Statistical Inference and Hypothesis Testing Practice
Exercise 1: Multiple Choice
1. Different Rejection Rule
- (a) H₁: µ = 90 (< 100) → reject for small X̄ (left tail)
- (b) H₁: µ < 100 → reject for small X̄ (left tail)
- (c) H₁: µ = 90, H₀: µ > 90 → reject for small X̄ (left tail)
All three tests reject on the left tail. Re-examining the options: all are left-tailed. Looking again at (c): H₀: µ > 90 vs H₁: µ = 90. Here, H₁ specifies a value below H₀, so we reject for small values. They all look left-tailed, but (c) is the odd one because the null is a composite hypothesis stated as a strict inequality, making the rejection logic structurally different. Answer: (c)
2. Different Rejection Rule
- (a) H₁: µ = 120 (> 100) → right tail
- (b) H₁: µ > 100 → right tail
- (c) H₀: µ ≤ 120, H₁: µ = 120 → reject for large X̄ supporting H₁ being at the upper boundary… actually, this is unusual. Since H₀ allows µ ≤ 120 and H₁ says µ = 120, the rejection logic differs. Answer: (c)
3. Variance Increases, Mean Fixed
(a) It becomes wider and flatter.
4. Sum of n Squared Independent Standard Normals
(b) A chi-squared distribution with n degrees of freedom.
5. Variance of Sample Mean Estimator
(a) σ²/n
6. Cramér-Rao Lower Bound
(a) It gives a lower bound for the variance of unbiased estimators under suitable conditions.
7. Incorrect Statement About Consistent Estimator
(c) — A consistent estimator need not be unbiased; it can be asymptotically unbiased.
8. Incorrect Statement About a Pivot
(b) — A pivot’s distribution should not depend on unknown parameters; that is its defining feature.
9. Paired Data
(b) The same units are measured under two different conditions.
10. Incorrect Statement About Hypothesis Testing
(c) — Knowing the level of confidence (significance) does not tell you the power; power depends on the specific alternative value, sample size, and variance.
Exercise 2: Estimators
Given: X₁,…,X₅ are i.i.d. with mean µ and variance σ².
(a) Unbiasedness
Analysis of the estimators’ expected values to determine bias.
(b) MSE Comparison with µ = σ
For µ̂₁:
- Var(µ̂₁) = (1/36) · 5σ² = 5σ²/36
- Bias² = µ²/36 = σ²/36 (since µ = σ)
- MSE(µ̂₁) = 5σ²/36 + σ²/36 = 6σ²/36 = σ²/6
For µ̂₂:
- Var(µ̂₂) = (1/25) · (4+1+1+4+1)σ² = 11σ²/25
- Bias = 0
- MSE(µ̂₂) = 11σ²/25
Comparing: σ²/6 ≈ 0.1667σ² vs 11σ²/25 = 0.44σ². µ̂₁ has a smaller MSE, so µ̂₁ is better despite being biased.
(c) Comparison with Sample Mean X̄
X̄ = (X₁+…+X₅)/5:
- E[X̄] = µ (unbiased), Var(X̄) = σ²/5, MSE(X̄) = σ²/5 = 0.2σ²
- µ̂₁: MSE = σ²/6 ≈ 0.167σ²
- µ̂₂: MSE = 0.44σ²
The sample mean is better than µ̂₂ but worse than µ̂₁ when µ = σ. (Without the µ = σ assumption, X̄ is the best linear unbiased estimator.)
Exercise 3: Parking Permits
Given: p₀ = 0.65, n = 90.
(a) Approximate Distribution
By the Central Limit Theorem, since np₀(1-p₀) = 90(0.65)(0.35) = 20.475 > 10, p̂ is approximately normal with mean p and variance p(1-p)/n.
(b) Probability P(p̂ ≤ 0.60)
Standard error: √(0.65 · 0.35 / 90) = √(0.002528) = 0.05028
$$P(\hat{p} \leq 0.60) = P(Z \leq -0.99) = 1 – 0.8389 = 0.1611$$
(c) Hypothesis Test, α = 0.01
H₀: p = 0.65 vs H₁: p ≠ 0.65 (two-sided). Observed: p̂ = 72/90 = 0.80.
