# Quick Practice Drills — English
## t-Tests, Confidence Intervals, Hypothesis Testing

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## Drill 1: Simple t-Test

**Given:**
- Sample size: n = 38
- Estimated coefficient: β̂ = 2.4
- Standard error: SE = 0.9

**Test:** H₀: β = 0 vs H₁: β ≠ 0 at α = 0.05

**Your tasks:**
1. Calculate the t-statistic
2. Find degrees of freedom
3. Find critical value (two-tailed, α = 0.05)
4. Make your decision: Reject or fail to reject H₀?
5. Calculate the p-value range using t-table

<details>
<summary>Answers</summary>

1. t = (2.4 - 0) / 0.9 = 2.667
2. df = 38 - 2 = 36 (simple regression)
3. Critical value ≈ 2.028 (or use 2.042 for df=30, 2.021 for df=40)
4. |2.667| > 2.028 → **Reject H₀**
5. From t-table: 2.434 < 2.667 < 2.750 → **0.01 < p < 0.02**

</details>

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## Drill 2: One-Tailed Test

**Given:**
- n = 55
- β̂ = -1.8, SE = 0.7

**Test:** H₀: β ≥ 0 vs H₁: β < 0 at α = 0.01

**Your tasks:**
1. Calculate t-statistic
2. Find critical value (one-tailed, α = 0.01)
3. Decision?

<details>
<summary>Answers</summary>

1. t = (-1.8 - 0) / 0.7 = -2.571 → |t| = 2.571
2. df = 53, one-tailed critical at α=0.01: ≈ 2.404
3. 2.571 > 2.404 → **Reject H₀** (evidence that β < 0)

</details>

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## Drill 3: 95% Confidence Interval

**Given:**
- n = 42
- β̂ = 3.6, SE = 1.2

**Your tasks:**
1. Construct 95% confidence interval
2. Interpret the interval
3. Does this interval contain 2.0? What does that tell you?

<details>
<summary>Answers</summary>

1. df = 40, t₀.₀₂₅ = 2.021
   Margin = 2.021 × 1.2 = 2.425
   CI: [3.6 - 2.425, 3.6 + 2.425] = **[1.175, 4.025]**

2. We are 95% confident the true β lies between 1.175 and 4.025

3. Yes, 2.0 is in the interval → We **cannot reject** H₀: β = 2.0 at α=0.05

</details>

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## Drill 4: Multiple Regression Test

**Regression output:**
| Variable | Coefficient | Std. Error |
|----------|-------------|------------|
| Intercept | 5.2 | 2.1 |
| X₁ | **1.5** | **0.4** |
| X₂ | -0.8 | 0.6 |
| X₃ | 2.1 | 0.9 |

n = 65

**Your tasks:**
1. Test each slope coefficient at α = 0.05
2. Which variables are significant?
3. Construct 90% CI for X₁

<details>
<summary>Answers</summary>

1. df = 65 - 3 - 1 = 61, critical value ≈ 1.96 (or 2.000 for df=60)

   - X₁: t = 1.5/0.4 = 3.75 → **Significant** ✓
   - X₂: t = -0.8/0.6 = -1.33 → |1.33| < 2.0 → **Not significant** ✗
   - X₃: t = 2.1/0.9 = 2.33 → **Significant** ✓

2. X₁ and X₃ are significant at 5% level

3. 90% CI for X₁: t₀.₀₅,₆₁ ≈ 1.671
   Margin = 1.671 × 0.4 = 0.668
   CI: [1.5 - 0.668, 1.5 + 0.668] = **[0.832, 2.168]**

</details>

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## Drill 5: CI for Hypothesis Test

**Given:**
- 95% CI for β: [0.5, 3.2]

**Test:** H₀: β = 4.0 vs H₁: β ≠ 4.0 at α = 0.05

**Your task:** Use the CI method to test this hypothesis.

<details>
<summary>Answer</summary>

4.0 lies **outside** the 95% CI [0.5, 3.2]

→ **Reject H₀** at α = 0.05

The hypothesized value 4.0 is not a plausible value for β.

