AP Statistics 3.10 Constructing a Confidence Interval for the Difference Between Two Population Proportions- Exam Style Questions - FRQs - New Syllabus
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(3.3\) — Constructing a Confidence Interval for a Population Proportion (Parts \( \mathrm{a} \), \( \mathrm{b} \))
• Topic \(3.4\) — Justifying a Claim Based on a Confidence Interval for a Population Proportion (Part \( \mathrm{a} \))
• Topic \(3.10\) — Constructing a Confidence Interval for the Difference Between Two Population Proportions (Part \( \mathrm{c} \))
▶️ Answer/Explanation
(a)
The appropriate procedure is a one-sample \(z\)-interval for a population proportion.
Let \(p\) = the true proportion of all adults in the United States who would have chosen the economy statement.
From the sample: \(\hat{p} = 0.37\), \(n = 1{,}048\), and the critical value for 95% confidence is \(z^* = 1.96\).
The confidence interval formula is:
\( \hat{p} \pm z^* \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}} \)
Substituting the values:
\( 0.37 \pm 1.96\sqrt{\dfrac{(0.37)(0.63)}{1{,}048}} \)
\( 0.37 \pm 1.96\sqrt{0.000222} \)
\( 0.37 \pm 1.96(0.0149) \)
\( 0.37 \pm 0.03 = (0.34,\ 0.40) \)
Interpretation: We are 95 percent confident that the interval from 0.34 to 0.40 captures the true proportion of all adults in the United States who would have chosen the economy statement.
(b)
This condition is necessary because the confidence interval formula for a population proportion relies on approximating the binomial distribution with a normal distribution. This approximation causes the sampling distribution of \(\hat{p}\) to be approximately normal, which is what allows us to use the \(z\)-critical value and construct a valid interval.
However, the normal approximation to the binomial works well only when both \(n\hat{p}\) and \(n(1-\hat{p})\) are at least 10. If either of these quantities is too small, the sampling distribution of \(\hat{p}\) will be noticeably skewed rather than approximately normal, and the confidence interval will not be reliable.
(c)
No, the two-sample \(z\)-interval for a difference between proportions is not an appropriate procedure here.
A key requirement for the two-sample \(z\)-interval is that the two proportions come from two independent samples. In this study, however, both proportions — the proportion choosing the environment statement and the proportion choosing the economy statement — come from the same single sample of 1,048 adults. Because each person was forced to choose between the two statements (or express no preference), the two proportions are not independent: knowing one proportion directly constrains the other. Since the independence condition is violated, the two-sample \(z\)-interval is not appropriate for this situation.
Question
Most-appropriate topic codes (AP Statistics):
• Topic \(3.10\) — Constructing a Confidence Interval for the Difference Between Two Population Proportions (Part \(\mathrm{b}\))
• Topic \(3.11\) — Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions (Parts \(\mathrm{b}\), \(\mathrm{e}\))
• Topic \(3.1\) — Estimators (Part \(\mathrm{c}\): estimating relative risk using \(\hat{p}_A/\hat{p}_B\))
• Topic \(3.11\) — Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions (Part \(\mathrm{d}\): interpreting the confidence interval for relative risk)
▶️ Answer/Explanation
(a)
Yes, this experiment can be performed as a double-blind experiment by introducing placebos for each treatment group.
Patients assigned to treatment A (pill) would also receive a placebo injection. Patients assigned to treatment B (injection) would also receive a placebo pill. This way, every patient receives both a pill and an injection, but one of the two is a placebo.
Since neither the patients nor the physicians administering the treatments know which is the real treatment and which is the placebo, neither group is aware of the treatment assignment — making the experiment double-blind.
(b)
First, compute the sample proportions:
\(\hat{p}_A = \frac{38}{154} \approx 0.2468, \qquad \hat{p}_B = \frac{16}{164} \approx 0.0976\)
The 95% confidence interval for \(p_A – p_B\) is:
\(\hat{p}_A – \hat{p}_B) \pm z^* \sqrt{\frac{\hat{p}_A(1-\hat{p}_A)}{n_A} + \frac{\hat{p}_B(1-\hat{p}_B)}{n_B}}\)
\(0.2468 – 0.0976) \pm 1.96\sqrt{\frac{(0.2468)(0.7532)}{154} + \frac{(0.0976)(0.9024)}{164}}\)
\(0.1492 \pm 1.96(0.0418)\)
\(0.1492 \pm 0.0818\)
\(\boxed{(0.0674,\ 0.2310)}\)
We are 95% confident that the true difference in 15-year survival rates \((p_A – p_B)\) is between \(0.0674\) and \(0.2310\). Because the entire interval lies above zero, this provides evidence that treatment A has a higher 15-year survival rate than treatment B.
