Question

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.

Short answer

Answers:

1. Therefore, the least squares prediction equation is$\hat{\mathbf{y}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{265955}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{007914}\mathit{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{00000425}{\mathit{x}}^{\mathbf{2}}$
2. At 99% significance level, it can be concluded that${\mathit{\beta }}_{\mathbf{1}}\mathbf{\ne }{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }\mathbf{0}$
3. The value of${{\mathit{R}}_{\mathbf{a}}}^{\mathbf{2}}$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.
4. The null hypothesis is whether ${\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$while the alternate checks if the value of${\mathit{\beta }}_{\mathbf{2}}\mathbf{<}\mathbf{0}$ .
5. Mathematically,${\mathit{H}}_{\mathbf{0}}\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$ ,while ${\mathit{H}}_a\mathbf{}\mathbf{:}{\mathit{\beta }}_{\mathbf{2}}\mathbf{<}\mathbf{0}$.
6. At 99% confidence level,${\mathit{\beta }}_{\mathbf{2}}\mathbf{=}\mathbf{0}$.This means that the parabola doesn’t have a curvature and it essentially is a straight line.
7. 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)

Step-by-step solution

Least Squares prediction equation

Ratio of repair to replacement cost (y)	Pipe diameter (x)	${\mathit{x}}^{\mathbf{2}}$
6.58	80	6400
6.97	100	10000
7.39	125	15625
7.61	150	22500
7.78	200	40000
7.92	250	62500
8.2	300	90000
8.42	350	122500
8.6	400	160000
8.97	450	202500
9.31	500	250000
9.47	600	360000
9.72	700	490000

Excel summary output

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.988335
R Square	0.976805
Adjusted R Square	0.972166
Standard Error	0.162211
Observations	13
ANOVA
	df	SS	MS	F	Significance F
Regression	2	11.08098	5.540491	210.5646	6.71E-09
Residual	10	0.263125	0.026313
Total	12	11.34411
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	6.265955	0.155438	40.31149	2.11E-12	5.919617	6.612294	5.919617	6.612294
Pipe diameter (x)	0.007914	0.000997	7.934744	1.26E-05	0.005692	0.010137	0.005692	0.010137
$x^2$	-4.3E-06	1.32E-06	-3.22867	0.009041	-7.2E-06	-1.3E-06	-7.2E-06	-1.3E-06

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

Overall goodness of fit

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

${\mathit{H}}_{\mathbf{a}}\mathbf{}\mathbf{:}$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, ${\mathit{\beta }}_1\mathbf{\ne }{\mathit{\beta }}_{\mathbf{2}}\mathbf{\ne }\mathbf{0}$

Interpretation of Ra2

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.

Null and alternate hypothesis

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

Significance of β2

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

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

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.

Prediction value

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).