Question

Surface roughness of pipe. Refer to the Anti-corrosion Methods and Materials (Vol. 50, 2003) study of the surface roughness of coated interior pipe used in oil fields, Exercise 2.46 (p. 96). The data (in micrometers) for sampled pipe sections are reproduced in the accompanying table; a Minitab analysis of the data appears below.

a.Locate a $\mathbf{95}\mathit{\%}$confidence interval for the mean surface roughness of coated interior pipe on the accompanying Minitab printout.

b.Would you expect the average surface roughness to beas high as $\mathbf{2}\mathbf{.}\mathbf{5}$micrometres? Explain.

Short answer

1. For confidence interval, the mean surface area is $1.8$.
2. The average surface roughness is lower than$2.5$ micrometers.

Step-by-step solution

Given information

Given that the data for $20$sampled pipe sections are reproduced.

(a) Find 95%a confidence interval for the mean surface roughness of coated interior pipe

The Minitab result is not provided here, as well as the variance of surface quality is unknown. The t-test is suitable under the normal test assumptions. This test is run in R, as well as the results are shown below to replicate the Minitab results. They obtain the following R output.

$\begin{array}{l}\text{The95\%confidenceintervalforμis:}\\ \text{LowerLimit:Mean}-\text{tcriticalvalue}\times \text{SD/sqrt}\left(\text{n}\right)\text{=1.881}-\text{2.093}\times \text{0.524/sqrt}\left(\text{20}\right)\text{=1.6358}\\ \text{UpperLimit:Mean+tcriticalvalue}\times \text{SD/sqrt}\left(\text{n}\right)\text{=1.881+2.093*0.524/sqrt}\left(\text{20}\right)\text{=2.1262}\\ \text{Therefore,the95\%confidenceintervalis1.636<μ<2.126}\\ \text{meanofχ1.881}\end{array}$

The $95\%$confidence interval of the$z$score will be given by $\mathrm{z}=\overline{\mathrm{X}}\pm 1.96\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}$

The total mean is given by:

$\begin{array}{c}\overline{\mathrm{X}}=\frac{\begin{array}{l}1.72+2.50+2.16+2.13+1.06+2.24+2.31+2.03+1.09+1.40+2.57+2.64+1.26+2.05\\ +1.19+2.13+1.27+1.51+2.41+1.95\end{array}}{20}\\ =1.8\end{array}$

$\text{Variance=}\mathrm{E}\left({\mathrm{x}}^2\right)-{\mathrm{E}}^2\left(\mathrm{x}\right)$

Then

$\begin{array}{c}E\left(x^2\right)=\frac{1}{20}\times \left[\begin{array}{l}{\left(1.72\right)}^2+{\left(2.50\right)}^2+{\left(2.16\right)}^2+{\left(2.13\right)}^2+{\left(1.06\right)}^2+{\left(2.24\right)}^2+\left(2.31\right){}^2+{\left(2.03\right)}^2\\ +{\left(1.09\right)}^2+{\left(1.40\right)}^2+{\left(2.57\right)}^2+{\left(2.64\right)}^2+{\left(1.26\right)}^2+{\left(2.05\right)}^2+{\left(1.19\right)}^2+{\left(2.13\right)}^2\\ +{\left(1.27\right)}^2+{\left(1.51\right)}^2+{\left(2.41\right)}^2+{\left(1.95\right)}^2\end{array}\right]\\ =3.79\end{array}$

Then the variance will be given by:

$\begin{array}{c}\mathrm{Variance}=\mathrm{E}\left({\mathrm{x}}^2\right)-{\mathrm{E}}^2\left(\mathrm{x}\right)\\ =3.79-{\left(1.8\right)}^2\\ =3.79-3.24\\ =0.55\end{array}$

Then the $z$score will be given by:

$\begin{array}{c}\mathrm{z}=\overline{\mathrm{X}}\pm 1.96\times \frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\\ =1.8\pm 1.96\times \frac{0.55}{\sqrt{20}}\\ =1.8\pm 0.241\\ =2.041\end{array}$

$\begin{array}{c}\mathrm{z}=1.8-0.241\\ =1.559\end{array}$

(b) Find the average surface roughness

The average surface roughness is $1.8$ , and it is lower than$2.5$ micrometers.