INF-6 Solutions: Hypothesis Testing for Small Sample Mean and Proportion

Problem 1 — t-Test Generator

Generated by generateTTest(). The approach is always the same five steps:

1. H₀ and H_a: State H₀: μ = μ₀ and the appropriate H_a (two-tailed ≠ μ₀, since the generator uses two-tailed tests).
2. Conditions: σ unknown (s given) and n < 30 → use t-test. Assume approximately normal population.
3. Test statistic: \(\text{SE} = s/\sqrt{n}\), then \(t = (\bar{x} - \mu_0)/\text{SE}\). Record df = n − 1.
4. p-value: Locate |t| in the df row of the t-table. Find the two critical values that bracket |t| (e.g., t* = 2.093 and t* = 2.539 for df = 19). The p-value is bounded between the corresponding two-tail α values (e.g., 0.02 < p < 0.05).
5. Conclusion: If p < α = 0.05, reject H₀ and state in context. If p ≥ 0.05, fail to reject H₀ and state in context.

Key reminder — bounding the p-value: The t-table gives critical values for specific α levels, not exact p-values. You will always report a range such as "0.02 < p < 0.05." This is sufficient to make the reject/fail-to-reject decision: if the entire range is below α, reject; if the range straddles α or is entirely above, fail to reject.

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Problem 2 — One-Tailed Proportion Test (Variants 0–2)

Variant 0 (Medication; n = 100, x = 58, p₀ = 0.50, right-tailed, α = 0.05):

• H₀: p = 0.50; H_a: p > 0.50 (right-tailed).
• Conditions: \(np_0 = 50 \geq 5\) ✓; \(n(1-p_0) = 50 \geq 5\) ✓.
• \(\hat{p} = 58/100 = 0.58\)
• \(\text{SE} = \sqrt{0.50 \times 0.50 / 100} = \sqrt{0.0025} = 0.05\)
• \(z = (0.58 - 0.50) / 0.05 = 0.08 / 0.05 = 1.60\)
• p-value (right-tailed): \(p = P(Z > 1.60) = 1 - \Phi(1.60) = 1 - 0.9452 = 0.0548\)
• Decision: \(p = 0.0548 > 0.05\). Fail to reject H₀. Insufficient evidence that the medication reduces blood pressure in more than 50% of patients.

Variant 1 (Customer satisfaction; n = 120, x = 30, p₀ = 0.30, left-tailed, α = 0.05):

• H₀: p = 0.30; H_a: p < 0.30 (left-tailed).
• Conditions: \(np_0 = 36 \geq 5\) ✓; \(n(1-p_0) = 84 \geq 5\) ✓.
• \(\hat{p} = 30/120 = 0.25\)
• \(\text{SE} = \sqrt{0.30 \times 0.70 / 120} \approx 0.04183\)
• \(z = (0.25 - 0.30) / 0.04183 \approx -1.195\)
• p-value (left-tailed): \(p = P(Z < -1.195) \approx 1 - \Phi(1.195) \approx 0.116\)
• Decision: \(p \approx 0.116 > 0.05\). Fail to reject H₀. Insufficient evidence that fewer than 30% of customers are satisfied.

Variant 2 (Social media; n = 150, x = 117, p₀ = 0.70, right-tailed, α = 0.01):

• H₀: p = 0.70; H_a: p > 0.70 (right-tailed).
• Conditions: \(np_0 = 105 \geq 5\) ✓; \(n(1-p_0) = 45 \geq 5\) ✓.
• \(\hat{p} = 117/150 = 0.78\)
• \(\text{SE} = \sqrt{0.70 \times 0.30 / 150} \approx 0.03742\)
• \(z = (0.78 - 0.70) / 0.03742 \approx 2.138\)
• p-value (right-tailed): \(p = 1 - \Phi(2.138) \approx 0.0163\)
• Decision: \(p \approx 0.0163 > 0.01\). Fail to reject H₀ at α = 0.01. (Would reject at α = 0.05.)

