Topic Review — Statistical Inference — Solutions

← Statistical Inference

This review covers all four lessons: sampling distributions and the Central Limit Theorem, confidence intervals for population mean, hypothesis testing for population mean, and Type I and Type II errors. Click each answer to reveal the worked solution.

Review Questions

1. A random sample of size n = 36 is drawn from a population with mean μ = 120 and standard deviation σ = 18. State the distribution of the sample mean &x̄, giving its mean and standard deviation. ▶ Show Answer
   By the Central Limit Theorem, &x̄ ~ N(μ, σ²/n) = N(120, 18²/36) = N(120, 9).
   So &x̄ ~ N(120, 9), with mean 120 and standard deviation (SE) = 3.
2. Using the distribution from Q1, find P(&x̄ > 124). ▶ Show Answer
   Z = (124 − 120) / 3 = 4/3 ≈ 1.333
   P(&x̄ > 124) = P(Z > 1.333) = 1 − Φ(1.333) ≈ 1 − 0.9088 ≈ 0.0912
3. A population has σ = 15 (known). A sample of n = 25 gives &x̄ = 84. Construct a 95% confidence interval for μ. ▶ Show Answer
   SE = σ/√n = 15/5 = 3
   95% CI: &x̄ ± 1.96 × SE = 84 ± 1.96 × 3 = 84 ± 5.88
   95% CI: (78.12, 89.88)
4. Interpret the confidence interval (78.12, 89.88) in the context of Q3. ▶ Show Answer
   We are 95% confident that the true population mean μ lies between 78.12 and 89.88.
   This means that if we repeated this sampling procedure many times, approximately 95% of the resulting confidence intervals would contain the true population mean.
5. A researcher wants a 99% confidence interval for μ with σ = 20 and margin of error no larger than 4. What is the minimum sample size required? ▶ Show Answer
   z* for 99% CI = 2.576
   ME = z* × σ/√n ≤ 4
   2.576 × 20/√n ≤ 4
   √n ≥ 2.576 × 20 / 4 = 12.88
   n ≥ 12.88² ≈ 165.9
   Minimum n = 166
6. A hypothesis test has H₀: μ = 50 vs H₁: μ ≠ 50, with σ = 10 known, n = 100. A sample gives &x̄ = 52. Perform the test at α = 0.05 and state your conclusion. ▶ Show Answer
   SE = 10/√100 = 1
   Test statistic: z = (&x̄ − μ₀)/SE = (52 − 50)/1 = 2
   Two-tailed critical value: z* = 1.96
   Since |z| = 2 > 1.96, we reject H₀.
   There is sufficient evidence at the 5% level that μ ≠ 50.
7. For the test in Q6, calculate the p-value and use it to confirm your conclusion. ▶ Show Answer
   p-value = 2 × P(Z > 2) = 2 × (1 − 0.9772) = 2 × 0.0228 = 0.0456
   Since p-value = 0.0456 < α = 0.05, we reject H₀. This confirms the conclusion in Q6.
8. A one-tailed test has H₀: μ = 200 vs H₁: μ < 200, σ = 25, n = 49, α = 0.01. The critical value is z = −2.326. Find the rejection region (critical value of &x̄). ▶ Show Answer
   SE = 25/√49 = 25/7 ≈ 3.571
   Critical &x̄ = μ₀ + z* × SE = 200 + (−2.326) × 3.571 = 200 − 8.30 ≈ 191.70
   Reject H₀ when &x̄ < 191.70.
9. A factory claims its bolts have mean diameter μ = 12 mm. Quality control suspects the bolts are undersize. With σ = 0.8 mm and a sample of 64 bolts giving &x̄ = 11.85 mm, test at α = 0.05. ▶ Show Answer
   H₀: μ = 12    H₁: μ < 12 (one-tailed, left)
   SE = 0.8/√64 = 0.1
   z = (11.85 − 12)/0.1 = −1.5
   Critical value (one-tailed, α = 0.05): z* = −1.645
   Since z = −1.5 > −1.645, we fail to reject H₀.
   There is insufficient evidence at 5% that the bolts are undersize.
10. In Q9, describe the Type II error that could have been made. What would need to happen to reduce the risk of this error? ▶ Show Answer
    A Type II error here would mean failing to reject H₀ (μ = 12) when the bolts are actually undersize (H₀ is false). The factory would continue producing bolts that are too small, thinking the process is fine.
    
