Topic Review — Interval Estimates for Proportions — Solutions

← Interval Estimates for Proportions

This review covers all lessons in this topic: confidence intervals for proportions, margin of error and sample size, and applications. Click each answer to reveal the worked solution.

Review Questions

1. Define the margin of error in the context of a confidence interval for a proportion. ▶ Show Answer
   The margin of error (ME) is the maximum expected difference between the sample proportion p̂ and the true population proportion p, at a given confidence level. It equals the half-width of the confidence interval:
   ME = z* × √(p̂(1−p̂)/n)
   The CI can be written as p̂ − ME < p < p̂ + ME.
2. A sample of 300 people finds 192 prefer online shopping. Construct a 95% confidence interval for the true proportion. ▶ Show Answer
   p̂ = 192/300 = 0.64
   Conditions: np̂ = 192 ≥ 5 ✓; n(1−p̂) = 108 ≥ 5 ✓
   SE = √(0.64 × 0.36 / 300) = √(0.000768) ≈ 0.02771
   ME = 1.96 × 0.02771 ≈ 0.05431
   95% CI: (0.586, 0.694)
   We are 95% confident the true proportion who prefer online shopping is between 58.6% and 69.4%.
3. Explain the correct interpretation of a 95% confidence interval. What does 95% refer to? ▶ Show Answer
   A 95% CI does not mean there is a 95% probability that the true p lies in this specific interval. Once computed, the interval either contains p or it doesn’t.
   The 95% refers to the procedure: if we took many repeated random samples and constructed a CI from each, approximately 95% of those intervals would contain the true p.
4. State the three critical values z* and their associated confidence levels. ▶ Show Answer
   90% confidence: z* = 1.645
   95% confidence: z* = 1.96
   99% confidence: z* = 2.576
   These values come from the standard normal distribution: z* is the value such that P(−z* < Z < z*) equals the confidence level.
5. A 95% CI for p is (0.33, 0.49). Find the sample proportion p̂, the margin of error, and back-calculate the approximate sample size. ▶ Show Answer
   p̂ = (0.33 + 0.49)/2 = 0.41
   ME = (0.49 − 0.33)/2 = 0.08
   Back-calculating n:
   0.08 = 1.96 × √(0.41 × 0.59/n)
   0.04082 = √(0.2419/n)
   0.001666 = 0.2419/n
   n ≈ 145.2 → n ≈ 146
6. Check whether the normality conditions are met for n = 80 and p̂ = 0.06. ▶ Show Answer
   Conditions: np̂ ≥ 5 and n(1−p̂) ≥ 5
   np̂ = 80 × 0.06 = 4.8 < 5 ✗
   Conditions are NOT met. The expected number of successes (4.8) is below 5, so the normal approximation to the distribution of p̂ is unreliable. A CI based on the normal formula should not be used; a larger sample or alternative method is needed.
7. A researcher wants ME ≤ 0.04 with 95% confidence. No prior estimate of p is available. Find the minimum sample size. ▶ Show Answer
   Use conservative p̂ = 0.5:
   n ≥ (1.96)² × 0.5 × 0.5 / (0.04)²
   n ≥ 3.8416 × 0.25 / 0.0016
   n ≥ 0.9604 / 0.0016 = 600.25
   Minimum sample size: n = 601
8. A manufacturer says its product has a 15% return rate. A random sample of 200 purchases finds 24 returns. Construct a 95% CI. Is the manufacturer’s claim consistent with the data? ▶ Show Answer
   p̂ = 24/200 = 0.12
   SE = √(0.12 × 0.88 / 200) = √(0.000528) ≈ 0.02298
   ME = 1.96 × 0.02298 ≈ 0.04503
   95% CI: (0.075, 0.165)
   The claimed value of 15% (0.15) lies inside the CI (0.075, 0.165). The data are consistent with the manufacturer’s claim.
9. Explain why using p̂ = 0.5 in the sample size formula gives the most conservative (largest) estimate of n. ▶ Show Answer
   The required sample size formula is n ≥ (z*)² × p̂(1−p̂) / ME². The factor p̂(1−p̂) is maximised when p̂ = 0.5, where p̂(1−p̂) = 0.25. Any other value of p̂ gives a smaller product (e.g., p̂ = 0.3 gives 0.21, p̂ = 0.1 gives 0.09). Using p̂ = 0.5 therefore produces the largest n, guaranteeing that the desired ME is achieved regardless of the actual value of p.
10. A pilot study estimates p̂ ≈ 0.72. A researcher wants a 99% CI with ME ≤ 0.03. Find the required sample size. ▶ Show Answer
    z* = 2.576, ME = 0.03, p̂ = 0.72
    n ≥ (2.576)² × 0.72 × 0.28 / (0.03)²
    n ≥ 6.6355 × 0.2016 / 0.0009
    n ≥ 1.3377 / 0.0009
    n ≥ 1486.3
    Minimum sample size: n = 1487
11. Two schools are surveyed about homework preferences. School 1: n = 150, 87 prefer less homework. School 2: n = 150, 72 prefer less homework. Construct 95% CIs for each and determine whether there is evidence of a genuine difference. ▶ Show Answer
    School 1: p̂₁ = 87/150 = 0.58
    SE₁ = √(0.58 × 0.42/150) = √(0.001624) ≈ 0.04030
    95% CI: (0.58 − 0.079, 0.58 + 0.079) = (0.501, 0.659)
    
