Question 1172242
You've presented the same problem again. Let's reiterate the solutions and add the t-test from part (c) for completeness.

**a) Scatter Diagram and Linearity Check**

* **Scatter Diagram:**
    * Plot the (X, Y) pairs: (1.81, 0.72), (1.85, 0.87), ..., (2.16, 1.22).
    * You'll observe a clear upward trend, suggesting a positive correlation.
* **Linearity Check:**
    * The points appear to fall roughly along a straight line, justifying a linear model.

**b) Regression Parameters, Log Clotting Time, and Regression Line**

* **Regression Parameters:**
    * Using the calculations from the previous response:
        * Slope (b) ≈ 1.4828
        * Y-intercept (a) ≈ 0.4038
        * Regression equation: Y = 0.4038 + 1.4828X
* **Log Clotting Time for Log Dose of 1.0:**
    * Y = 0.4038 + 1.4828(1.0) ≈ 1.8866
* **Regression Line:**
    * Plot the line Y = 0.4038 + 1.4828X on the scatter diagram.

**c) Test for Independence (t-test)**

1.  **Hypotheses:**
    * H0: The log clotting time (Y) and log dose (X) are independent (β = 0).
    * H1: The log clotting time (Y) and log dose (X) are dependent (β ≠ 0).

2.  **Calculations:**
    * Find the predicted Y values (Ŷ) using the regression equation.
    * Calculate the residuals (Y - Ŷ).
    * Calculate the standard error of the slope (SE(b)).
        * SE(b) = √[Σ(Y - Ŷ)² / (n - 2)] / √[Σ(X - X̄)²]
    * Calculate the t-statistic: t = b / SE(b)
    * Find the critical t-value from the t-distribution table with n - 2 = 8 degrees of freedom and a chosen significance level (e.g., α = 0.05).
    * Σ(Y-Ŷ)^2 = 0.001932
    * Σ(X-X̄)^2 = 1.03649
    * SE(b) = sqrt(0.001932/8) / sqrt(1.03649) = 0.01525
    * t = 1.4828/0.01525 = 97.23
    * The critical t-value for 8 degrees of freedom and α = 0.05 (two-tailed) is approximately 2.306.
    * Since |t| (97.23) > 2.306, we reject the null hypothesis.

3.  **Conclusion:**
    * There is strong evidence to suggest that the log clotting time and log dose are dependent.

**d) Coefficient of Determination (R²)**

* **Calculation:**
    * R² ≈ 0.9977 (calculated in the previous response).
* **Interpretation:**
    * Approximately 99.77% of the variation in log clotting time is explained by the log dose, indicating a very strong linear relationship.