Practice Problems: Hypothesis Testing

Solution:

• H₀: μ = 3000 (average consumption is 3000 calories)
• Hₐ: μ > 3000 (average consumption is more than 3000 calories)

This is a one-tailed test (specifically, a right-tailed test) because we're only interested in whether consumption is greater than 3000, not different in either direction.

Type I Error (α): Conclude the drug has side effects when it actually doesn't.

Consequence: A safe drug is rejected and doesn't reach patients who could benefit.

Type II Error (β): Conclude the drug has no side effects when it actually does.

Consequence: A dangerous drug is approved and patients are harmed.

Which is worse? Type II error is more serious in this case because patient safety is at risk. It's better to be overly cautious (reject some safe drugs) than to approve dangerous drugs.

Is power = 0.65 acceptable? No. This means only a 65% chance of detecting a real effect, and a 35% chance of missing it (Type II error). The standard recommendation is power ≥ 0.80 (80%).

Three ways to increase power to 0.80:

1. Increase sample size (n) — Most practical and common approach
2. Increase significance level (α) — Use α = 0.10 instead of 0.05 (though this increases Type I error risk)
3. Reduce variability — Use more precise measurement instruments or more homogeneous sample

Best choice: Increasing sample size is usually the best option because it doesn't involve tradeoffs like the other methods.

Use a t-test.

Reason: The population standard deviation (σ) is unknown. We only have the sample standard deviation (s = 4 inches). When σ is unknown, we must use the t-test with degrees of freedom df = n - 1 = 50 - 1 = 49.

General rule: In real-world scenarios, we almost always use t-tests for means because we rarely know the true population standard deviation.

Decision at α = 0.05:

p-value (0.038) < α (0.05) → Reject H₀

There is sufficient evidence to reject the null hypothesis at the 0.05 significance level.

Decision at α = 0.01:

p-value (0.038) > α (0.01) → Fail to reject H₀

There is not sufficient evidence to reject the null hypothesis at the 0.01 significance level.

Plain English interpretation:

If the null hypothesis were true, there would be only a 3.8% chance of obtaining results as extreme as (or more extreme than) what we observed. This is fairly unlikely, which suggests the null hypothesis may not be true.

Step 1: Hypotheses

• H₀: μ = 500
• Hₐ: μ ≠ 500 (two-tailed)
• α = 0.05

Step 2: Test statistic

z = (485 - 500) / (50/√40) = -15 / 7.906 = -1.90

Step 3: Critical value

Two-tailed, α = 0.05: ±1.96

Step 4: Decision

|z| = 1.90 < 1.96 → Fail to reject H₀

Conclusion: There is not sufficient evidence to reject the manufacturer's claim that batteries last 500 hours on average. The observed difference could be due to sampling variability.

Step 1: Hypotheses

• H₀: μ = 30
• Hₐ: μ > 30 (right-tailed)
• α = 0.05

Step 2: Test statistic

z = (33 - 30) / (8/√35) = 3 / 1.352 = 2.22

Step 3: P-value

P(Z > 2.22) = 1 - 0.9868 = 0.0132

Step 4: Decision

p-value (0.0132) < α (0.05) → Reject H₀

Conclusion: There is strong evidence (p = 0.0132) that the program increases strength by more than 30 pounds.

Step 1: Hypotheses

• H₀: μ = 15
• Hₐ: μ < 15 (left-tailed)
• α = 0.01

Step 2: Test statistic

z = (13.5 - 15) / (5/√60) = -1.5 / 0.6455 = -2.32

Step 3: Critical value

Left-tailed, α = 0.01: -2.326

Step 4: Decision

z = -2.32 < -2.326 → Reject H₀ (barely!)

Conclusion: There is sufficient evidence at the 0.01 level that the average weight loss is less than the claimed 15 pounds. The z-value is very close to the critical value, making this a borderline decision.

