91影视

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.

$\begin{array}{l}\text{罢丑别95\%肠辞苍蹿颈诲别苍肠别颈苍迟别谤惫补濒蹿辞谤渭颈蝉:}\\ \text{LowerLimit:Mean}鈭�\text{tcriticalvalue}脳\text{SD/sqrt}\left(\text{n}\right)\text{=1.881}鈭�\text{2.093}脳\text{0.524/sqrt}\left(\text{20}\right)\text{=1.6358}\\ \text{UpperLimit:Mean+tcriticalvalue}脳\text{SD/sqrt}\left(\text{n}\right)\text{=1.881+2.093*0.524/sqrt}\left(\text{20}\right)\text{=2.1262}\\ \text{罢丑别谤别蹿辞谤别,迟丑别95\%肠辞苍蹿颈诲别苍肠别颈苍迟别谤惫补濒颈蝉1.636\&濒迟;渭\&濒迟;2.126}\\ \text{尘别补苍辞蹿蠂1.881}\end{array}$

The $95\%$confidence interval of the$z$score will be given by $\mathrm{z}=\stackrel{炉}{\mathrm{X}}卤1.96脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}$

The total mean is given by:

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

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

Then

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

Then the variance will be given by:

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

Then the $z$score will be given by:

$\begin{array}{c}\mathrm{z}=\stackrel{炉}{\mathrm{X}}卤1.96脳\frac{\mathrm{蟽}}{\sqrt{\mathrm{n}}}\\ =1.8卤1.96脳\frac{0.55}{\sqrt{20}}\\ =1.8卤0.241\\ =2.041\end{array}$

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

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