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Problem 3 — Proportion Test Generator

Generated by generateProportionTest(). The approach is always the same five steps:

1. H₀ and H_a: State H₀: p = p₀ and the H_a matching the scenario direction (left, right, or two-tailed).
2. Conditions: Compute \(np_0\) and \(n(1-p_0)\). Both must be ≥ 5.
3. Test statistic: \(\hat{p} = x/n\), then \(\text{SE} = \sqrt{p_0(1-p_0)/n}\), then \(z = (\hat{p} - p_0)/\text{SE}\). Note: SE uses p₀, not \(\hat{p}\).
4. p-value: Match tail to H_a: right-tailed → p = 1 − Φ(z); left-tailed → p = Φ(z); two-tailed → p = 2(1 − Φ(|z|)).
5. Conclusion: If p < α, reject H₀ and state in context. If p ≥ α, fail to reject H₀ and state in context.

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Problem 4 — Find the Error

Generated by generateProportionFindTheError(). The generator rotates through three possible error types. All three share a common theme: the proportion test mechanics are partially correct, but one critical step contains a flaw.

Error Type 1 — p̂ in the denominator:

The student uses \(\hat{p}\) in the SE formula instead of \(p_0\). The correct SE is \(\sqrt{p_0(1-p_0)/n}\); the incorrect SE is \(\sqrt{\hat{p}(1-\hat{p})/n}\). This produces a different z-value and can change the conclusion, especially when \(\hat{p}\) and \(p_0\) are far apart.

Error Type 2 — Wrong p-value direction:

For a two-tailed test, the student reports only the one-tail area instead of multiplying by 2. For example, if z = 2.10, the one-tail area is 0.018, but the correct two-tailed p-value is 0.036. Failing to double the area underestimates p and may lead to a spurious rejection.

Error Type 3 — "Accept H₀" language:

The numerical work is correct, but the conclusion states "accept H₀" or "the data prove H₀." This is never valid. When p ≥ α, the correct statement is "we fail to reject H₀" — meaning only that the data are not surprising enough under H₀ to reject it. Failing to reject does not prove H₀ is true; it means there is insufficient evidence against it.

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Problem 5 — Multi-Step Synthesis: Hospital Quality Audit

Data: n = 20 patients; x̄ = 4.1 days, s = 1.2 days; 7 readmissions (\(\hat{p} = 0.35\)); claimed μ₀ = 3.5 days and p₀ = 0.25. Test both at α = 0.05.

Part (a) — t-test for mean post-operative stay:

• H₀: μ = 3.5; H_a: μ ≠ 3.5 (two-tailed — auditor checks both directions).
• σ unknown, n = 20 < 30 → t-test, df = 19. Assume approximately normal.
• \(\text{SE} = 1.2/\sqrt{20} \approx 1.2/4.472 \approx 0.2683\) days
• \(t = (4.1 - 3.5)/0.2683 = 0.6/0.2683 \approx 2.24\)
• t-table, df = 19: \(t^* = 2.093\) (two-tail 0.05); \(t^* = 2.539\) (two-tail 0.02). Since \(2.093 < 2.24 < 2.539\): \(0.02 < p < 0.05\).
• Decision: p < 0.05. Reject H₀. Sufficient evidence at α = 0.05 that the mean post-operative stay differs from 3.5 days.

Part (b) — z-test for readmission proportion:

• H₀: p = 0.25; H_a: p ≠ 0.25 (two-tailed).
• Conditions: \(np_0 = 20(0.25) = 5 \geq 5\) ✓ (barely); \(n(1-p_0) = 15 \geq 5\) ✓.
• \(\text{SE} = \sqrt{0.25 \times 0.75/20} = \sqrt{0.009375} \approx 0.09682\)
• \(z = (0.35 - 0.25)/0.09682 = 0.10/0.09682 \approx 1.033\)
• \(p = 2P(Z > 1.033) \approx 2(1 - 0.8493) = 2(0.1507) \approx 0.301\)
• Decision: p ≈ 0.301 > 0.05. Fail to reject H₀. Insufficient evidence the readmission proportion differs from 25%.

