Solutions — Applications of Confidence Intervals

Fluency A 95% CI for the proportion of students who pass a test is (0.72, 0.84). A teacher claims the pass rate is 90%. Is this claim consistent with the data? Explain.

The claimed value p₀ = 0.90 lies outside the 95% CI (0.72, 0.84).

The data are not consistent with the teacher’s claim. The sample evidence suggests the true pass rate is between 72% and 84%, which is below the claimed 90%. At the 95% confidence level, we have evidence to doubt the teacher’s claim.

Fluency A survey of 250 adults finds that 140 exercise regularly. Construct a 95% CI and state whether a claim of p = 0.50 is supported.

p̂ = 140/250 = 0.56
Conditions: np̂ = 140 ≥ 5 ✓; n(1−p̂) = 110 ≥ 5 ✓

SE = √(0.56 × 0.44 / 250) = √(0.0009856) ≈ 0.03139
ME = 1.96 × 0.03139 ≈ 0.06153
95% CI: (0.498, 0.622)

The claimed value p = 0.50 lies just inside the interval (0.498 < 0.50 < 0.622), so the claim is consistent with the data (barely). We cannot rule out p = 0.50 based on this sample.

Fluency In a quality control check of 400 units, 32 are faulty. Construct a 99% CI for the true defect proportion. The company claims its defect rate is under 10% — is this consistent with the data?

p̂ = 32/400 = 0.08
Conditions: np̂ = 32 ≥ 5 ✓; n(1−p̂) = 368 ≥ 5 ✓

SE = √(0.08 × 0.92 / 400) = √(0.000184) ≈ 0.01356
ME = 2.576 × 0.01356 ≈ 0.03493
99% CI: (0.045, 0.115)

The upper bound of the interval is 11.5%, which exceeds 10%. Since the interval extends above 10%, the data do not definitively confirm the defect rate is under 10% at the 99% confidence level. The claim is borderline — it is possible but not certain that the true rate is under 10%.

Fluency Write a complete contextual interpretation of the CI (0.61, 0.73) constructed from a survey of 500 people asked whether they support a new road.

We are 95% confident that the true proportion of the population who support the new road is between 61% and 73%.

This means: if this same survey were repeated many times, approximately 95% of the resulting confidence intervals would contain the true population proportion. Based on this sample of 500 people, support for the new road appears to be a clear majority — the entire interval lies well above 50%.

Understanding A government department conducts surveys in two cities. In City X (n = 400), 236 residents support a new park. In City Y (n = 400), 212 support it. Construct 95% CIs for each city and determine whether there is evidence of a difference in support levels.

City X: p̂_X = 236/400 = 0.59
SE_X = √(0.59 × 0.41 / 400) = √(0.000605) ≈ 0.02460
ME_X = 1.96 × 0.02460 ≈ 0.04822
95% CI for City X: (0.542, 0.638)

City Y: p̂_Y = 212/400 = 0.53
SE_Y = √(0.53 × 0.47 / 400) = √(0.000623) ≈ 0.02495
ME_Y = 1.96 × 0.02495 ≈ 0.04890
95% CI for City Y: (0.481, 0.579)

The intervals overlap (City X lower: 0.542; City Y upper: 0.579). There is insufficient evidence to conclude the cities have significantly different support levels at the 95% confidence level.

Understanding A researcher claims that a new teaching method produces a pass rate of at least 80%. A trial with 120 students has 88 passing. Construct a 95% CI. Does the interval support the researcher’s claim?

p̂ = 88/120 ≈ 0.7333
Conditions: np̂ = 88 ≥ 5 ✓; n(1−p̂) = 32 ≥ 5 ✓

SE = √(0.7333 × 0.2667 / 120) = √(0.001629) ≈ 0.04036
ME = 1.96 × 0.04036 ≈ 0.07910
95% CI: (0.654, 0.812)

The claimed pass rate of 80% (p = 0.80) lies inside the interval. However, much of the interval is below 80% (the lower bound is 65.4%). The data are consistent with a pass rate of 80%, but also consistent with rates as low as 65% — so the evidence in favour of the claim is not strong. A larger trial would be needed to confirm the claim convincingly.

Understanding A news article states: “A poll of 1000 voters gives candidate A 53% support, with a margin of error of ±3.1%.” Verify the margin of error and explain whether candidate A can claim majority support.

