91Ó°ÊÓ

We are given the following information: $$ n = 10, \mathrm{SSE} = 24, \Sigma x_i = 59, \Sigma x_i^2 = 397 $$ The regression line equation is: $$ \hat{y} = 0.074 + 0.46x $$ and the value for \(x_0\) is \(5\).

Calculate the mean of the x values, denoted by \(\bar{x}\). To do this, divide the sum of the x values by the total number of data points: $$ \bar{x} = \frac{\Sigma x_i}{n} = \frac{59}{10} = 5.9 $$

Compute the standard error of estimate, denoted by \(s\): $$ s = \sqrt{\frac{SSE}{n - 2}} = \sqrt{\frac{24}{10 - 2}} = \sqrt{\frac{24}{8}} = 1.732 $$

Now, calculate the sum of squares of deviations (SS_x), using the following formula: $$ SS_x = \Sigma x_i^2 - \frac{(\Sigma x_i)^2}{n} = 397 - \frac{(59)^2}{10} = 397 - 348.1 = 48.9 $$

Now, we will find the prediction interval using the following formula: $$ \hat{y} \pm t_{\alpha / 2, n - 2} \cdot s \cdot \sqrt{1 + \frac{1}{n} + \frac{(x_0 - \bar{x})^2}{SS_x}} $$ where \(t_{\alpha / 2, n - 2}\) is the t-score with \(\alpha / 2\) and \(n - 2\) degrees of freedom. We are given that the 90% prediction interval is wanted, which means \(\alpha = 0.1\). From a t-table, we can find that \(t_{0.05, 8} \approx 2.306\). Now, let's plug in all the values: $$ \hat{y} \pm 2.306 \cdot 1.732 \cdot \sqrt{1 + \frac{1}{10} + \frac{(5 - 5.9)^2}{48.9}} $$ To find \(\hat{y}\), we need to plug in \(x_0\) into the regression equation: $$ \hat{y} = 0.074 + 0.46(5) = 2.374 $$ Now we can substitute this value back into the prediction interval formula: $$ 2.374 \pm 2.306 \cdot 1.732 \cdot \sqrt{1 + \frac{1}{10} + \frac{(5 - 5.9)^2}{48.9}} $$ This simplifies to: $$ 2.374 \pm 3.974 $$ So the \(90\%\) prediction interval for \(y\) when \(x = 5\) is \((2.374 - 3.974, 2.374 + 3.974) = (-1.6, 6.348)\).

The given question asks if this interval is wider than the corresponding confidence interval from Exercises 3-4. We can see that the width of this interval is \(3.974 * 2 = 7.948\). Now refer back to Exercises 3-4 where the confidence interval is computed and compare the widths of the intervals. If the confidence interval width is smaller than 7.948, then the prediction interval is wider; otherwise, the prediction interval is not wider.