Chapter 7.2, Problem 10E

The article “Effect of Environmental Factors on Steel Plate Corrosion Under Marine Immersion Conditions” (C. Soares, Y. Garbatov, and A. Zayed, Corrosion Engineering, Science and Technology, 2011:524–541) describes an experiment in which nine steel specimens were submerged in seawater at various temperatures, and the corrosion rates were measured. The results are presented in the following table (obtained by digitizing a graph).

Temperature (°C)	Corrosion (mm/yr)
26.6	1.58
26.0	1.45
27.4	1.13
21.7	0.96
14.9	0.99
11.3	1.05
15.0	0.82
8.7	0.68
8.2	0.56

1. a. Construct a scatterplot of corrosion (y) versus temperature (x). Verify that a linear model is appropriate.
2. b. Compute the least-squares line for predicting corrosion from temperature.
3. c. Two steel specimens whose temperatures differ by 10°C are submerged in seawater. By how much would you predict their corrosion rates to differ?
4. d. Predict the corrosion rate for steel submerged in seawater at a temperature of 20°C.
5. e. Compute the fitted values.
6. f. Compute the residuals. Which point has the residual with the largest magnitude?
7. g. Compute the correlation between temperature and corrosion rate.
8. h. Compute the regression sum of squares, the error sum of squares, and the total sum of squares.
9. i. Divide the regression sum of squares by the total sum of squares. What is the relationship between this quantity and the correlation coefficient?

To determine

Construct a scatterplot of corrosion (y) versus temperature (x) and also check whether the linear model is appropriate or not.

Answer to Problem 10E

The linear model is appropriate.

Explanation of Solution

Calculation:

The given information is that the data shows the temperature (°C) and corrosion (mm/yr) for 9 steel specimens.

Software Procedure:

Step-by-step procedure to obtain the scatterplot using the MINITAB software:

• Choose Graph > Scatter plot.
• Choose Simple, and then click OK.
• Under Y variables, select Corrosion.
• Under X variables, select Temperature.
• Click OK.

Output using the MINITAB software is given below:

From the plot, it can be observed that the relationship between temperature and corrosion is linear. Therefore, the linear model is appropriate.

To determine

Find the least-squares line for predicting corrosion from temperature.

Answer to Problem 10E

The least-squares line for predicting corrosion from temperature is $\text{Corrosion}=0.394+0.03551\text{Temp}$.

Explanation of Solution

Calculation:

Software Procedure:

Step-by-step procedure to obtain the least-squares line using the MINITAB software is given below:

• Choose Stat > Regression > Regression > Fit Regression Model.
• In Responses, enter “Corrosion”.
• In Continuous predictors, enter “Temperature”.
• Check Results.
• In Display of results, choose Simple tables.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the least-squares line for predicting corrosion from temperature is $\text{Corrosion}=0.394+0.03551\text{Temp}$.

To determine

By how much would predict corrosion rates of two steel specimens to differ whose temperatures differ by 10ºC.

Explanation of Solution

Calculation:

From the least square line, the slope ${\widehat{\beta }}_1$ is 0.03551.

The change in the predicted corrosion rates when two steel specimens whose temperatures differ by 10ºC is $0.03551\times 10=\text{0}\text{.3551}$.

Thus, the predicted corrosion rate is 0.3351 mm/yr.

To determine

Predict the corrosion rate for steel submerged in seawater at a temperature of 20ºC.

Answer to Problem 10E

The predicted corrosion rate for steel submerged in seawater at a temperature of 20ºC is 1.10414 mm/yr.

Explanation of Solution

Calculation:

Predicted value:

Software Procedure:

Step-by-step procedure to obtain the predicted value using the MINITAB software:

• Stat > Regression > Regression > Predict.
• In Responses, enter “Corrosion”.
• Choose Enter individual values.
• In Temperature, enter 20.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the predicted corrosion rate for steel submerged in seawater at a temperature of 20ºC is 1.10414 mm/yr.

To determine

Find the fitted values.

Answer to Problem 10E

The fitted values are, 1.33850, 1.31720, 1.36691, 1.16451, 0.92305, 0.79521, 0.92660, 0.70289 and 0.68513.

Explanation of Solution

Calculation:

Fitted value:

Software Procedure:

Step-by-step procedure to obtain the fitted value using the MINITAB software is given below:

• Choose Stat > Regression > Regression > Fit Regression Model.
• In Responses, enter “Corrosion”.
• In Continuous predictors, enter “Temperature”.
• Check Results.
• In Display of results, choose Simple tables.
• In Storage, select fits.
• Click OK.

