Question

Let a linear model \( Y = \beta_0 + \beta_1 X + \epsilon \) be fitted to the following data, where \( \epsilon \) is normally distributed with mean 0 and unknown variance \( \sigma^2>0 \): \[ \begin{array}{|c|c|c|c|c|c|} \hline x_i & 0 & 1 & 2 & 3 & 4 \\ \hline y_i & 3 & 4 & 5 & 6 & 7 \\ \hline \end{array} \] Let \( \hat{Y}_0 \) denote the ordinary least-square estimator of \( Y \) at \( X = 6 \), and the variance of \( \hat{Y}_0 \) be \( c\sigma^2 \). Then the value of the real constant \( c \) (rounded off to one decimal place) is equal to ________

Hint:

The variance of prediction increases with the distance between \( X_0 \) and \( \bar{X} \); this reflects the uncertainty of extrapolation in regression.

Answer 1

Correct Answer: 1.8

Given \( Y = \beta_0 + \beta_1 X + \epsilon \), we can calculate the fitted regression line parameters. First, compute the sample means: \[ \bar{X} = \frac{0 + 1 + 2 + 3 + 4}{5} = 2, \quad \bar{Y} = \frac{3 + 4 + 5 + 6 + 7}{5} = 5. \] The slope is given by: \[ \hat{\beta}_1 = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{\sum (X_i - \bar{X})^2}. \] \[ \sum (X_i - \bar{X})(Y_i - \bar{Y}) = ( -2)(-2) + (-1)(-1) + 0(0) + 1(1) + 2(2) = 10, \] \[ \sum (X_i - \bar{X})^2 = (-2)^2 + (-1)^2 + 0^2 + 1^2 + 2^2 = 10. \] Thus, \( \hat{\beta}_1 = 1 \). Intercept: \[ \hat{\beta}_0 = \bar{Y} - \hat{\beta}_1 \bar{X} = 5 - 1(2) = 3. \] So the fitted model is: \[ \hat{Y} = 3 + X. \] The variance of the predicted value at \( X_0 = 6 \) is given by: \[ \text{Var}(\hat{Y}_0) = \sigma^2 \left[ \frac{1}{n} + \frac{(X_0 - \bar{X})^2}{\sum (X_i - \bar{X})^2} \right]. \] Substitute values: \[ \text{Var}(\hat{Y}_0) = \sigma^2 \left[ \frac{1}{5} + \frac{(6 - 2)^2}{10} \right] = \sigma^2 \left[ 0.2 + 1.6 \right] = 1.8\sigma^2. \] Hence, \( c = \boxed{1.8} \).