Let's break down the solution step by step for each part: --- ### **Step 1: Fit the Quadratic Model** Given data: \[ \begin{array}{c|cccc} x & -1 & 0 & 1 & 2 \\ \hline Y & -5 & -2 & 1 & 3 \\ \end{array} \] Model: \(\hat{Y} = \hat{\beta}_0 + \hat{\beta}_1 x + \hat{\beta}_2 x^2\) Let's solve for the coefficients: Set up the system using the points: For \(x = -1\): \(-5 = \hat{\beta}_0 - \hat{\beta}_1 + \hat{\beta}_2\) For \(x = 0\): \(-2 = \hat{\beta}_0\) For \(x = 1\): \(1 = \hat{\beta}_0 + \hat{\beta}_1 + \hat{\beta}_2\) For \(x = 2\): \(3 = \hat{\beta}_0 + 2\hat{\beta}_1 + 4\hat{\beta}_2\) From \(x=0\), \(\hat{\beta}_0 = -2\). Plug \(\hat{\beta}_0 = -2\) into the others: - For \(x = -1\): \(-5 = -2 - \hat{\beta}_1 + \hat{\beta}_2 \implies -3 = -\hat{\beta}_1 + \hat{\beta}_2\) \(\implies \hat{\beta}_1 - \hat{\beta}_2 = 3\) ... (1) - For \(x = 1\): \(1 = -2 + \hat{\beta}_1 + \hat{\beta}_2 \implies 3 = \hat{\beta}_1 + \hat{\beta}_2\) ... (2) Add (1) and (2): \[ (\hat{\beta}_1 - \hat{\beta}_2) + (\hat{\beta}_1 + \hat{\beta}_2) = 3 + 3 \implies 2\hat{\beta}_1 = 6 \implies \hat{\beta}_1 = 3 \] Substitute into (2): \(3 = 3 + \hat{\beta}_2 \implies \hat{\beta}_2 = 0\). So, coefficients: \[ \boxed{\hat{\beta}_0 = -2, \hat{\beta}_1 = 3, \hat{\beta}_2 = 0} \] **Explanation:** We solved the system using the \(x\) values to find the coefficients for the quadratic model. --- ### **Step 2: Use the Model to Find Confidence Intervals** #### (a) **90% CI for \(E(Y)\) when \(x^* = 4\)** Plug in \(x^* = 4\): \[ \hat{Y} = -2 + 3(4) + 0(16) = -2 + 12 = 10 \] Now, the 90% confidence interval for the mean is: \[ \hat{Y} \pm t_{n-3,\,0.95} \cdot s_{\hat{Y}} \] - \(n = 4\) data points, so \(n-3 = 1\) degree of freedom. - \(t_{1, 0.95} \approx 6.314\) (from t-tables). - The standard error (\(s_{\hat{Y}}\)) can be calculated, but since the data fits a quadratic perfectly (no residuals), \(s^2 = 0\). So, the confidence interval is just [10, 10]. **Explanation:** With only 4 points and a perfect fit, the standard error is zero, so the CI collapses to the point estimate. --- #### (b) **90% CI for \(Y^*\) when \(x^* = 4\)** For predicting a new value (\(Y^*\)), the formula is: \[ \hat{Y} \pm t_{n-3,\,0.95} \cdot s_{pred} \] where \(s_{pred}^2 = s^2 \left[1 + h^*\right]\), and \(h^*\) is the leverage at \(x^*\). But again, \(s^2 = 0\) here, so the prediction interval is also just [10, 10]. **Explanation:** With a perfect fit (no residual error), the model predicts with certainty at \(x^* = 4\). --- ## **Final Answers** **(a) The 90% confidence interval for \(E(Y)\) when \(x^* = 4\) is [10, 10].** **(b) The 90% confidence interval for \(Y^*\) when \(x^* = 4\) is [10, 10].** --- > **Note:** In real data (with nonzero residuals), you would compute \(s^2\) from residuals and use the standard formula for the confidence and prediction intervals. Here, the fit is exact, so the intervals reduce to a single point.