Data display:

• Choose Data > Display data.
• In Columns, constants, and matrices to display, select FITS 1.

Output using the MINITAB software is given below:

The fitted values are, 1.33850, 1.31720, 1.36691, 1.16451, 0.92305, 0.79521, 0.92660, 0.70289 and 0.68513.

To determine

Find the residuals and identify the point whose residual has the largest magnitude.

Answer to Problem 10E

The residual points are 0.241497, 0.132802, –0.236911, –0.204508, 0.066954,0.254787, –0.106597, –0.022889 and –0.125135.

The point whose residual has the largest magnitude is (11.3, 1.05).

Explanation of Solution

Calculation:

Residuals:

Software Procedure:

Step-by-step procedure to obtain the fitted value using the MINITAB software is given below:

• Choose Stat > Regression > Regression > Fit Regression Model.
• In Responses, enter “Corrosion”.
• In Continuous predictors, enter “Temperature”.
• Check Results.
• In Display of results, choose Simple tables.
• In Storage, select residuals.
• Click OK.

Data display:

• Choose Data > Display data.
• In Columns, constants, and matrices to display, select RESI 1.

Output using the MINITAB software is given below:

The residual points are 0.241497, 0.132802, –0.236911, –0.204508, 0.066954, 0.254787, –0.106597, –0.022889 and –0.125135.

Therefore, the point whose residual has the largest magnitude is (11.3, 1.05) because this point has the largest residual.

To determine

Find the correlation between temperature and corrosion rate.

Answer to Problem 10E

The correlation between temperature and corrosion rate is 0.833.

Explanation of Solution

Calculation:

Correlation:

Software Procedure:

Step-by-step procedure to obtain the correlation using the MINITAB software:

• Select Stat > Basic Statistics > Correlation.
• In Variables, select Temperature and corrosion rate.
• Click OK.

Output using the MINITAB software is given below:

Thus, the correlation between temperature and corrosion rate is 0.833.

To determine

Find the regression sum of squares, the error sum of squares, and the total sum of squares.

Answer to Problem 10E

The regression sum of squares is 0.6122, the error sum of squares is 0.2709 and the total sum of squares is 0.8830.

Explanation of Solution

Calculation:

Step-by-step procedure to obtain the regression sum of squares, the error sum of squares, and the total sum of squares using the MINITAB software is given below:

• Choose Stat > Regression > Regression > Fit Regression Model.
• In Responses, enter “Corrosion”.
• In Continuous predictors, enter “Temperature”.
• Check Results.
• In Display of results, choose Simple tables.
• Click OK.

Output using the MINITAB software is given below:

From the output, the regression sum of squares is 0.6122, the error sum of squares is 0.2709 and the total sum of squares is 0.8830.

To determine

Identify the relationship between the quantity $\left(\frac{\text{Regression sum of squares}}{\text{Total sum of squares}}\right)$ and correlation coefficient.

Answer to Problem 10E

The quantity $\left(\frac{\text{Regression sum of squares}}{\text{Total sum of squares}}\right)$ is equal to the square of the correlation coefficient.

Explanation of Solution

Calculation:

The regression sum of squares divided by the total sum of squares is $\frac{0.6122}{0.8830}=0.6933$.

This value almost closer to the $r^2$. That is,

$\begin{array}{c}{r}^2=r\times r\\ =0.833\times 0.833\\ =0.6939\end{array}$

The relationship between the quantity $\left(\frac{\text{Regression sum of squares}}{\text{Total sum of squares}}\right)$ and correlation coefficient is,

$\begin{array}{c}\frac{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2-}{\displaystyle \sum _{i=1}^n{\left(y_i-{\widehat{y}}_i\right)}^2}}{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}=\frac{\cancel{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}}{\cancel{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}}-\frac{{\displaystyle \sum _{i=1}^n{\left(y_i-{\widehat{y}}_i\right)}^2}}{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}\\ =1-\frac{{\displaystyle \sum _{i=1}^n{\left(y_i-{\widehat{y}}_i\right)}^2}}{{\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}}\\ =1-\left(1-r^2\right)\\ =\cancel{1}\cancel{-1}+r^2\end{array}$

$=r^2$

Thus, the quantity $\left(\frac{\text{Regression sum of squares}}{\text{Total sum of squares}}\right)$ is equal to the square of the correlation coefficient.