$ . $S E = \overset{p}{^} (1 - \overset{p}{^}) / n = (0.40) (0.60) /400 \approx 0.0245$ . 95% CI: $0.40 \pm 1.96 (0.0245) \approx 0.40 \pm 0.048$ , giving $(0.352, 0.448)$ .

Key takeaway
One-proportion z-interval: $\hat{p} \pm z^* \sqrt{\hat{p}(1-\hat{p})/n}$; use $z^* = 1.96$ for 95% confidence.
Question 2 · Easy
A poll finds that 54 of 200 surveyed voters support a ballot measure. A reporter states: "We are 95% confident that between 20.5% and 33.5% of all voters support the measure." How should this interval be correctly interpreted?
- A
  There is a 95% probability that the true population proportion is between 20.5% and 33.5%.
  Why not A: A frequentist confidence interval does not assign probability to a fixed (but unknown) parameter.
- B
  We are 95% confident that the interval (20.5%, 33.5%) captures the true proportion of all voters who support the measure.Correct
- C
  95% of voters support the measure at a level between 20.5% and 33.5%.
  Why not C: This misinterprets the interval as describing individual voter behavior.
- D
  The interval was constructed by a method that is correct 95% of the time in any given poll.
  Why not D: The 95% confidence level means 95% of intervals constructed by this method will capture $p$ in the long run, not that this specific interval is correct 95% of the time.
Explanation
A confidence interval is interpreted as: we are 95% confident this specific interval captures the true population parameter. The "95%" refers to the long-run success rate of the method — if we built many such intervals, about 95% would contain $p$ . We do not assign probability to whether a specific interval contains $p$ (it either does or doesn't).

Key takeaway
Confidence interval interpretation: 'We are C% confident the interval captures the true parameter.' Do not say 'there is a C% probability the parameter is in the interval.'
Question 3 · Easy
A company claims that 80% of its customers are satisfied. A consumer group surveys 100 customers and finds 72 are satisfied. Which hypothesis should be used to test whether the satisfaction rate is lower than claimed?
- A
  $H_{0} : p = 0.72$ , $H_{a} : p < 0.72$
  Why not A: Using the sample proportion as the null hypothesis value instead of the claimed population value.
- B
  $H_{0} : p = 0.80$ , $H_{a} : p \neq = 0.80$
  Why not B: A two-sided alternative is used when testing whether the proportion differs in either direction; here we only care about lower.
- C
  $H_{0} : p = 0.80$ , $H_{a} : p < 0.80$ Correct
- D
  $H_{0} : p < 0.80$ , $H_{a} : p = 0.80$
  Why not D: The null hypothesis always states equality; the alternative is the claim being tested.
Explanation
The null hypothesis states what is claimed ( $p = 0.80$ ). The consumer group is testing whether satisfaction is lower than claimed, so the alternative is one-sided: $H_{a} : p < 0.80$ . Always state $H_{0}$ with equality.

Key takeaway
$H_0$ states equality to the claimed value; $H_a$ reflects the direction of the research question (less than, greater than, or not equal).
Question 4 · Easy
For the test $H_{0} : p = 0.80$ vs. $H_{a} : p < 0.80$ using the data from the previous question ( $n = 100$ , $\overset{p}{^} = 0.72$ ), the $z$ -test statistic is approximately $- 2.0$ and the $p$ -value is 0.023. At $α = 0.05$ , what is the correct conclusion?
- A
  Fail to reject $H_{0}$ ; there is insufficient evidence that fewer than 80% of customers are satisfied.
  Why not A: $p$ -value = 0.023 < $α$ = 0.05, so we reject $H_{0}$ .
- B
  Reject $H_{0}$ ; there is convincing evidence that fewer than 80% of customers are satisfied.Correct
- C
  Reject $H_{0}$ ; we have proved the true satisfaction rate is exactly 0.72.
  Why not C: Rejecting $H_{0}$ does not prove the specific value of the parameter; it only provides evidence against $H_{0}$ .
- D
  Accept $H_{0}$ ; the satisfaction rate is 80%.
  Why not D: We never 'accept' the null hypothesis; failing to reject $H_{0}$ is different from proving it true.
Explanation
Since $p$ -value = 0.023 < $α = 0.05$ , we reject $H_{0}$ . There is statistically significant evidence that the true proportion satisfied is less than 80%. We conclude there is convincing evidence the claim is overstated.

Key takeaway
Reject $H_0$ when $p$-value < $\alpha$. State conclusion in context and direction; never 'accept' $H_0$.
Question 5 · Easy
A researcher increases the confidence level from 90% to 99% while keeping the sample size constant. What happens to the confidence interval?
- A
  The interval becomes narrower because higher confidence means more precision.
  Why not A: Higher confidence requires a wider interval to capture the parameter with greater certainty.
- B
  The interval becomes wider because a higher $z^{*}$ is used.Correct
- C
  The interval stays the same because the sample data do not change.
  Why not C: The critical value $z^{*}$ changes with the confidence level, so the interval changes.
- D
  The interval becomes wider and the point estimate changes.
  Why not D: The point estimate $\overset{p}{^}$ depends only on the sample data, not on the confidence level.
Explanation
A higher confidence level requires a larger $z^{*}$ (1.645 for 90%, 2.576 for 99%), which multiplies the standard error to give a larger margin of error. To be more confident of capturing the parameter, the interval must be wider.

