Chapter 3.2, Problem 45E

To determine

To find and interpret: the residuals for the week.

Answer to Problem 45E

Residual for the week is $-\text{0}\text{.085}$

The line predicted a pH value for that week that was 0.085 too large.

Explanation of Solution

Given:

The actual pH measurement for week 50 was 5.08.

Calculation:

Consider the case acid rain, regression line is:

  $pH=5.43-0.0053(weeks)$

Since we assume that number of weeks will help explain pH value. So, week is the explanatory variable and pH value is the response variable. Predicted pH measurement for week 50 is

  $\begin{array}{l}pH=5.43-0.0053\times 50\\ \text{ = 5}\text{.165}\end{array}$

The predicted pH value for week 50 is 5.165 and actual (observed) pH measurement for week 50 was 5.08.

  $\begin{array}{l}\text{residual = observed Y}-\text{predectedY}\\ \text{ = Y}-\text{<mover accent="true"><mo>Y</mo><mo>^</mo></mover>}\\ \text{ = 5}\text{.08}-\text{5}\text{.165}\\ \text{ =}-\text{0}\text{.085}\end{array}$

This residuals is negative; this shows the data point lies below the line.

The line predicted a pH value for that week that was 0.085 too large.

Conclusion:

Hence, the line predicted a pH value for that week that was 0.085 too large.

Residual for the week is $-\text{0}\text{.085}$