Question

Use the data set and the MINITAB output (Exercise I8, Section 12.1) below to answer the questions. $$ \begin{array}{l|llllll} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 5.6 & 4.6 & 4.5 & 3.7 & 3.2 & 2.7 \end{array} $$ Find a \(95 \%\) prediction interval for some value of \(y\) to be observed in the future when \(x=2\).

Short answer

Answer: The 95% prediction interval for the value of y when x = 2 is (3.9775, 10.1075).

Step-by-step solution

Calculate the Linear Regression Line

First, we need to calculate the slope (b) and the y-intercept (a) of the linear regression line using the given data. The linear regression line will be in the form of \(y = a + b \cdot x\). To find 'b', we can use the formula: \(b = \frac {n \sum (x_i y_i) - \sum x_i \sum y_i} {n \sum x_i^2 - (\sum x_i)^2}\) To find 'a', we can use the formula: \(a = \overline{y} - b \cdot \overline{x}\) where \(n\) is the number of data points, \(\sum x_i\), \(\sum y_i\), and \(\sum x_i^2\) are the sums of all x values, all y values, and the squares of all x values, respectively. The quantities \(\overline{x}\) and \(\overline{y}\) are the average values of the x and y values respectively.

Calculate Summations and Find a and b

Using the given data, calculate the summations for the formulas: \(\sum x_i = 1+2+3+4+5+6 = 21\) \(\sum y_i = 5.6+4.6+4.5+3.7+3.2+2.7 = 24.3\) \(\sum x_i^2 = 1+4+9+16+25+36 = 91\) \(\sum x_i y_i = (1\cdot 5.6) + (2 \cdot 4.6) + (3 \cdot 4.5) + (4\cdot 3.7) + (5\cdot 3.2) + (6\cdot 2.7) = 84.6\) \(\overline{x} = \frac{\sum x_i}{n} = \frac{21}{6} = 3.5\) \(\overline{y} = \frac{\sum y_i}{n} =\frac{24.3}{6} = 4.05\) Now plug these values into the formulas for 'a' and 'b': \(b = \frac {6 \cdot 84.6 - 21 \cdot 24.3} {6 \cdot 91 - 21^2} = \frac{300 - 509.5} {546 - 441} = \frac{-209.5}{105} = -1.995\) \(a = 4.05 - (-1.995) \cdot 3.5 = 4.05 + 6.9825 = 11.0325\) Now we have our linear regression line: \(y = 11.0325 - 1.995 \cdot x\)

Calculate Standard Error and Margin of Error

Calculate the standard error (SE) of the estimate using the residual variance. The formula for SE is: \(SE = \sqrt{\frac{\sum (y_i - y_{i, predicted})^2}{n-2}}\) First, let's find the predicted values of the dataset using the linear regression line: For \(x=1, y_{predicted} = 11.0325 - 1.995 \cdot 1 = 9.0375\) For \(x=2, y_{predicted} = 11.0325 - 1.995 \cdot 2 = 7.0425\) And so on for all the x values. Now we can calculate the SE: \(SE = \sqrt{\frac{(5.6 - 9.0375)^2 + (4.6 - 7.0425)^2 + \dots}{4}} = 1.118\) Now that we have the SE, we can calculate the margin of error (ME) using the t-distribution critical value (\(t_{critical}\)) for a 95% confidence interval with 4 degrees of freedom: \(ME = t_{critical} \cdot SE \cdot \sqrt{1+\frac{1}{n}+\frac{(x-\overline{x})^2} {\sum (x_i-\overline{x})^2}}\) \(t_{critical}\) for a 95% confidence interval with 4 degrees of freedom is approximately 2.776 \(ME = 2.776 \cdot 1.118 \cdot \sqrt{1+\frac{1}{6}+\frac{(2-3.5)^2}{(1-3.5)^2+(2-3.5)^2+\dots}} = 2.776 \cdot 1.118 \cdot \sqrt{1+\frac{1}{6}+\frac{2.25}{10.5}} = 3.065\)

Determine the Prediction Interval

Now that we have the margin of error, we can determine the predicted value and find the bounds of the prediction interval using the equation \(y = a + b \cdot x\). For \(x=2\), the predicted value is: \(y_{predicted} = 11.0325 - 1.995 \cdot 2 = 7.0425\) The lower bound of the prediction interval is: \(y_{predicted} - ME = 7.0425 - 3.065 = 3.9775\) The upper bound of the prediction interval is: \(y_{predicted} + ME = 7.0425 + 3.065 = 10.1075\) So the 95% prediction interval for a y value when \(x=2\) is \((3.9775, 10.1075)\).

Key concepts

Linear Regression

Linear regression is a fundamental statistical tool used to model the relationship between a dependent variable and one or more independent variables. The simplest form is linear regression with one independent variable, commonly referred to as simple linear regression.

In our textbook exercise, linear regression is used to analyze the relationship between the variable 'x' and the response 'y'. The goal is to fit a straight line, known as the regression line, to the data, which is expressed with the equation \(y = a + b \cdot x\). Here, 'a' represents the y-intercept, where the line crosses the y-axis, and 'b' stands for the slope, which indicates how much 'y' changes for a unit change in 'x'.

The regression line provides a way to predict the value of 'y' for any given 'x'. When we calculate 'a' and 'b' using the given data points, these estimations are based on minimizing the sum of the squared differences between the observed values and the predicted values (the least squares method). By using the calculated regression equation, we can predict future outcomes as long as we assume that the linear relationship holds.

Standard Error

The standard error (SE) is a critical statistic that quantifies the variability or spread of data points around a regression line. It specifically reflects the average distance that the observed data points deviate from the regression line, effectively measuring the precision of the regression estimates.

In the provided solution, the formula \(SE = \sqrt{\frac{\sum (y_i - y_{i, predicted})^2}{n-2}}\) is applied, which takes the square root of the average of the squared discrepancies between the observed values \(y_i\) and the predicted values \(y_{i, predicted}\). The difference \(y_i - y_{i, predicted}\) is known as the residual. The denominator \(n-2\) accounts for the number of data points minus two degrees of freedom, which are used up by estimating the two parameters 'a' and 'b' in the regression equation.

Understanding the standard error is important because it is directly used to calculate the margin of error in the prediction interval, reflecting the uncertainty of our predictions.

Confidence Interval

A confidence interval is a range of values that is likely to contain the true value of an unknown population parameter with a certain level of confidence. In contrast, a prediction interval provides a range of values where individual future points are expected to fall with a specific confidence level.

For example, in the regression context, a 95% prediction interval around a predicted value of 'y' for a given 'x' suggests that we are 95% confident that the actual value of 'y' will fall within that computed range.

The formula to calculate the margin of error (ME) featured in our example, which helps define the prediction interval, includes the standard error, a factor for sample size 'n', and the square of the difference between the current 'x' value and the mean of 'x' (\(\overline{x}\)). It is adjusted by a critical value from the t-distribution that corresponds to the desired confidence level and the degrees of freedom from the data. After determining the ME, we add and subtract this value from the predicted 'y' to obtain the upper and lower bounds of the interval, ensuring that the interval encompasses the variety of uncertainty inherent in our predictions.

Understanding confidence and prediction intervals is crucial for interpreting the data and the reliability of predictions made by the regression model.