</details>

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## Drill 6: Interpretation Check

**Given:** β̂ = 2.3, p-value = 0.08

**Which statements are correct?**

- [ ] The effect is not statistically significant at 5% level
- [ ] There is an 8% probability that β = 0
- [ ] If H₀ were true, there's 8% chance of seeing this result
- [ ] We are 92% confident there is an effect
- [ ] At 10% level, the effect would be significant

<details>
<summary>Answers</summary>

Correct: ✓✗✓✗✓

- ✓ Not significant at 5% (0.08 > 0.05)
- ✗ P-value is NOT probability H₀ is true
- ✓ Correct interpretation of p-value
- ✗ Confidence and significance are different concepts
- ✓ Would be significant at 10% (0.08 < 0.10)

</details>

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## Drill 7: Full Problem — Coffee Shop Prices

**Scenario:** A researcher studies coffee shop prices in 28 cities.

**Model:** Priceᵢ = β₀ + β₁Rentᵢ + β₂Wageᵢ + uᵢ

**Output:**
| Variable | Coefficient | Std. Error |
|----------|-------------|------------|
| Intercept | 1.50 | 0.80 |
| Rent (€100/m²) | **0.25** | **0.08** |
| Wage (€/hour) | 0.15 | 0.12 |

**Questions:**
1. Test if rent affects price at 5% level
2. Test if wage affects price at 5% level
3. Construct 95% CI for rent coefficient
4. A politician claims each €100 rent increase raises price by €0.40. Test this claim.

<details>
<summary>Answers</summary>

1. Rent: t = 0.25/0.08 = 3.125, df = 25, critical ≈ 2.060
   3.125 > 2.060 → **Significant** ✓

2. Wage: t = 0.15/0.12 = 1.25, |1.25| < 2.060 → **Not significant** ✗

3. 95% CI for rent: t₀.₀₂₅,₂₅ = 2.060
   Margin = 2.060 × 0.08 = 0.165
   CI: [0.25 - 0.165, 0.25 + 0.165] = **[0.085, 0.415]**

4. Politician claims β₁ = 0.40. Is 0.40 in the CI [0.085, 0.415]?
   Yes! (0.40 is inside the interval)
   → **Cannot reject** the politician's claim at 5% level.

</details>

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## Drill 8: Quick Calculations

**Calculate mentally or with scratch paper:**

| β̂ | SE | n | k | t-stat | Significant at 5%? |
|----|----|---|---|--------|-------------------|
| 4.0 | 1.5 | 30 | 1 | ? | ? |
| -2.5 | 1.0 | 50 | 2 | ? | ? |
| 0.8 | 0.3 | 100 | 3 | ? | ? |
| 1.2 | 0.9 | 25 | 1 | ? | ? |

<details>
<summary>Answers</summary>

| β̂ | SE | n | k | t-stat | df | critical | Significant? |
|----|----|---|---|--------|-----|----------|--------------|
| 4.0 | 1.5 | 30 | 1 | 2.67 | 28 | 2.048 | **Yes** |
| -2.5 | 1.0 | 50 | 2 | 2.50 | 47 | 2.012 | **Yes** |
| 0.8 | 0.3 | 100 | 3 | 2.67 | 96 | 1.985 | **Yes** |
| 1.2 | 0.9 | 25 | 1 | 1.33 | 23 | 2.069 | **No** |

</details>

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## Formula Sheet

**t-statistic:**
$$t = \frac{\hat{\beta} - \beta_0}{SE(\hat{\beta})}$$

**Confidence Interval:**
$$CI = \hat{\beta} \pm t_{\alpha/2, df} \times SE(\hat{\beta})$$

**Degrees of freedom:**
- Simple regression: df = n - 2
- Multiple regression: df = n - k - 1 (where k = number of X variables)

**Common critical values (two-tailed):**

| df | α = 0.10 | α = 0.05 | α = 0.01 |
|----|----------|----------|----------|
| 25 | 1.708 | 2.060 | 2.787 |
| 30 | 1.697 | 2.042 | 2.750 |
| 40 | 1.684 | 2.021 | 2.704 |
| 60 | 1.671 | 2.000 | 2.660 |
| 120 | 1.658 | 1.980 | 2.617 |
| ∞ | 1.645 | 1.960 | 2.576 |

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*Practice these until you can do them in your sleep! 🎯*