Question: Estimating repair and replacement costs of water pipes. Refer to the IHS Journal of Hydraulic Engineering (September, 2012) study of the repair and replacement of water pipes, Exercise 11.21 (p. 655). Recall that a team of civil engineers used regression analysis to model y = the ratio of repair to replacement cost of commercial pipe as a function of x = the diameter (in millimeters) of the pipe. Data for a sample of 13 different pipe sizes are reproduced in the accompanying table. In Exercise 11.21, you fit a straight-line model to the data. Now consider the quadratic model,$\mathit{E}\mathbf{\left(}\mathit{y}\mathbf{\right)}\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}$. A Minitab printout of the analysis follows (next column).

1. Give the least squares prediction equation relating ratio of repair to replacement cost (y) to pipe diameter (x).
2. Conduct a global F-test for the model using$\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$. What do you conclude about overall model adequacy?
3. Evaluate the adjusted coefficient of determination,${{\mathit{R}}_{\mathbf{a}}}^{\mathbf{2}}$, for the model.
4. Give the null and alternative hypotheses for testing if the rate of increase of ratio (y) with diameter (x) is slower for larger pipe sizes.
5. Carry out the test, part d, using $\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$.
6. Locate, on the printout, a 95% prediction interval for the ratio of repair to replacement cost for a pipe with a diameter of 240 millimeters. Interpret the result.

Excel summary output

Therefore, the least-squares prediction equation is$\hat{\mathbf{y}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{265955}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{007914}\mathit{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{00000425}{\mathit{x}}^{\mathbf{2}}$

${\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_1\mathbf{=}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

${\mathit{H}}_{\mathbf{a}}\mathbf{}\mathbf{:}$At least one of the parameters${\mathit{\beta }}_1$ or${\mathit{\beta }}_{\mathbf{2}}$ is non zero

Here, F test statistic $\mathbf{=}\frac{\mathbf{S}\mathbf{S}\mathbf{E}}{\mathbf{n}\mathbf{-}\mathbf{k}\mathbf{+}\mathbf{1}}\mathbf{=}\mathbf{210}\mathbf{.}\mathbf{5646}$.Here, p-value = 6.71E-09

${\mathit{H}}_{\mathbf{0}}\mathbf{}$is rejected if p-value < $\mathit{\alpha }$. For$\mathit{\alpha }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$ , Since p-value <0.01

Sufficient evidence to reject${\mathit{H}}_{\mathbf{0}}\mathbf{}$ at 99% confidence interval.

Hence, ${\mathit{\beta }}_1\mathbf{\ne }{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }\mathbf{0}$

The value of ${R_a}^2$in the excel output is 0.972166. This means that around 97% of the variation in the variables is explained by the model. The higher the value, the better the model is for fitting to the data. 97% indicates the model is an ideal fit for the data.

To test whether the rate of increase of ratio (y) with a diameter (x) is slower for larger diameter pipe sizes, we want to check whether the curvature of the parabola is the negative meaning if the parabola is downward sloping.

The null hypothesis is whether${\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$ while the alternate checks if the value of${\mathit{\beta }}_{\mathbf{2}}\mathbf{<}\mathbf{0}$ .

Mathematically, ${\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$while${\mathit{H}}_a\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{<}\mathbf{0}$ .

${\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$

${\mathit{H}}_a\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{<}\mathbf{0}$

Here, t-test statistic$\mathbf{=}\frac{\widehat{{\mathbf{\beta }}_{\mathbf{2}}}}{\mathbf{s}{\mathbf{\beta }}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{000004}}{\mathbf{0}\mathbf{.}\mathbf{000001}}\mathbf{=}\mathbf{-}\mathbf{4}$

Value of${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{12}}$ is 2.681

${\mathbf{H}}_{\mathbf{0}}$is rejected if t statistic > ${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{12}}$. For$\alpha =0.01$ , since t <${\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{12}}$

Not sufficient evidence to reject ${\mathbf{H}}_{\mathbf{0}}$at a 99% confidence interval.

Thus,${\mathbf{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$.

This means that the parabola doesn’t have a curvature and it essentially is a straight line.

The 95% prediction interval for the ratio of repair to replacement cost for a pipe with a diameter of 250 mm is given as (7.59472, 8.36237). This means that the ratio of repair to replacement can be between 7.59472 and 8.36237 for a given value of the diameter of the pipe (here 240 mm).