Part (c) — Interpreting two simultaneous test results:

Rejecting the stay claim (mean differs from 3.5 days) provides statistical evidence that the hospital's stated average is inaccurate. However, it does not prove the hospital is negligent — it only shows a detectable deviation from 3.5 days. Whether a difference of 0.6 days is practically significant depends on clinical context and resource implications, not just on the p-value.

Failing to reject the readmission claim does not confirm the 25% rate is correct. With n = 20, the test has very low power — a true rate of 35% or even 40% could easily go undetected. The auditor should flag this as a limitation and recommend a larger follow-up sample for the readmission claim.

One test rejecting while another does not is not a contradiction: each test addresses a separate claim about a separate parameter, with its own sampling variability. Both conclusions are about evidence, not proof.

Part (d) — "Accept H₀" reasoning error:

The administrator's argument is a classic "fail to reject = accept" fallacy. "Fail to reject H₀" means only that the sample data are not surprising enough under H₀ to reject it at the chosen α. It does not mean H₀ is proven true, confirmed, or verified.

With n = 20, the test has extremely low power. If the true readmission rate were 35%, there would be only a modest probability of detecting it. The administrator is treating the absence of statistical evidence as positive evidence — the same logical error as concluding "no news is good news" when the test was underpowered to begin with.

The correct statement: "We do not have sufficient evidence at the 5% level to conclude the readmission proportion differs from 25%. A larger sample is needed before any affirmative conclusion can be drawn."

Question 1 — Feynman Test: Why p₀ (not p̂) in the Denominator

Model answer:

The denominator of the z test statistic is the standard error of \(\hat{p}\) under H₀ — that is, the spread we'd expect in sample proportions if the null hypothesis were exactly right. H₀ says the true proportion is p₀. If H₀ is true, then \(\hat{p}\) varies around p₀ with standard deviation \(\sqrt{p_0(1-p_0)/n}\). That is the SE we use.

If instead we used \(\hat{p}\) in the denominator, we'd be computing the SE as if the true proportion equaled our sample estimate — circular reasoning. We'd be assuming our sample is exactly right in order to determine whether our sample is surprising. That defeats the entire purpose of the test.

The rule: SE uses p₀ because we assume H₀ is true when computing the p-value. This is the defining feature of the proportion test and the single most common error in this lesson.

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Question 2 — Apply: Placement Test

Scenario: H₀: p = 0.40; H_a: p ≠ 0.40 (two-tailed); n = 120, x = 54, α = 0.05.

Part A: H₀: p = 0.40; H_a: p ≠ 0.40 (two-tailed). The null uses the claimed proportion (0.40), not the sample estimate.

Part B (conditions):

• \(np_0 = 120(0.40) = 48 \geq 5\) ✓
• \(n(1-p_0) = 120(0.60) = 72 \geq 5\) ✓
• Conditions are met. Use the z-test for proportions.

Part C (test statistic):

\[ \hat{p} = 54/120 = 0.45 \]

\[ \text{SE} = \sqrt{\frac{0.40 \times 0.60}{120}} = \sqrt{0.002} \approx 0.04472 \]

\[ z = \frac{0.45 - 0.40}{0.04472} = \frac{0.05}{0.04472} \approx 1.118 \]

Part D (full conclusion):

\[ p = 2P(Z > 1.118) = 2(1 - \Phi(1.118)) \approx 2(1 - 0.8682) = 2(0.1318) = 0.264 \]

p ≈ 0.264 > 0.05. We fail to reject H₀.

There is insufficient evidence at the 5% significance level to conclude that the true passage rate for the advanced mathematics placement test differs from 40%. The sample proportion of 45% is consistent with random variation from a true rate of 40%.

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Question 3 — Error Analysis: df Off by One

Setup: n = 10, t = 2.50, two-tailed test. Researcher uses df = 10 and finds t* = 2.228 → rejects H₀.

The error: The researcher used df = n = 10 instead of df = n − 1 = 9. For a one-sample t-test, degrees of freedom are always n − 1. Using df = 10 yields a critical value of t* = 2.228 (two-tail α = 0.05); the correct df = 9 gives t* = 2.262.