Verification (assuming 95% CI):
ME = 1.96 × √(0.53 × 0.47 / 1000)
ME = 1.96 × √(0.0002491)
ME = 1.96 × 0.01578 ≈ 0.0309 ≈ 3.1% ✓

95% CI: (0.499, 0.561)

Majority claim: The lower bound of the CI is approximately 49.9%, which is just below 50%. The interval barely includes values below the majority threshold. Strictly speaking, the data are not conclusive evidence of majority support — 50% is within (or right on the edge of) the margin of error. Candidate A leads in the poll but cannot definitively claim majority support based on this sample alone.

Understanding Two surveys about internet usage are conducted. Survey 1: n = 200, p̂ = 0.74. Survey 2: n = 800, p̂ = 0.68. Construct 95% CIs for both. Do the intervals overlap? What does this suggest?

Survey 1 (n = 200):
SE₁ = √(0.74 × 0.26 / 200) = √(0.000962) ≈ 0.03102
ME₁ = 1.96 × 0.03102 ≈ 0.06080
95% CI: (0.679, 0.801)

Survey 2 (n = 800):
SE₂ = √(0.68 × 0.32 / 800) = √(0.000272) ≈ 0.01650
ME₂ = 1.96 × 0.01650 ≈ 0.03233
95% CI: (0.648, 0.712)

The intervals overlap (Survey 1 lower bound 0.679 < Survey 2 upper bound 0.712). There is insufficient evidence to conclude the true internet usage proportions differ between the two groups. The apparent difference in sample proportions (0.74 vs 0.68) may be due to sampling variability, particularly given Survey 1’s smaller and less precise estimate.

Problem Solving A pharmaceutical company claims its new drug relieves symptoms in more than 60% of patients. A clinical trial of 180 patients finds 114 experience relief.

(a) Construct a 95% CI for the true proportion.
(b) Does the data support the company’s claim? Justify.
(c) Regulators require strong evidence (99% CI). Does the claim still hold?

p̂ = 114/180 ≈ 0.6333
Conditions: np̂ = 114 ≥ 5 ✓; n(1−p̂) = 66 ≥ 5 ✓

SE = √(0.6333 × 0.3667 / 180) = √(0.001289) ≈ 0.03590

(a) 95% CI:
ME = 1.96 × 0.03590 ≈ 0.07036
95% CI: (0.563, 0.704)

(b) The claimed threshold of 60% (p = 0.60) lies inside the 95% CI. The data are consistent with a relief rate of 60% or more — the lower bound (56.3%) is below 60%, so the data do not conclusively prove >60% relief. The sample proportion (63.3%) is above 60%, but uncertainty in the estimate means we cannot be certain the true rate exceeds 60%.

(c) 99% CI:
ME = 2.576 × 0.03590 ≈ 0.09247
99% CI: (0.541, 0.726)

At the 99% level, the lower bound drops to 54.1% — even further below 60%. The claim that “more than 60%” are relieved is not supported at the 99% confidence level. The trial would need a larger sample to provide strong evidence for the company’s claim.

Problem Solving A sports analyst compares the proportion of home wins in two football leagues. League A: 145 home wins out of 240 games. League B: 198 home wins out of 340 games.

(a) Construct 95% CIs for the home win proportion in each league.
(b) Do the intervals overlap? What conclusion can you draw about whether the leagues have different home-ground advantages?
(c) Both leagues have a long-standing belief that home advantage produces wins “more than half the time.” Does the data from each league support this belief?

(a)
League A: p̂_A = 145/240 ≈ 0.6042
SE_A = √(0.6042 × 0.3958 / 240) = √(0.000996) ≈ 0.03156
ME_A = 1.96 × 0.03156 ≈ 0.06185
95% CI for League A: (0.542, 0.666)

League B: p̂_B = 198/340 ≈ 0.5824
SE_B = √(0.5824 × 0.4176 / 340) = √(0.000715) ≈ 0.02675
ME_B = 1.96 × 0.02675 ≈ 0.05242
95% CI for League B: (0.530, 0.635)

(b) The intervals overlap (League A lower: 0.542; League B upper: 0.635). There is no strong evidence of a difference in home-ground advantage between the leagues at the 95% confidence level. Both leagues appear to have similar home win rates.

(c) League A CI: (0.542, 0.666) — entire interval above 0.50. Yes, the data support the belief for League A.
League B CI: (0.530, 0.635) — entire interval above 0.50. Yes, the data support the belief for League B.

Both leagues have CIs entirely above 50%, providing strong evidence that home teams win more than half their games in both competitions.