Recalculation with x̄ = 13.6:

z = (13.6 - 15) / (5/√60) = -1.4 / 0.6455 = -2.17

Decision:

z = -2.17 > -2.326 → Fail to reject H₀

Discussion:

A change of just 0.1 pounds in the sample mean (from 13.5 to 13.6) completely reverses our conclusion! This illustrates important points:

• When p-values are close to α, results are borderline and sensitive to small changes
• Statistical significance doesn't mean practical significance
• We should be cautious about making strong claims from borderline results
• Replication studies are important to confirm findings

Step 1: Hypotheses

• H₀: μ = 25000
• Hₐ: μ ≠ 25000 (two-tailed)
• α = 0.05

Step 2: Test statistic

t = (27400 - 25000) / (6000/√25) = 2400 / 1200 = 2.0

df = 25 - 1 = 24

Step 3: Critical value

Two-tailed, α = 0.05, df = 24: ±2.064

Step 4: Decision

|t| = 2.0 < 2.064 → Fail to reject H₀

Conclusion: There is not sufficient evidence to conclude the average student debt differs from $25,000. Though the sample mean is higher, the difference is not statistically significant.

Step 1: Hypotheses

• H₀: μ = 75
• Hₐ: μ > 75 (right-tailed)
• α = 0.05

Step 2: Test statistic

t = (79 - 75) / (8/√18) = 4 / 1.886 = 2.12

df = 18 - 1 = 17

Step 3: Critical value

Right-tailed, α = 0.05, df = 17: 1.740

Step 4: Decision

t = 2.12 > 1.740 → Reject H₀

Conclusion: There is sufficient evidence that the new teaching method increases test scores above 75. The improvement is statistically significant.

Step 1: Hypotheses

• H₀: μ = 30
• Hₐ: μ < 30 (left-tailed)
• α = 0.01

Step 2: Test statistic

t = (27.5 - 30) / (6/√22) = -2.5 / 1.279 = -1.95

df = 22 - 1 = 21

Step 3: Critical value and p-value

Critical value (α = 0.01, df = 21): -2.518

From t-table (df = 21): For t = -1.95, p-value is between 0.025 and 0.05 (one-tailed)

Step 4: Decision

t = -1.95 > -2.518 → Fail to reject H₀

p-value > α (0.01) → Fail to reject H₀

Conclusion: There is not sufficient evidence at the 0.01 level that the average commute time is less than 30 minutes.

Study A (n = 20):

t = (52000 - 50000) / (8000/√20) = 2000 / 1789 = 1.12

df = 19, critical value (α = 0.05, two-tailed) ≈ 2.093

Decision: Fail to reject H₀ (1.12 < 2.093)

Study B (n = 100):

t = (52000 - 50000) / (8000/√100) = 2000 / 800 = 2.5

df = 99, critical value (α = 0.05, two-tailed) ≈ 1.984

Decision: Reject H₀ (2.5 > 1.984)

Analysis: Study B is more likely to reject H₀ because:

• Larger sample size reduces standard error
• Same difference (x̄ - μ₀) produces larger test statistic with bigger sample
• This shows why sample size matters: with enough data, even small differences can be statistically significant
• This also illustrates the difference between statistical significance (Study B) and practical significance ($2,000 difference may not be practically meaningful)

Check conditions:

np₀ = 80(0.15) = 12 ≥ 10

n(1-p₀) = 80(0.85) = 68 ≥ 10

Conclusion: Both conditions are satisfied. We can use the normal approximation (z-test) for this proportion test.

Step 1: Hypotheses

• H₀: p = 0.80
• Hₐ: p ≠ 0.80 (two-tailed)
• α = 0.05

Step 2: Check conditions

np₀ = 200(0.80) = 160 ≥ 10

n(1-p₀) = 200(0.20) = 40 ≥ 10

Step 3: Test statistic

p̂ = 150/200 = 0.75

z = (0.75 - 0.80) / √(0.80(0.20)/200) = -0.05 / 0.0283 = -1.77

Step 4: Critical value

Two-tailed, α = 0.05: ±1.96

Step 5: Decision

|z| = 1.77 < 1.96 → Fail to reject H₀

Conclusion: There is not sufficient evidence that the employment rate differs from 80%.

Step 1: Hypotheses

• H₀: p = 0.55
• Hₐ: p > 0.55 (right-tailed)
• α = 0.01

Step 2: Check conditions

np₀ = 300(0.55) = 165 ≥ 10

n(1-p₀) = 300(0.45) = 135 ≥ 10

Step 3: Test statistic

p̂ = 180/300 = 0.60

z = (0.60 - 0.55) / √(0.55(0.45)/300) = 0.05 / 0.0287 = 1.74

Step 4: Critical value

Right-tailed, α = 0.01: 2.326

Step 5: Decision

z = 1.74 < 2.326 → Fail to reject H₀

Conclusion: There is not sufficient evidence at the 0.01 level that more than 55% support the bill.