Key takeaway
Higher confidence level → larger $z^*$ → wider interval. There is a trade-off between confidence and precision.
Question 6 · Easy
A researcher wants to estimate a population proportion to within ±0.04 with 95% confidence, but has no prior estimate of $p$ . What minimum sample size is needed?
- A
  385
  Why not A: Using $z^{*} = 1.645$ (90% confidence) instead of $z^{*} = 1.96$ (95%).
- B
  601Correct
- C
  25
  Why not C: Computing $n = 1/ (M E)^{2}$ without the $z^{* 2}$ and $p^{*} (1 - p^{*})$ factors.
- D
  2401
  Why not D: Using $M E = 0.02$ instead of $M E = 0.04$ in the formula.
Explanation
With no prior estimate, use $p^{*} = 0.50$ (maximizes required $n$ ). $n = \frac{z ^{* 2} p ^{*} ( 1 - p ^{*} )}{M E ^{2}} = \frac{( 1.96 ) ^{2} ( 0.25 )}{( 0.04 ) ^{2}} = \frac{3.8416 \times 0.25}{0.0016} = \frac{0.9604}{0.0016} = 600.25$ . Round up to $n = 601$ .

Key takeaway
Sample size formula: $n = z^{*2} p^*(1-p^*) / ME^2$. When $p$ is unknown, use $p^* = 0.5$ to get the most conservative (largest) $n$.
Question 7 · Easy
A random sample of 150 teenagers found 90 who own a smartphone. A random sample of 120 adults found 60 who own a smartphone. Researchers test $H_{0} : p_{teen} = p_{adult}$ vs. $H_{a} : p_{teen} \neq = p_{adult}$ . What is the pooled proportion $\overset{p}{^}_{c}$ ?
- A
  $\overset{p}{^}_{c} = 0.575$
  Why not A: Averaging the two sample proportions (0.60 + 0.50)/2 = 0.55 is not the pooled estimate when sample sizes differ.
- B
  $\overset{p}{^}_{c} = \frac{90 + 60}{150 + 120} = \frac{150}{270} \approx 0.556$ Correct
- C
  $\overset{p}{^}_{c} = \frac{90}{150} = 0.60$
  Why not C: This is only the teen sample proportion, not the combined pooled proportion.
- D
  $\overset{p}{^}_{c} = \frac{0.60 + 0.50}{2} = 0.55$
  Why not D: Simple average ignores that the sample sizes differ; pooled proportion weights by sample size.
Explanation
The pooled proportion combines all successes and all observations: $\overset{p}{^}_{c} = (90 + 60) / (150 + 120) = 150/270 \approx 0.556$ . This is used as the best estimate of the common $p$ under $H_{0}$ .

Key takeaway
Pooled proportion for two-sample test: $\hat{p}_c = (x_1 + x_2)/(n_1 + n_2)$; do not average the two sample proportions.
Question 8 · Easy
A 95% confidence interval for the proportion of adults who read daily is $(0.31, 0.47)$ . A school administrator says, "95% of adults read daily between 31% and 47% of the time." What is wrong with this interpretation?
- A
  The interval should be centered at the mean, not the proportion.
  Why not A: For proportions, the center is $\overset{p}{^}$ , which is the sample proportion. This is correct.
- B
  The interval estimates the population proportion, not the frequency of individual behavior; 'between 31% and 47%' refers to confidence about a fixed parameter.Correct
- C
  The interval is too wide; a 95% interval should be narrower.
  Why not C: Width depends on sample size and $\overset{p}{^}$ ; there is no inherent standard for what is 'too wide'.
- D
  The confidence level should be stated as a range, not a single value.
  Why not D: Confidence levels are always stated as a single value (e.g., 95%), not a range.
Explanation
The confidence interval estimates the true population proportion of adults who read daily. It does not mean individuals read 31%–47% of the time. The correct statement: we are 95% confident the true proportion of adults who read daily is between 31% and 47%.