(c)
The estimated relative risk is:
\(\frac{\hat{p}_A}{\hat{p}_B} = \frac{38/154}{16/164} = \frac{0.2468}{0.0976} \approx 2.53\)
\(\boxed{\text{Estimated relative risk} \approx 2.53}\)
This means patients receiving treatment A are estimated to be about 2.53 times as likely to survive at least 15 years as patients receiving treatment B.
(d)
A 95% confidence interval for \(\ln\!\left(\dfrac{p_A}{p_B}\right)\) is given as \((0.3868,\ 1.4690)\).
To convert this to a confidence interval for the relative risk \(\dfrac{p_A}{p_B}\), apply the exponential (inverse of the natural log) to each endpoint:
\(e^{0.3868} \approx 1.47 \qquad \text{and} \qquad e^{1.4690} \approx 4.34\)
\(\boxed{\left(1.47,\ 4.34\right)}\)
We are 95% confident that patients receiving treatment A are between 1.47 and 4.34 times as likely to survive at least 15 years compared to patients receiving treatment B.
(e)
When the survival proportions are small (as here, approximately 0.25 and 0.10), the confidence interval for the relative risk is more informative and practically meaningful than the confidence interval for the difference in proportions.
Knowing that a patient’s chance of survival is between 1.47 and 4.34 times greater with treatment A is more vivid and easier to interpret clinically than knowing the absolute difference in proportions is somewhere between 0.07 and 0.23 — a range that may sound small even though it represents a substantial relative advantage.
Question
Most-appropriate topic codes (AP Statistics):
• Topic 3.11 — Justifying a Claim Based on a Confidence Interval for the Difference Between Two Population Proportions (Part \(\mathrm{b}\))
▶️ Answer/Explanation
(a)
Step 1: Identify the procedure and check conditions.
We will use a two-sample \(z\) confidence interval for \(p_D – p_N\), the difference in the proportions of parts meeting specifications for the day shift and night shift.
Let \(\hat{p}_D\) = proportion of day shift parts meeting specifications, and \(\hat{p}_N\) = proportion of night shift parts meeting specifications.
\(\hat{p}_D = \frac{188}{200} = 0.94 \qquad \hat{p}_N = \frac{180}{200} = 0.90\)
Conditions:
1. Independent random samples: The problem states that random samples of parts were selected from each shift. We assume that day shift and night shift production are independent — that is, each part is produced by one shift only and machine quality does not vary over time.
2. Large sample sizes (normal approximation valid):
\(n_D \hat{p}_D = 200(0.94) = 188 > 10\)
\(n_D(1-\hat{p}_D) = 200(0.06) = 12 > 10\)
\(n_N \hat{p}_N = 200(0.90) = 180 > 10\)
\(n_N(1-\hat{p}_N) = 200(0.10) = 20 > 10\)
All conditions are satisfied. It is reasonable to use the large-sample normal procedure.
Step 2: Compute the confidence interval.
The formula for the 96% confidence interval for \(p_D – p_N\) is:
\((\hat{p}_D – \hat{p}_N) \pm z^* \sqrt{\frac{\hat{p}_D(1-\hat{p}_D)}{n_D} + \frac{\hat{p}_N(1-\hat{p}_N)}{n_N}}\)
For a 96% confidence level, the critical value is \(z^* = 2.0537\).
\((0.94 – 0.90) \pm 2.0537\sqrt{\frac{(0.94)(0.06)}{200} + \frac{(0.90)(0.10)}{200}}\)
\(= 0.04 \pm 2.0537\sqrt{0.000282 + 0.00045}\)
\(= 0.04 \pm 2.0537\sqrt{0.000732}\)
\(= 0.04 \pm 2.0537(0.02705)\)
\(= 0.04 \pm 0.0556\)
\(\boxed{(-0.0156,\ 0.0956)}\)
Step 3: Interpret the interval.
Based on these samples, we are 96 percent confident that the true difference in the proportions of parts meeting specifications for the day shift and night shift is between \(-0.0156\) and \(0.0956\).
(b)
Since \(0\) is contained within the 96 percent confidence interval \((-0.0156,\ 0.0956)\), zero is a plausible value for the difference \(p_D – p_N\).
This means that at the \(\alpha = 0.04\) significance level, we do not have sufficient evidence to support the manager’s belief that there is a statistically significant difference between the proportions of parts meeting specifications for the two shifts.
\(\boxed{\text{No statistically significant difference detected at } \alpha = 0.04}\)