Step 1: Hypotheses

• H₀: p = 0.05
• Hₐ: p < 0.05 (left-tailed)
• α = 0.05

Step 2: Check conditions

np₀ = 400(0.05) = 20 ≥ 10

n(1-p₀) = 400(0.95) = 380 ≥ 10

Step 3: Test statistic

p̂ = 15/400 = 0.0375

z = (0.0375 - 0.05) / √(0.05(0.95)/400) = -0.0125 / 0.0109 = -1.15

Step 4: P-value

P(Z < -1.15) = 0.1251

Step 5: Decision

p-value (0.1251) > α (0.05) → Fail to reject H₀

Conclusion: There is not sufficient evidence that the defect rate is lower than 5%.

Check conditions:

np₀ = 60(0.02) = 1.2 < 10 (FAILS!)

n(1-p₀) = 60(0.98) = 58.8 ≥ 10

Conclusion: NO, they cannot use the normal approximation because np₀ < 10.

What to do instead:

• Option 1: Increase sample size. Need n ≥ 10/0.02 = 500 to meet the condition
• Option 2: Use exact binomial test (not covered in this course, but the proper method when normal approximation fails)
• Option 3: Use a different statistical approach designed for rare events

Key lesson: Always check conditions before conducting a test. If conditions aren't met, the test results are not valid!

Step 1: Hypotheses

• H₀: p = 0.02 (defect rate is 2%)
• Hₐ: p > 0.02 (defect rate has increased) — right-tailed
• α = 0.05

Step 2: Check conditions

np₀ = 500(0.02) = 10 ≥ 10 (just barely)

n(1-p₀) = 500(0.98) = 490 ≥ 10

Step 3: Test statistic

p̂ = 15/500 = 0.03

z = (0.03 - 0.02) / √(0.02(0.98)/500) = 0.01 / 0.00626 = 1.60

Step 4: Critical value and p-value

Critical value (α = 0.05, right-tailed): 1.645

z = 1.60 < 1.645 → Fail to reject H₀

P-value: P(Z > 1.60) = 0.0548

Step 5: Statistical conclusion

There is not quite enough evidence at the 0.05 level to conclude the defect rate has increased above 2%.

Practical recommendation:

While the result is not statistically significant (p = 0.055 is very close to 0.05), the observed 3% defect rate is 50% higher than the claimed 2%. From a quality control perspective:

• Recommendation: Investigate the production process
• The result is borderline—very close to significance
• Cost of investigating < cost of shipping defective chips
• Better safe than sorry in quality control
• Monitor next batches closely

This illustrates that statistical decisions should be combined with practical considerations!

(a) Hypothesis Test:

H₀: p = 0.60, Hₐ: p ≠ 0.60 (two-tailed), α = 0.01

Check conditions: np₀ = 90 ≥ 10 , n(1-p₀) = 60 ≥ 10

p̂ = 105/150 = 0.70

z = (0.70 - 0.60) / √(0.60(0.40)/150) = 0.10 / 0.04 = 2.5

Critical values: ±2.576

|z| = 2.5 < 2.576 → Fail to reject H₀

Statistical conclusion: Not sufficient evidence at α = 0.01 that cure rate differs from 60%.

(b) Type I Error:

Conclude the new drug has a different cure rate when it's actually the same as the historical 60%.

Consequence: Company invests in a drug that's no better than existing treatment.

(c) Type II Error:

Conclude the drug's cure rate is not different from 60% when it actually is different (better or worse).

Consequence: If better, an improved treatment is rejected. If worse, a harmful drug continues in development.

(d) Which error is worse?

Type II error is worse if the drug is actually worse (lower cure rate) because patients could be harmed. However, Type II error is also bad if the drug is actually better—patients miss out on an improved treatment.

In medical contexts, both errors have serious consequences, but patient safety (avoiding harmful treatments) is typically prioritized.

(e) Choice of α:

Recommended: α = 0.05

Reasoning:

• α = 0.01 is very conservative (harder to detect real differences)
• With α = 0.05, our z = 2.5 > 1.96 would reject H₀, suggesting the drug is better
• In this case, the drug appears to improve cure rate from 60% to 70%—that's clinically meaningful
• Using α = 0.05 balances Type I and Type II error risks
• Further testing (Phase III trials) would confirm findings before approval

Note: This demonstrates that α = 0.05 would detect the improvement while α = 0.01 would not. The choice of α matters!