Key takeaway
A CI estimates a population proportion, not an individual's frequency of behavior; always restate the parameter in context.
Question 9 · Medium
The $p$ -value for a one-sided test is 0.03. Which of the following is a correct interpretation of this $p$ -value?
- A
  There is a 3% probability that $H_{0}$ is true.
  Why not A: $H_{0}$ is either true or false (fixed); the $p$ -value is not the probability $H_{0}$ is true.
- B
  If $H_{0}$ were true, there is a 3% probability of obtaining a test statistic at least as extreme as the one observed.Correct
- C
  There is a 97% probability that the alternative hypothesis is correct.
  Why not C: The $p$ -value says nothing about the probability that $H_{a}$ is true.
- D
  The data support $H_{0}$ with 97% confidence.
  Why not D: A small $p$ -value is evidence against $H_{0}$ , not for it.
Explanation
The $p$ -value is the probability of observing a test statistic as extreme as or more extreme than the observed one, assuming $H_{0}$ is true. It measures the compatibility of the data with $H_{0}$ . A small $p$ -value means the observed data would be unlikely if $H_{0}$ were true.

Key takeaway
$p$-value = $P(\text{data this extreme or more extreme} \mid H_0 \text{ true})$. It is not the probability $H_0$ is true.
Question 10 · Medium
Researchers conduct a two-sample $z$ -test for proportions comparing recovery rates between two treatments. They obtain $p$ -value = 0.08 with $α = 0.05$ . A skeptic says the treatment difference is statistically significant. Is the skeptic correct?
- A
  Yes; $p$ -value = 0.08 is close to $α = 0.05$ , which is nearly significant.
  Why not A: 'Nearly significant' is not a formal statistical conclusion; the standard is whether $p$ -value < $α$ .
- B
  No; since $p$ -value = 0.08 > $α = 0.05$ , we fail to reject $H_{0}$ and cannot conclude a significant difference.Correct
- C
  Yes; any $p$ -value below 0.10 indicates significance.
  Why not C: Significance is determined by comparison to the pre-set $α$ , not to 0.10 as a universal threshold.
- D
  No; a two-sided test should be used, which would double the $p$ -value to 0.16.
  Why not D: We cannot retroactively change the test direction; the $p$ -value is compared to $α$ as stated.
Explanation
Since $p$ -value = 0.08 > $α = 0.05$ , we fail to reject $H_{0}$ . The result is not statistically significant at the 5% level. We cannot conclude the proportions differ. 'Borderline' $p$ -values do not justify claiming significance.

Key takeaway
Fail to reject $H_0$ when $p$-value ≥ $\alpha$; 'nearly significant' is not a valid conclusion.
Question 11 · Hard
A 90% confidence interval for the proportion of defective items is $(0.04, 0.12)$ . A quality engineer claims the defect rate is 15%. Is the claim consistent with the interval at the 10% significance level?
- A
  Yes; since 15% is close to 12%, it is consistent.
  Why not A: 15% falls outside the interval; 'close to' is not a valid statistical criterion.
- B
  No; since 15% falls outside the 90% CI, a two-sided test at $α = 0.10$ would reject $H_{0} : p = 0.15$ .Correct
- C
  Cannot determine without computing the $p$ -value.
  Why not C: For a two-sided test, there is a direct duality: a value outside a $(1 - α)$ CI corresponds to rejecting at significance level $α$ .
- D
  Yes; a 90% CI at $α = 0.10$ can only reject proportions below 0.04.
  Why not D: A two-sided interval rejects values outside both bounds (either tail).
Explanation
There is a duality between confidence intervals and two-sided hypothesis tests: if a value falls outside a $(1 - α)$ CI, the corresponding two-sided test at significance level $α$ would reject. $p = 0.15$ falls outside $(0.04, 0.12)$ , so a two-sided test at $α = 0.10$ would reject $H_{0} : p = 0.15$ .

Key takeaway
Duality: values outside a $(1-\alpha)$ CI would be rejected by a two-sided hypothesis test at level $\alpha$.
Question 12 · Hard
A large school district conducts a two-sample $z$ -test and finds a statistically significant difference in graduation rates between two schools ( $p$ -value = 0.002). The school with the lower rate has $\overset{p}{^}_{1} = 0.91$ vs. $\overset{p}{^}_{2} = 0.93$ . What is the most important caveat when reporting these results?
- A
  The test should not have been performed because both rates are above 0.90.
  Why not A: There is no restriction on performing tests when rates are high.
- B
  Statistical significance does not imply practical significance; a 2-percentage-point difference may not be educationally meaningful.Correct
- C
  A $p$ -value of 0.002 proves the schools are different in graduation policy.
  Why not C: Statistical significance in rates does not reveal policy causes, and 'proves' is too strong.
- D
  With a large enough sample, any small difference will become statistically significant, so no conclusion is possible.
  Why not D: Large samples do enable detecting small differences, but conclusions remain valid — they just need to distinguish statistical from practical significance.
Explanation
With very large sample sizes, even tiny differences become statistically significant. Here, $\overset{p}{^}_{1} = 0.91$ vs. $\overset{p}{^}_{2} = 0.93$ is a 2-percentage-point gap. A small $p$ -value confirms this is not due to chance, but does not tell us whether a 2% difference in graduation rates is meaningful for policy or practice. Always distinguish statistical and practical significance.

Key takeaway
Statistical significance ≠ practical importance. Large samples detect tiny differences; always evaluate effect size alongside the $p$-value.