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Module 6 Practice Problems

20 Comprehensive Problems • Covers All Module 6 Topics

How to Use These Practice Problems

Problems 1-5: CI Interpretation & Concepts

Problem 1Interpreting Confidence IntervalsEasy

A 95% CI for mean commute time is (28.3, 35.7) minutes. What is the margin of error?

Hint: Margin of error = (upper - lower) / 2

Solution:

Margin of error = (35.7 - 28.3) / 2 = 7.4 / 2 = 3.7 minutes

Alternatively: Center = (28.3 + 35.7) / 2 = 32, so E = 35.7 - 32 = 3.7

Problem 2CI Correct InterpretationEasy

A poll reports: "We are 95% confident that between 52% and 58% of voters support the measure." Which interpretation is correct?

A: There's a 95% chance the true proportion is between 0.52 and 0.58.

B: If we repeated this process many times, 95% of intervals would contain the true proportion.

Hint: Confidence refers to the method, not a specific interval.

Solution:

Answer: B is correct.

Explanation: The parameter (true proportion) is fixed. The 95% refers to confidence in the method: 95% of intervals constructed this way will capture the true parameter. Interpretation A incorrectly treats the parameter as random.

Problems 6-10: CIs for Means

Problem 6CI for Mean CalculationMedium

A sample of n = 20 students has x̄ = 85 and s = 10. Construct a 95% CI for μ. (Use t* = 2.093 for df = 19)

Hint: CI = x̄ ± t* × (s/√n)

Solution:

Given: x̄ = 85, s = 10, n = 20, df = 19, t* = 2.093

Standard error: SE = s/√n = 10/√20 = 10/4.472 = 2.236

Margin of error: E = t* × SE = 2.093 × 2.236 = 4.68

CI: 85 ± 4.68 = (80.32, 89.68)

Interpretation: We are 95% confident that the true mean score is between 80.32 and 89.68.

Problems 11-15: CIs for Proportions

Problem 11CI for ProportionMedium

In a survey of 400 voters, 228 support Candidate A. Construct a 95% CI for the proportion of all voters who support A. (Use z* = 1.96)

Hint: First check success-failure condition. Then use p̂ ± z* × √(p̂(1-p̂)/n)

Solution:

Calculate p̂: p̂ = 228/400 = 0.57

Check conditions: np̂ = 400(0.57) = 228 ≥ 10 , n(1-p̂) = 400(0.43) = 172 ≥ 10

SE: √(0.57 × 0.43 / 400) = √(0.0006128) = 0.02475

E: 1.96 × 0.02475 = 0.0485

CI: 0.57 ± 0.049 = (0.521, 0.619) or (52.1%, 61.9%)

Problems 16-20: Sample Size

Problem 16Sample Size for MeanHard

How many students must be sampled to estimate mean GPA within ±0.1 with 95% confidence if σ ≈ 0.5? (Use z* = 1.96)

Hint: n = (z*σ / E)²

Solution:

Given: E = 0.1, z* = 1.96, σ = 0.5

Calculate: n = (1.96 × 0.5 / 0.1)² = (9.8)² = 96.04

Round up: n = 97 students

Problem 20Sample Size for Proportion (Conservative)Hard

A company wants to estimate customer satisfaction proportion within ±4% with 95% confidence. They have no prior estimate. How many customers should they survey?

Hint: Use p̂ = 0.5 (conservative), E = 0.04, z* = 1.96

Solution:

Conservative approach: p̂ = 0.5

Given: E = 0.04, z* = 1.96

Calculate: n = 0.5(0.5) × (1.96/0.04)² = 0.25 × (49)² = 0.25 × 2401 = 600.25

Round up: n = 601 customers

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