Does the conclusion change?

• With df = 10 (wrong): t* = 2.228; t = 2.50 > 2.228 → reject H₀.
• With df = 9 (correct): t* = 2.262; t = 2.50 > 2.262 → still reject H₀.
• In this case, the conclusion does not change. The df error was present but inconsequential here.

When would it matter? If t were between 2.228 and 2.262 — for example, t = 2.24 — the wrong df (10) would give a critical value of 2.228, leading to rejection; the correct df (9) would give 2.262, leading to failure to reject. The off-by-one df error is consequential near the boundary of the critical region. It is worth correcting even when the conclusion happens not to change.

The habit to build: Always write df = n − 1 before opening the t-table. Circle it. Then look up the correct row.

Path A — The Health Researcher

Data: n = 18, x̄ = 5.1 days, s = 1.2 days, μ₀ = 4.5 days, α = 0.05 (two-tailed).

Task 1 — H₀, H_a, and test choice:

• H₀: μ = 4.5 days; H_a: μ ≠ 4.5 days (two-tailed).
• Why two-tailed: The researcher is a skeptic — she wants to know if the true mean differs from the hospital's claim in either direction. Without a prior directional suspicion ("I believe recovery times are longer"), two-tailed is correct.
• Why t-test: σ is unknown (only s = 1.2 days given); n = 18 < 30; population of recovery times stated to be approximately normal. Use t with df = 17.

Task 2 — Compute t and bound p:

\[ \text{SE} = s/\sqrt{n} = 1.2/\sqrt{18} = 1.2/4.243 \approx 0.2828 \text{ days} \]

\[ t = (5.1 - 4.5)/0.2828 = 0.6/0.2828 \approx 2.121 \]

t-table, df = 17: t* = 2.110 (two-tail 0.05); t* = 2.567 (two-tail 0.02). Since 2.110 < 2.121 < 2.567: 0.02 < p < 0.05. (Since t = 2.121 > 2.110, p < 0.05 strictly.)

Task 3 — Conclusion and Type II error:

p < 0.05 = α. Reject H₀. There is sufficient evidence at the 5% level to conclude that the mean recovery time differs from 4.5 days. The sample data suggest recovery times may be longer than the hospital claims.

Type II error in context: A Type II error (failing to reject H₀ when it is false) would mean concluding the hospital's claim of 4.5 days is consistent with the data when the true mean is actually higher. The practical consequence: patients receive inaccurate discharge expectations, and policymakers may under-allocate post-operative care resources. This is a real-world harm arising from insufficient statistical evidence.

Task 4 — z vs. t argument:

The flaw in the colleague's suggestion: the criterion for using z vs. t is not sample size but whether σ is known. Here σ is unknown; only s is given. The t-distribution is required regardless of whether n is "close to 30."

Using z* = 1.96 (the two-tailed critical value at α = 0.05) vs. t* = 2.110 (df = 17, two-tail α = 0.05): since t = 2.121 > 2.110, both reject H₀ in this case. But using z when σ is unknown produces a smaller critical value (1.96 < 2.110), making it easier to reject — this is anti-conservative. In borderline cases (e.g., if t = 2.00), z would reject (2.00 > 1.96) while t would fail to reject (2.00 < 2.110), leading to a Type I error rate above the nominal α = 0.05.

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Path B — The Policy Analyst

Data: n = 150, x = 63, p₀ = 0.35, α = 0.05 (two-tailed).

Task 1 — H₀, H_a, and conditions:

• H₀: p = 0.35; H_a: p ≠ 0.35 (two-tailed).
• Why two-tailed: The group wants to know if the true proportion at their university differs from the government figure in any direction — no prior directional claim is made.
• Conditions: \(np_0 = 150(0.35) = 52.5 \geq 5\) ✓; \(n(1-p_0) = 150(0.65) = 97.5 \geq 5\) ✓; random sample ✓. Use the z-test for proportions.