    School 2: p̂₂ = 72/150 = 0.48
    SE₂ = √(0.48 × 0.52/150) = √(0.001664) ≈ 0.04080
    95% CI: (0.48 − 0.080, 0.48 + 0.080) = (0.400, 0.560)
    
    The intervals overlap (School 1 lower: 0.501; School 2 upper: 0.560). There is insufficient evidence to conclude a genuine difference at the 95% level.
12. A current 95% CI uses n = 500. How large a sample is needed to reduce the margin of error by one third (to 2/3 of its current value)? ▶ Show Answer
    ME ∝ 1/√n
    To get ME_new = (2/3) ME_old:
    1/√n_new = (2/3) × 1/√n_old
    √n_new = (3/2) × √n_old
    n_new = (3/2)² × n_old = 2.25 × 500 = 1125
    
    A sample of 1125 is needed to reduce the margin of error to two-thirds of the current value.
13. A survey of 600 Australians finds 420 support a new environmental policy.
    • (a) Construct a 90% CI.
    • (b) Construct a 99% CI.
    • (c) Compare the widths. What trade-off does this illustrate?
    ▶ Show Answer
    p̂ = 420/600 = 0.70
    SE = √(0.70 × 0.30 / 600) = √(0.00035) ≈ 0.01871
    
    (a) 90% CI: ME = 1.645 × 0.01871 ≈ 0.03078
    CI: (0.669, 0.731)
    
    (b) 99% CI: ME = 2.576 × 0.01871 ≈ 0.04819
    CI: (0.652, 0.748)
    
    (c) Width of 90% CI ≈ 0.062; Width of 99% CI ≈ 0.096
    The 99% CI is about 55% wider. This illustrates the trade-off between confidence and precision: higher confidence requires a wider interval for the same sample data.
14. An online news poll of 2000 respondents reports 58% support for a proposition with ME = ±2.2%. A statistician says the poll is unreliable. Give one reason why. ▶ Show Answer
    An online news poll suffers from voluntary response bias (self-selection bias). Only people who choose to respond participate — typically those with strong opinions. This makes the sample unrepresentative of the general population. The margin of error calculation assumes a random sample; applying it to a non-random sample produces a misleading measure of precision. The stated ±2.2% refers only to sampling variability, not to bias from non-random sampling.
15. A 95% CI based on p̂ = 0.48 and n = 400 is constructed. Without recalculating, what would happen to this interval if the confidence level were changed to 90%? To 99%? Describe the effect on both the margin of error and the interval width. ▶ Show Answer
    The SE = √(0.48 × 0.52/400) ≈ 0.02497 is fixed (depends only on p̂ and n).
    
    90% CI: z* decreases from 1.96 to 1.645. ME decreases from 1.96 × 0.02497 ≈ 0.04894 to 1.645 × 0.02497 ≈ 0.04108. The interval narrows (less certain, more precise).
    
    99% CI: z* increases from 1.96 to 2.576. ME increases to 2.576 × 0.02497 ≈ 0.06432. The interval widens (more certain, less precise).
    
    Summary: lower confidence → narrower interval; higher confidence → wider interval.