Step 1: Set up hypotheses Claim: μ = 500 mL Suspect: μ < 500 mL (underfilling)

H₀: μ = 500 Hₐ: μ < 500 (one-tailed, left)

Step 2: Check conditions n = 25

RANDOM: Assume random sample ✓ NORMAL: Population normal (given) ✓

• With n = 25 < 30, need this assumption INDEPENDENT: Assume 25 ≤ 0.10N ✓

Use t-test (σ unknown)

Step 3: Calculate test statistic df = n - 1 = 24

SE = s/√n = 6/√25 = 6/5 = 1.2

t = (x̄ - μ₀)/SE = (497 - 500)/1.2 = -3/1.2 = -2.50

Step 4: Find p-value Left-tailed test df = 24, t = -2.50

From t-table: P(t < -2.50) ≈ 0.01

p-value ≈ 0.01

Step 5: Make decision p-value ≈ 0.01 α = 0.05

Is 0.01 < 0.05? YES

REJECT H₀

Step 6: State conclusion At the α = 0.05 significance level, there is sufficient evidence that the mean fill is less than 500 mL. The machine appears to be underfilling bottles.

Step 7: Practical interpretation Sample mean: 497 mL Target: 500 mL Difference: -3 mL

This 3 mL shortage is:

• Statistically significant
• Not just random variation
• Machine needs adjustment

Answer: H₀: μ = 500, Hₐ: μ < 500 Test statistic: t = -2.50 (df = 24) P-value: 0.01 Decision: Reject H₀ at α = 0.05 Conclusion: Mean fill is significantly less than 500 mL

Step 1: The key difference Z-TEST: Population SD (σ) is KNOWN T-TEST: Population SD (σ) is UNKNOWN (use s)

Step 2: When to use Z-TEST for mean Conditions:

1. σ is known (rare!)
2. Random sample
3. Normal population OR n ≥ 30

Test statistic: z = (x̄ - μ₀)/(σ/√n)

Use z-table for p-value

Step 3: When to use T-TEST for mean Conditions:

1. σ is unknown (almost always!)
2. Random sample
3. Normal population (if n < 30) OR n ≥ 30 (can use CLT)
4. Independent observations

Test statistic: t = (x̄ - μ₀)/(s/√n)

Use t-table with df = n - 1

Step 4: Why σ is rarely known In practice:

• If we knew σ, we'd probably know μ
• Population parameters rarely known
• Must estimate from sample
• Use s as estimate of σ

Real-world: Almost always use t-test!

Step 5: Small vs large samples

SMALL SAMPLE (n < 30): Must use T-TEST:

• Need normality assumption
• T-distribution accounts for uncertainty in s
• Heavier tails than z
• More conservative

LARGE SAMPLE (n ≥ 30): Can use T-TEST (preferred):

• CLT applies
• t-distribution → normal
• Still use t because σ unknown

Could use Z-TEST if σ known (very rare)

Step 6: T-distribution properties Depends on df = n - 1:

• Heavier tails when df small
• More probability in extremes
• Approaches normal as df → ∞

Examples: df = 5: Very heavy tails df = 30: Close to normal df = 100: Essentially normal

Step 7: Comparison for 95% critical values Z*: Always 1.96

T* depends on df: df = 5: t* = 2.571 (much larger!) df = 10: t* = 2.228 df = 20: t* = 2.086 df = 30: t* = 2.042 df = ∞: t* → 1.96

Step 8: Decision flowchart for means Is σ known? ├─ YES → Z-test (rare) │ Need: random, normal or n≥30 │ └─ NO → T-test (almost always) Need: random, normal (if n<30), independent

Step 9: Common scenarios

SCENARIO 1: n = 20, s = 5, σ unknown → T-TEST (df = 19) Need normality assumption

SCENARIO 2: n = 100, s = 12, σ unknown → T-TEST (df = 99) CLT applies, t ≈ z

SCENARIO 3: n = 50, σ = 8 known → Z-TEST But this is very rare!