Task 2 — p̂, z, and p-value:

\[ \hat{p} = 63/150 = 0.42 \]

\[ \text{SE} = \sqrt{\frac{0.35 \times 0.65}{150}} = \sqrt{\frac{0.2275}{150}} = \sqrt{0.001517} \approx 0.03894 \]

\[ z = (0.42 - 0.35)/0.03894 = 0.07/0.03894 \approx 1.797 \]

\[ p = 2P(Z > 1.797) \approx 2(1 - \Phi(1.80)) = 2(1 - 0.9641) = 2(0.0359) = 0.0718 \]

Task 3 — Decision and CI verification:

p ≈ 0.0718 > 0.05. Fail to reject H₀. Insufficient evidence that the proportion of students working more than 20 hours per week at this university differs from 35%.

95% CI for p (using p̂ in SE, not p₀ — CIs use the sample estimate):

\[ \text{SE}_\text{CI} = \sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{0.42 \times 0.58/150} \approx 0.04030 \]

\[ \text{CI} = 0.42 \pm 1.96 \times 0.04030 = 0.42 \pm 0.0790 = (0.341,\; 0.499) \]

p₀ = 0.35 falls inside (0.341, 0.499) → fail to reject H₀ at α = 0.05. ✓ Both methods agree.

Key distinction: The hypothesis test uses p₀ in the SE; the confidence interval uses p̂. This is the only case where the two-formula distinction matters — they serve different purposes.

Task 4 — Practical vs. statistical significance:

Statistical significance: The test failed to reject H₀ at α = 0.05. The data are not statistically surprising under H₀: p = 0.35. The advocacy group cannot claim "dramatically higher" rates — the evidence does not support even a statistically detectable difference from 35%.

Practical significance: Even if the test had rejected H₀, a difference of p̂ − p₀ = 0.42 − 0.35 = 0.07 (7 percentage points) may or may not be "dramatic" by policy standards. Whether 7 more students per 100 working over 20 hours per week warrants major policy intervention is a domain judgment, not a statistical one. Statistical significance does not imply large or important effects.

Responsible communication: "Our sample shows 42% — 7 percentage points above the government's 35% — but we do not have sufficient statistical evidence at the 5% level to conclude the true rate at our university differs from the national figure. A larger sample is needed to draw firmer conclusions."

Challenge 1 — Paired Data: t-Test on Differences

The key insight for paired data: treat the differences d_i = after − before as a single sample, then apply the standard one-sample t-test to the d_i values. This eliminates between-subject variability and increases power.

Variant 0 (Typing speed; 7 participants; H₀: μ_d = 0 vs. H_a: μ_d ≠ 0):

Participant	Before	After	di = After − Before
1	52	58	6
2	45	50	5
3	60	65	5
4	38	42	4
5	55	62	7
6	48	51	3
7	50	55	5

\(\bar{d} = (6 + 5 + 5 + 4 + 7 + 3 + 5)/7 = 35/7 = 5.00\) wpm

Deviations from \(\bar{d} = 5\): 1, 0, 0, −1, 2, −2, 0. Sum of squared deviations: \(1 + 0 + 0 + 1 + 4 + 4 + 0 = 10\).

\(s_d^2 = 10/(7-1) = 10/6 \approx 1.667\); \(s_d \approx 1.291\) wpm

\(\text{SE} = s_d/\sqrt{n} = 1.291/\sqrt{7} = 1.291/2.646 \approx 0.4879\)

\(t = \bar{d}/\text{SE} = 5.00/0.4879 \approx 10.25\), df = 6

t-table, df = 6: t* = 3.707 at two-tail α = 0.01. Since 10.25 ≫ 3.707: p < 0.01. Reject H₀.

Conclusion: There is very strong evidence (p < 0.01) that the training program improves typing speed.

Variant 1 (Reaction time; 6 athletes; H₀: μ_d = 0 vs. H_a: μ_d < 0, left-tailed, α = 0.05):

Athlete	Before	After	di = After − Before
1	280	265	−15
2	310	295	−15
3	295	290	−5
4	320	305	−15
5	275	260	−15
6	300	285	−15

\(\bar{d} = (-15 - 15 - 5 - 15 - 15 - 15)/6 = -80/6 \approx -13.33\) ms

Deviations from −13.33: −1.67, −1.67, 8.33, −1.67, −1.67, −1.67.