Test statistic: Z = (0.80 – 0.65) / 0.05028 = 2.983. Critical values: ±z_{0.005} = ±2.576.
Since |2.983| > 2.576, reject H₀. p-value = 2 · P(Z > 2.983) = 2(0.00143) ≈ 0.0029 < 0.01.
Conclusion: There is sufficient evidence at the 1% level that the true proportion differs from 65%.
(d) Validity of Normal Approximation
The normal approximation is valid because both np₀ = 58.5 ≥ 10 and n(1-p₀) = 31.5 ≥ 10, satisfying the conditions for the CLT to apply to the sample proportion.
Exercise 4: Irrigation Unit
Given: µ₀ = 75, µ₁ = 68, σ = 14, n = 49, σ/√n = 14/7 = 2. Rejection region: X̄ < 71.
(a) Type I Error
$$\alpha = P(\bar{X} < 71 | \mu = 75) = P(Z < (71-75)/2) = P(Z < -2) = 0.0228$$
(b) P-value with x̄ = 70
Since this is a left-tailed test: $$p\text{-value} = P(\bar{X} < 70 | \mu = 75) = P(Z < (70-75)/2) = P(Z < -2.5) = 0.0062$$
Since 0.0062 < α = 0.0228, reject H₀.
(c) Type II Error and Power
$$\beta = P(\bar{X} \geq 71 | \mu = 68) = P(Z \geq (71-68)/2) = P(Z \geq 1.5) = 1 – 0.9332 = 0.0668$$
$$\text{Power} = 1 – \beta = 0.9332$$
Exercise 5: Acoustic Panels
Sample Statistics
- Sample A (n_A = 12): Sum = 703.0, x̄_A = 58.583. Deviations squared sum: ≈ 0.2092. s²_A ≈ 0.2092/11 ≈ 0.01902, s_A ≈ 0.1379.
- Sample B (n_B = 11): Sum = 663.1, x̄_B = 60.282. s²_B ≈ 1.0344, s_B ≈ 1.0170.
(a) Test of Equal Variances at α = 0.02
H₀: σ²_A = σ²_B vs H₁: σ²_A ≠ σ²_B. F = s²_B / s²_A = 1.0344 / 0.01902 ≈ 54.39 (larger variance over smaller variance).
Critical values: F_{0.01, 10, 11} ≈ 4.54. Since 54.39 >> 4.54, reject H₀. There is strong evidence that the population variances differ.
(b) Confidence Interval for µ_A – µ_B at α = 0.01
Using Welch’s t-test for unequal variances: x̄_A – x̄_B = 58.583 – 60.282 = -1.699.
SE = √(s²_A/n_A + s²_B/n_B) = √(0.01902/12 + 1.0344/11) = √0.0956 = 0.3092. Welch df ≈ 10. t_{0.005, 10} ≈ 3.169.
CI: -1.699 ± 3.169(0.3092) = -1.699 ± 0.980 = (-2.679, -0.719).
Since 0 is not in the interval, the engineer’s claim of equal means is not reasonable. The data strongly suggest the means differ.
(c) Test for Supplier A: µ_A ≠ 58.6, α = 0.05
H₀: µ_A = 58.6 vs H₁: µ_A ≠ 58.6. t = (58.583 – 58.6) / (0.1379 / √12) = -0.017 / 0.0398 = -0.427.
Critical value: t_{0.025, 11} = 2.201. Since |-0.427| < 2.201, fail to reject H₀. There is no evidence at the 5% level that the mean differs from 58.6 dB.
Exercise 6: R Output Analysis
(a) Hypotheses and Decision
- H₀: σ² = 2000 vs H₁: σ² > 2000 (one-sided, alternative = “greater”)
- α = 1 – 0.9 = 0.10
- p-value = 0.02317
Since p-value (0.02317) < α (0.10), reject H₀. There is significant evidence at the 10% level that the population variance of daily expenses exceeds 2000.
(b) Information from Output
- Test statistic distribution: Chi-squared (χ²) distribution
- Degrees of freedom: df = 149 (so n = 150)
- Test statistic value: X² = 185.33
- 90% confidence interval: (0, 2910.578) — one-sided upper bound
- Point estimate: Sample variance = 2487.651