SCENARIO 4: n = 15, s = 3, σ unknown, skewed population → T-TEST not appropriate! n too small, population not normal

Step 10: Summary table Z-TEST T-TEST ┌────────────────┬─────────────────┐ When to use │ σ known (rare) │ σ unknown │ ├────────────────┼─────────────────┤ Test stat │ z=(x̄-μ₀)/(σ/√n)│ t=(x̄-μ₀)/(s/√n)│ ├────────────────┼─────────────────┤ Reference │ z-table │ t-table (df=n-1)│ ├────────────────┼─────────────────┤ Sample size │ n≥30 if not │ n≥30 OR normal │ requirement │ normal │ population │ └────────────────┴─────────────────┘

Answer: USE T-TEST when:

• σ is unknown (use sample s)
• Almost all real-world situations
• Need: random, normal (n<30) or n≥30, independent
• Test statistic: t = (x̄ - μ₀)/(s/√n)
• Use t-distribution with df = n-1

USE Z-TEST when:

• σ is KNOWN (very rare!)
• Population SD given in problem
• Test statistic: z = (x̄ - μ₀)/(σ/√n)
• Use standard normal distribution

In practice, almost always use t-test because σ is rarely known!

Step 1: Set up hypotheses Claim: μ = 20 hours Test: μ < 20 (students study less)

H₀: μ = 20 Hₐ: μ < 20 (one-tailed, left)

Step 2: Check conditions n = 36

RANDOM: Random sample (given) ✓ NORMAL: n = 36 ≥ 30, CLT applies ✓ INDEPENDENT: Assume 36 ≤ 0.10N ✓

Use t-test (σ unknown)

Step 3: Calculate SE SE = s/√n = 6/√36 = 6/6 = 1

Step 4: Calculate test statistic df = n - 1 = 35

t = (x̄ - μ₀)/SE = (18.5 - 20)/1 = -1.5/1 = -1.50

Step 5: Find p-value Left-tailed test df = 35, t = -1.50

From t-table: P(t < -1.50) ≈ 0.07

p-value ≈ 0.07

Step 6: Make decision p-value = 0.07 α = 0.01

Is 0.07 < 0.01? NO

FAIL TO REJECT H₀

Step 7: State conclusion At the α = 0.01 significance level, there is insufficient evidence that students study less than 20 hours per week.

The observed difference could reasonably occur by chance.

Step 8: Additional interpretation Sample mean: 18.5 hours Claimed mean: 20 hours Difference: -1.5 hours

While students in sample study less:

• Not statistically significant at α = 0.01
• p-value (0.07) >> α (0.01)
• Could be sampling variation
• Cannot reject claim

Step 9: What if α = 0.10? If we used α = 0.10: p = 0.07 < 0.10 Would reject H₀!

But with strict α = 0.01: Need stronger evidence This result not convincing enough

Answer: H₀: μ = 20, Hₐ: μ < 20 Test statistic: t = -1.50 (df = 35) P-value: 0.07 Decision: Fail to reject H₀ at α = 0.01 Conclusion: Insufficient evidence mean is less than 20 hours

Step 1: Identify test type H₀: μ = 100 Hₐ: μ ≠ 100 (TWO-TAILED!) σ unknown → use t-test

Step 2: Check conditions (assume met) RANDOM: Assume ✓ NORMAL: n = 16 < 30, need normality assumption ✓ INDEPENDENT: Assume ✓

Step 3: Calculate SE SE = s/√n = 8/√16 = 8/4 = 2

Step 4: Calculate test statistic df = n - 1 = 15

t = (x̄ - μ₀)/SE = (105 - 100)/2 = 5/2 = 2.50

Step 5: Find p-value (TWO-TAILED!) df = 15, t = 2.50

From t-table: P(t > 2.50) ≈ 0.012

TWO-TAILED p-value: p = 2 × 0.012 = 0.024

Step 6: Make decision p-value = 0.024 α = 0.05

Is 0.024 < 0.05? YES

REJECT H₀

Step 7: State conclusion At the α = 0.05 significance level, there is sufficient evidence that the true mean differs from 100.

The sample data suggests μ ≠ 100.

Step 8: Direction of difference x̄ = 105 > 100 Evidence suggests μ > 100

But we tested two-tailed:

• Could be μ > 100 or μ < 100
• Data suggests higher
• Significant in either direction

Step 9: Strength of evidence p = 0.024 fairly small

Interpretation:

• If μ really = 100
• Only 2.4% chance of x̄ this far from 100
• Fairly unlikely under H₀
• Moderate evidence against H₀

Step 10: What if one-tailed? If we had tested Hₐ: μ > 100: One-tailed p = 0.012 Even stronger evidence!

