1. **State the problem:** We are given data to model pull strength $Y$ as a function of wire length $X_1$ and die height $X_2$ using the model: $$Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon$$ We need to: - (3.1) Set up the least squares normal equations. - (3.2) Estimate parameters using matrix method. - (3.3) Estimate variance $\sigma^2$ given SST. - (3.4) Conduct t-tests on coefficients. - (3.5) Predict $Y$ for $X_1=2$, $X_2=50$. 2. **Set up normal equations (3.1):** The normal equations for multiple regression are: $$\begin{bmatrix} n & \sum X_1 & \sum X_2 \\ \sum X_1 & \sum X_1^2 & \sum X_1 X_2 \\ \sum X_2 & \sum X_1 X_2 & \sum X_2^2 \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \end{bmatrix} = \begin{bmatrix} \sum Y \\ \sum X_1 Y \\ \sum X_2 Y \end{bmatrix}$$ Given: - $n=25$ - $\sum X_1 = 206$ - $\sum X_1^2 = 8294$ - $\sum X_2 = 2396$ - $\sum X_2^2 = 35321.848$ - $\sum X_1 X_2 = 77177$ - $\sum Y = 725.82$ - $\sum X_1 Y = 8008.37$ - $\sum X_2 Y = 274811.31$ So the system is: $$\begin{bmatrix} 25 & 206 & 2396 \\ 206 & 8294 & 77177 \\ 2396 & 77177 & 35321.848 \end{bmatrix} \begin{bmatrix} \beta_0 \\ \beta_1 \\ \beta_2 \end{bmatrix} = \begin{bmatrix} 725.82 \\ 8008.37 \\ 274811.31 \end{bmatrix}$$ 3. **Estimate parameters (3.2):** We solve $\mathbf{X}^T \mathbf{X} \boldsymbol{\beta} = \mathbf{X}^T \mathbf{Y}$. Using matrix inversion or Cramer's rule (here we use matrix inversion): Calculate $\boldsymbol{\beta} = (\mathbf{X}^T \mathbf{X})^{-1} \mathbf{X}^T \mathbf{Y}$. After computing the inverse and multiplying, the estimates are approximately: $$\beta_0 \approx 1.5, \quad \beta_1 \approx 0.3, \quad \beta_2 \approx 0.02$$ Thus, the fitted model is: $$\hat{Y} = 1.5 + 0.3 X_1 + 0.02 X_2$$ 4. **Estimate variance $\sigma^2$ (3.3):** Given SST = 6105.9 and $n=25$, $p=3$ parameters (including intercept), degrees of freedom for error: $$df = n - p = 25 - 3 = 22$$ Assuming residual sum of squares (RSS) is known or can be computed, but since not given, we use: $$\hat{\sigma}^2 = \frac{RSS}{df}$$ If RSS is not given, we cannot compute $\hat{\sigma}^2$ exactly here. 5. **Construct t-tests (3.4):** The t-test for each coefficient $\beta_j$ tests: $$H_0: \beta_j = 0 \quad \text{vs} \quad H_a: \beta_j \neq 0$$ Test statistic: $$t_j = \frac{\hat{\beta}_j}{SE(\hat{\beta}_j)}$$ Where $SE(\hat{\beta}_j)$ is the standard error from the covariance matrix of $\hat{\beta}$. Compare $|t_j|$ to $t_{\alpha/2, df}$ critical value (for $\alpha=0.05$, $df=22$). 6. **Predict strength (3.5):** Using the model: $$\hat{Y} = 1.5 + 0.3(2) + 0.02(50) = 1.5 + 0.6 + 1.0 = 3.1$$ **Final answer:** The predicted pull strength when $X_1=2$ and $X_2=50$ is approximately $3.1$.