91Ó°ÊÓ

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

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.

$\begin{array}{l}\text{°Õ³ó±ð95\%³¦´Ç²Ô´Ú¾±»å±ð²Ô³¦±ð¾±²Ô³Ù±ð°ù±¹²¹±ô´Ú´Ç°ùμ¾±²õ:}\\ \text{LowerLimit:Mean}−\text{tcriticalvalue}×\text{SD/sqrt}\left(\text{n}\right)\text{=1.881}−\text{2.093}×\text{0.524/sqrt}\left(\text{20}\right)\text{=1.6358}\\ \text{UpperLimit:Mean+tcriticalvalue}×\text{SD/sqrt}\left(\text{n}\right)\text{=1.881+2.093*0.524/sqrt}\left(\text{20}\right)\text{=2.1262}\\ \text{°Õ³ó±ð°ù±ð´Ú´Ç°ù±ð,³Ù³ó±ð95\%³¦´Ç²Ô´Ú¾±»å±ð²Ô³¦±ð¾±²Ô³Ù±ð°ù±¹²¹±ô¾±²õ1.636\&±ô³Ù;μ\&±ô³Ù;2.126}\\ \text{³¾±ð²¹²Ô´Ç´Úχ1.881}\end{array}$

The $95\%$confidence interval of the$z$score will be given by $\mathrm{z}=\stackrel{¯}{\mathrm{X}}Â\pm 1.96×\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}$

The total mean is given by:

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

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

Then

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

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:

$\begin{array}{c}\mathrm{z}=\stackrel{¯}{\mathrm{X}}Â\pm 1.96×\frac{\mathrm{σ}}{\sqrt{\mathrm{n}}}\\ =1.8Â\pm 1.96×\frac{0.55}{\sqrt{20}}\\ =1.8Â\pm 0.241\\ =2.041\end{array}$

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

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