Sum of squared deviations: \(2.79 + 2.79 + 69.39 + 2.79 + 2.79 + 2.79 = 83.34\)

\(s_d^2 = 83.34/5 \approx 16.67\); \(s_d \approx 4.08\) ms

\(t = -13.33/(4.08/\sqrt{6}) = -13.33/(4.08/2.449) = -13.33/1.665 \approx -8.00\), df = 5

Left-tailed: |t| = 8.00 ≫ t* = 2.015 (one-tail α = 0.05, df = 5). p ≪ 0.05. Reject H₀.

Conclusion: Strong evidence that the conditioning drill reduces reaction time.

Variant 2 (Caloric intake; 8 patients; H₀: μ_d = 0 vs. H_a: μ_d ≠ 0, α = 0.05):

Patient	Before	After	di = After − Before
1	2400	2200	−200
2	2100	2050	−50
3	2800	2500	−300
4	2300	2250	−50
5	2600	2350	−250
6	1900	1950	+50
7	2200	2100	−100
8	2500	2300	−200

\(\bar{d} = (-200 - 50 - 300 - 50 - 250 + 50 - 100 - 200)/8 = -1100/8 = -137.5\) kcal

Deviations from −137.5: −62.5, 87.5, −162.5, 87.5, −112.5, 187.5, 37.5, −62.5.

Sum of squared deviations: \(3906.25 + 7656.25 + 26406.25 + 7656.25 + 12656.25 + 35156.25 + 1406.25 + 3906.25 = 98750\)

\(s_d^2 = 98750/7 \approx 14107\); \(s_d \approx 118.8\) kcal

\(t = -137.5/(118.8/\sqrt{8}) = -137.5/(118.8/2.828) = -137.5/42.01 \approx -3.27\), df = 7

Two-tailed, df = 7: t* = 2.365 (two-tail 0.05); t* = 2.998 (two-tail 0.02). Since |t| = 3.27 > 2.998: p < 0.02. Reject H₀.

Conclusion: Sufficient evidence at α = 0.05 that the dietary intervention changes mean caloric intake (specifically, reduces it).

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Challenge 2 — Power of the t-Test

Setup: μ₀ = 500 mL, true μ = 505 mL, n = 16, s = 20 mL, α = 0.05 (two-tailed), df = 15.

Part (a) — Approximate power:

Non-centrality parameter (standardized shift): \(\delta = (\mu - \mu_0)/(s/\sqrt{n}) = (505 - 500)/(20/\sqrt{16}) = 5/5 = 1.00\)

Critical value: t* = 2.131 (df = 15, two-tailed α = 0.05).

Under the true distribution shifted by δ = 1.00, the test rejects when t > 2.131. Using the standard normal approximation, the probability of rejection is approximately:

\[ P(Z > t^* - \delta) = P(Z > 2.131 - 1.00) = P(Z > 1.131) \approx 1 - \Phi(1.131) \approx 1 - 0.871 = 0.129 \]

Approximate power ≈ 13%. (The left-tail contribution is negligible here.)

Part (b) — What's needed for 80% power:

Power of 13% means that if the true mean has shifted by 5 mL from 500 mL, this test will detect it only about 1 time in 8. This is very low power — the test is designed to detect gross failures, not small shifts.

To achieve ≈80% power for a shift of 5 mL with σ ≈ 20: using the standard power formula \(n \approx (z_\alpha + z_\beta)^2 \times (\sigma/\Delta)^2\) where z_α = 1.96 (two-tailed α = 0.05) and z_β = 0.84 (80% power): \(n \approx (1.96 + 0.84)^2 \times (20/5)^2 = (2.80)^2 \times 16 = 7.84 \times 16 \approx 125\) observations. The current n = 16 is far too small to detect a 5-mL shift reliably.