But problem states ≠ (two-tailed) Must use p = 0.024

Step 11: Connection to CI 95% CI for μ: t* = 2.131 (df = 15) CI = 105 ± 2.131(2) = 105 ± 4.26 = (100.74, 109.26)

100 is NOT in this interval: Confirms rejection at α = 0.05!

Answer: Test statistic: t = 2.50 (df = 15) P-value: 0.024 (two-tailed) Decision: Reject H₀ at α = 0.05 Conclusion: Sufficient evidence that μ ≠ 100

The mean differs significantly from 100, appearing to be higher based on x̄ = 105.

Step 1: Find the p-value (same for both) df = 20, t = 2.15

Need to know: one-tailed or two-tailed? Assume TWO-TAILED (testing μ ≠ 50)

From t-table: P(t > 2.15) ≈ 0.022

Two-tailed p-value: p = 2 × 0.022 = 0.044

Step 2: Student A's decision (α = 0.05) p-value = 0.044 α = 0.05

Is 0.044 < 0.05? YES

Student A: REJECT H₀

Conclusion: Sufficient evidence at 0.05 level that μ ≠ 50

Step 3: Student B's decision (α = 0.01) p-value = 0.044 α = 0.01

Is 0.044 < 0.01? NO

Student B: FAIL TO REJECT H₀

Conclusion: Insufficient evidence at 0.01 level that μ ≠ 50

Step 4: Why different decisions? Same data, same p-value, different standards!

p = 0.044 is:

• Small enough for α = 0.05 (less stringent)
• NOT small enough for α = 0.01 (more stringent)

Step 5: Understanding α as threshold Think of α as "evidence requirement"

α = 0.05: Need p < 0.05

• Willing to accept 5% error rate
• Less strict
• Easier to reject H₀

α = 0.01: Need p < 0.01

• Want stronger evidence
• More strict
• Harder to reject H₀

Step 6: Interpret p-value p = 0.044 = 4.4%

Meaning: "If H₀ true, 4.4% chance of results this extreme"

Student A thinks:

• 4.4% is rare enough (< 5%)
• Unlikely under H₀
• Reject H₀

Student B thinks:

• 4.4% not rare enough (not < 1%)
• Not convincing enough
• Don't reject H₀

Step 7: Neither is "wrong"! Both are correct for their chosen α!

Different standards:

• Student A uses conventional α = 0.05
• Student B uses stricter α = 0.01

Appropriate α depends on context

Step 8: When to use different α levels α = 0.10 (liberal):

• Preliminary research
• Exploratory studies
• Don't want to miss potential effects

α = 0.05 (standard):

• Most common
• Good balance
• Convention in many fields

α = 0.01 (conservative):

• High-stakes decisions
• Medical treatments
• Want strong evidence

α = 0.001 (very conservative):

• Particle physics
• Exceptional claims
• Need extraordinary evidence

Step 9: This is a borderline case! p = 0.044 is close to 0.05

Barely significant at α = 0.05 Not significant at α = 0.01

Shows importance of:

1. Choosing α BEFORE seeing data
2. Reporting actual p-value
3. Not treating 0.05 as magic cutoff

Step 10: Better reporting Instead of just "significant" or "not":

Report: "t(20) = 2.15, p = 0.044"

Lets reader judge:

• Evidence is moderate
• Borderline significance
• Just below conventional α
• Not overwhelming evidence

Step 11: Practical advice When results are borderline:

• Report exact p-value
• Don't just say "significant"
• Consider practical significance
• May need more data
• Be cautious in conclusions

Step 12: What both students agree on Both agree:

• Data shows t = 2.15
• This is fairly unusual if H₀ true
• Evidence leans against H₀
• But evidence is not overwhelming

Disagree:

• Is evidence strong ENOUGH?
• Depends on chosen standard

Answer: STUDENT A (α = 0.05): REJECT H₀ p = 0.044 < 0.05 → Significant Sufficient evidence μ ≠ 50

STUDENT B (α = 0.01): FAIL TO REJECT H₀
p = 0.044 > 0.01 → Not significant Insufficient evidence at stricter standard

WHY DIFFERENT? α is the threshold for decision. Same p-value (0.044) meets less strict standard (0.05) but not stricter standard (0.01). Both decisions are correct for their chosen significance level. This shows that α = 0.05 is not a magic cutoff - it's a conventional standard that can be adjusted based on context and consequences of errors.