EN FR
← Course home

INF-4: Solutions — Confidence Intervals for a Proportion

Module 4 · Statistical Inference

How to use this page: Try each problem in the lesson before checking solutions here. If your answer doesn't match, read the solution carefully — especially the part that explains why common wrong answers are wrong. Understanding the error matters more than getting the right answer the first time.

← Back to Lesson INF-4

Section 5: Guided Practice Solutions

▾

Problem 1 — Compute and Check Conditions (Variants 0–2)

  • Variant 0 (budget app survey, n = 400, x = 148): (a) . (b) ✓; ✓ — z-interval valid.
  • Variant 1 (composting survey, n = 250, x = 75): (a) . (b) ✓; ✓.
  • Variant 2 (clinical trial, n = 40, x = 18): (a) . (b) ✓; ✓.

Common mistake: writing without dividing by n, or checking “n ≥ 30” instead of the actual conditions and .

Problem 2 — Standard Error and CI Endpoints (Variants 0–2)

  • Variant 0 (, n = 400, 95%): , → CI .
  • Variant 1 (, n = 250, 90%): , → CI .
  • Variant 2 (, n = 40, 99%): , → CI . Note how wide this is — small n (40) with high confidence (99%) spans over 40 percentage points.

Common mistakes: (1) using with the unknown — always use ; (2) adding/subtracting the SE directly instead of ; (3) using the wrong for the stated level.

Problem 3 — Interpretation (n = 500, x = 310, 95% CI = (0.577, 0.663))

Correct answer: “We are 95% confident that the true proportion of adults who support stricter labelling lies between 57.7% and 66.3%.”

Why the other options are wrong:

  • Option A (“95% probability that p is in this interval”) treats the fixed parameter p as random.
  • Option C (“95% of the 500 adults”) confuses sample members with the parameter.
  • Option D (“future sample proportions fall in this range”) describes the sampling distribution of , not a CI for p.

Problem 4 — Sample Size (vaccination coverage, E = 0.03, 95%, )

(a) Use — the prior estimate. Using would be more conservative but wastes sample size when a good prior is available.

(b) . Round up: .

Common mistakes: (1) using when a prior estimate is given; (2) rounding 1024.4 down — always round up; (3) forgetting to square before multiplying by .

Section 6: Independent Practice Solutions

▾

Problem 1 — Full CI Computation (Generator)

The approach is always the same:

  1. .
  2. Check and .
  3. .
  4. (1.645 / 1.96 / 2.576 for 90% / 95% / 99%).
  5. .
  6. Interpret: “We are [CL]% confident the true proportion lies between [lower] and [upper].”

Problem 2 — Margin of Error and Interpretation (Variants 0–2)

  • Variant 0 (Pharmacare poll, n = 900, x = 513): (a) ; → 3.3 percentage points. (b) “Probability 95%” implies p is random — it isn’t; say “95% confident the true proportion is between 53.7% and 60.3%.” (c) New : . To halve E, quadruple n.
  • Variant 1 (food insecurity, n = 500, x = 185): (a) → 3.5 pp. (b) “90% chance” wrongly treats p as random — say “90% confident.” (c) .
  • Variant 2 (burnout survey, n = 300, x = 81): (a) → 6.2 pp. (b) “99% sure” → “99% confident.” (c) .

Problem 3 — Sample Size (Generator)

  1. Identify E, confidence level (→ ), and (use 0.5 if no prior).
  2. .
  3. Round up to the next whole number.

Problem 4 — When Conditions Are Not Met (Variants 0–2)

  • Variant 0 (rare-book collectors, n = 30, x = 2): . — condition fails. Do not use the z-interval; use a larger sample, the Clopper–Pearson exact interval, or the Wilson interval.
  • Variant 1 (defective components, n = 20, x = 1): . — fails. Use an exact binomial method or collect more data.
  • Variant 2 (nesting sites, n = 15, x = 14): . — the failure count fails. The distribution is strongly left-skewed and the z-interval would give an upper bound above 1; use an exact or Wilson interval.

Key lesson: conditions can fail at either extreme — very small (too few successes) or very large (too few failures). Always check both.

Section 7: Mastery Check Solutions

▾

Problem 1 — Feynman Test: Why in the SE Formula

Since is unknown, we substitute into the SE formula. This works when the sample is large enough for to be close to (CLT). The conditions and check whether the approximation is reliable; if they fail, the SE estimate using may be too inaccurate to trust.

Problem 2 — Apply: 99% CI for Vegetable Consumption (n = 450, x = 180)

(a) Correct SE formula: .

(b) . Conditions: ✓; ✓. , → CI . We are 99% confident that between 34.1% and 45.9% of all adults consume fewer than two vegetable servings daily.

Problem 3 — Error Analysis (CI = (0.42, 0.58), n = 200)

Error 1: “95% probability that p is in the interval” — p is fixed; use “95% confident.”

Error 2: concluding p = 0.50 because 0.50 is inside the interval. The CI says 0.50 is plausible, not that it equals the true value.

Correct statement: “We are 95% confident the true proportion is between 0.42 and 0.58. We cannot rule out an even split, but we cannot conclude one exists either.”

Section 8: Boss Fight Solutions

▾

Path A — The Analyst: Restaurant Inspections (n = 120, x = 47)

Task 1: . Conditions: ✓; ✓.

Task 2: , → CI .

Task 3: 40% falls inside (0.304, 0.479) — the data do not support the city’s claim at 95% confidence.

Task 4: restaurants.

Path B — The Architect: CEGEP Tutoring Study

Task 1: → ±5.7 percentage points.

Task 2: With : — barely better. Since , the improvement over the worst case is tiny.

Task 3: . This exceeds the budget of 300 — the target precision is not achievable with available resources.

Section 9: Challenge Problem Solutions

▾

Challenge 1 — Wilson Interval

Full worked solutions are embedded in each variant’s “Show Solution” toggle within Section 9.

Key takeaway: the standard z-interval can produce bounds below 0 or above 1 when conditions fail. The Wilson interval always stays in [0, 1] and performs better for small samples or extreme proportions.

Challenge 2 — Margin of Error as a Function of

(b) is maximized at because is maximized there — a proportion near 0.5 carries maximum variability.

(c) Using is not conservative for all . For near 0.5, , so the required n is larger than predicts. The only safe universal choice is .

Challenge 3 — Two Polls (Generator)

  1. Compute and for each poll.
  2. Build a CI for each independently at the stated confidence level.
  3. Check overlap: if , the intervals overlap.
  4. Overlapping CIs are consistent with each other; non-overlapping suggest a real difference. Formally comparing two proportions requires a two-sample z-test, not just CI overlap.
← Return to Lesson INF-4