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Excel summary output

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

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

${\mathit{H}}_{\mathbf{a}}\mathbf{}\mathbf{:}$At least one of the parameters${\mathit{尾}}_1$ or${\mathit{尾}}_{\mathbf{2}}$ 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 < $\mathit{伪}$. For$\mathit{伪}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{01}$ , Since p-value <0.01

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

Hence, ${\mathit{尾}}_1\mathbf{鈮�}{\mathit{尾}}_{\mathbf{2}}\mathbf{鈮�}\mathbf{0}$

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${\mathit{尾}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$ while the alternate checks if the value of${\mathit{尾}}_{\mathbf{2}}\mathbf{<}\mathbf{0}$ .

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

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

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

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

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$伪=0.01$ , 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,${\mathbf{尾}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$.

This means that the parabola doesn鈥檛 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).