Question

Use the information given (reproduced below) to find a prediction interval for a particular value of \(y\) when \(x=x_{0} .\) Is the interval wider than the corresponding confidence interval from Exercises \(3-4 ?\) $$ \begin{array}{l} n=10, \mathrm{SSE}=24, \Sigma x_{i}=59, \Sigma x_{i}^{2}=397, \\ \hat{y}=.074+.46 x, x_{0}=5,90 \% \text { prediction interval } \end{array} $$

Short answer

Answer: To determine if the prediction interval for \(y\) when \(x = 5\) is wider than the corresponding confidence interval from Exercises 3-4, compare the widths of the intervals. The width of the prediction interval is \(7.948\). If the width of the confidence interval is smaller than \(7.948\), then the prediction interval is wider; otherwise, the prediction interval is not wider.

Step-by-step solution

Gather given information

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 \(5\).

Find the mean value of x

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: $$ \bar{x} = \frac{\Sigma x_i}{n} = \frac{59}{10} = 5.9 $$

Find the standard error of estimate

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

Calculate the sum of squares of deviations

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

Find the prediction interval

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 \((2.374 - 3.974, 2.374 + 3.974) = (-1.6, 6.348)\).

Compare prediction interval to confidence interval

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.

Key concepts

Understanding Regression Analysis

Regression analysis is a statistical method that allows us to understand the relationship between a dependent variable (often denoted as y) and one or more independent variables (denoted as x). In simple terms, it helps us to predict the value of y based on the values of x.

In the case of linear regression, as seen in the exercise, the relationship is modeled as a straight line, also known as the regression line. This line is represented by the equation ŷ = a + bx, where ŷ is the predicted value, a is the y-intercept, b is the slope, and x is the value of the independent variable. Here, the hat on y is used to denote that it is an estimate.

Regression analysis is commonly used in fields like economics, social sciences, and engineering for forecasting and determining the strength of predictors. It's important to note that while regression can give us a high-level view of potential trends, it doesn't prove causation; it simply suggests association between variables.

The Standard Error of Estimate

The standard error of estimate, often denoted as s, is a measure of the accuracy of predictions made with a regression line. It tells us how much the actual values deviate, on average, from the predicted values.

In other words, it quantifies the spread of the observed values around the regression line, providing insight into the precision of the regression model. The smaller the standard error, the closer the data points fall to the regression line, indicating that the model has a higher predictive accuracy.

To find this value, as outlined in the solution steps for the exercise, we compute s using the formula s = √(SSE/(n-2)), where SSE is the 'Sum of Squares due to Error' and n is the total number of observations. The SSE is a measure of the total deviation of the actual values from the predicted values, while n-2 accounts for the degrees of freedom in a bivariate regression analysis.

The Sum of Squares of Deviations

The sum of squares of deviations, commonly referred to as SS_x, is an essential part of calculating both the standard error of estimate and the prediction interval. It represents the amount of variation in the independent variable x.

To calculate SS_x, we subtract the square of the sum of all x-values divided by the number of observations from the sum of all x-values squared: SS_x = ∑x_i² - (∑x_i)²/n.

• The term ∑x_i² means that we sum up all the squared individual values of x.
• The term (∑x_i)²/n helps to adjust for the mean of x, avoiding overestimation of the variability.

Understanding SS_x is key because it directly affects the precision of the estimate. If SS_x is large, the data points are more spread out, and each individual point has less leverage over the regression line. Conversely, a small SS_x means the data points have a greater influence on the regression line's slope and position.