    To reduce the risk of a Type II error (increase power):
    • Increase the sample size n (reduces SE, making it easier to detect small deviations)
    • Increase α (e.g. use 0.10 instead of 0.05), accepting more Type I errors
    • Use more precise measuring equipment (reduces σ)
11. Define the power of a hypothesis test. A test has α = 0.05 and detects a particular alternative with power 0.75. What is the probability of a Type II error? ▶ Show Answer
    Power = P(reject H₀ | H₀ is false) = the probability of correctly detecting a false null hypothesis.
    Power = 1 − β
    0.75 = 1 − β
    β = 0.25
    There is a 25% chance of failing to detect the real effect (Type II error).
12. A 95% CI for μ (with known σ = 12, n = 36) gives the interval (58.08, 61.92). A researcher claims μ = 63. Is this claim consistent with the interval? What would the result of a two-tailed test at α = 0.05 be? ▶ Show Answer
    The value μ = 63 lies outside the 95% CI (58.08, 61.92).
    A 95% CI is equivalent to a two-tailed test at α = 0.05: we reject H₀ if and only if μ₀ lies outside the CI.
    Since 63 ∉ (58.08, 61.92), we reject H₀: μ = 63 at the 5% level.
    There is sufficient evidence that μ ≠ 63.
13. Increasing the confidence level from 95% to 99% (same n and σ) makes the confidence interval wider. Explain why, and state the trade-off involved. ▶ Show Answer
    The confidence interval is &x̄ ± z* × SE. The only thing that changes between 95% and 99% is the critical value z*.
    • For 95%: z* = 1.96
    • For 99%: z* = 2.576
    A larger z* means a wider interval. We need to be “more sure” of capturing μ, so we cast a wider net.
    Trade-off: Higher confidence comes at the cost of precision. A 99% CI is wider and less informative about where exactly μ is. To get a narrow interval with high confidence, you need a larger sample size n.
14. A random variable X ~ N(μ, 64). A sample of 16 gives &x̄ = 42. Assuming H₀: μ = 45 vs H₁: μ < 45 at α = 0.05:
    • (a) Carry out the test and state your conclusion.
    • (b) Suppose the true mean is μ = 40. Calculate β (P(Type II error)).
    ▶ Show Answer
    σ = √64 = 8, SE = 8/√16 = 2
    (a) z = (42 − 45)/2 = −1.5
    Critical value (one-tailed, left, α = 0.05): −1.645
    Since z = −1.5 > −1.645, fail to reject H₀. Insufficient evidence that μ < 45.
    
    (b) Rejection region: reject when &x̄ < 45 − 1.645 × 2 = 45 − 3.29 = 41.71
    β = P(&x̄ ≥ 41.71 | μ = 40) = P(Z ≥ (41.71 − 40)/2) = P(Z ≥ 0.855)
    β = 1 − Φ(0.855) ≈ 1 − 0.804 ≈ 0.196
15. A manufacturing company tests a new process with n = 100, σ = 5. The test is H₀: μ = 200 vs H₁: μ ≠ 200 at α = 0.02.
    • (a) Find the critical values for &x̄.
    • (b) A sample gives &x̄ = 198.9. State the conclusion.
    • (c) Comment on the possibility of each type of error given your conclusion.
    ▶ Show Answer
    SE = 5/√100 = 0.5
    (a) For α = 0.02, two-tailed: z* = ±2.326
    Critical &x̄ values: 200 ± 2.326 × 0.5 = 200 ± 1.163
    Reject H₀ when &x̄ < 198.837 or &x̄ > 201.163
    
    (b) &x̄ = 198.9 > 198.837, so we fail to reject H₀. There is insufficient evidence at the 2% level that μ ≠ 200.
    
    (c) Since we did not reject H₀:
    • A Type I error is impossible here (you can only make a Type I error when you reject H₀).
    • A Type II error is possible: if the true mean really is different from 200, we have incorrectly failed to detect this.