Question

Problem 6s.11ac Question Help Refer to Table $6.1 - Factors for Computing Control Chart Limits (3 sigma) for this problem. Twelve samples, each containing five parts, were taken from a process that produces steel rods at Emmanual Kodzi's factory. The length of each rod in the samples was determined. The results were tabulated and sample means and ranges were computed. The results were: Sample Sample Mean Range qe (in.) (in.) 9.402 0.033 9.404 0.041 9.391 0.034 9.408 0.051 9.399 0.031 9.401 0.036 Sample Sample Mean (in.) Range (in.) 9.403 0.041 9.403 0.034 9.393 0.027 0.029 9.401 0.039 9.406 0.047 401 For the given data, the x inches (round your response to four decimal places).

Answer 1

$\bar {\bar x}$ is the mean of all the sample means. Hence we have $\bar {\bar x}$ as below

$\bar {\bar x}$ = (9.402+9.404+9.391+9.408+9.399+9.401+9.403+9.403+9.393+9.401+9.401+9.406)/12 = 112.812/12 = 9.401

Hence $\bar {\bar x}$ = 9.401 inches

Also, ${\bar R}$ is the mean of all the sample ranges. Hence ${\bar R}$ is as below

${\bar R}$ = (0.033+0.041+0.034+0.051+0.031+0.036+0.041+0.034+0.027+0.029+0.039+0.047)/12 = 0.443/12 = 0.0369

Hence ${\bar R}$ is 0.0369 inches

Now for the x-bar chart the control limits are given as per below formula

Control Limits = ($\bar {\bar x}$) $\small \pm$ A₂*(${\bar R}$)

It is given that the sample size is 5. From the given table, for sample size of 5, A₂ is 0.577

Hence we have the below control limits

UCL_x = 9.401 + 0.577*0.0369 = 9.401 + 0.0213 = 9.4223 inches

LCL_x = 9.401 - 0.577*0.0369 = 9.401 - 0.0213 = 9.3797 inches

We can see that all the sample means lie within the control limits. Hence we have the below

Based on x-chart, is one or more samples beyond the control limits : No

For the given data, ${\bar R}$ is 0.0369 as calculated above

The control limits for R-chart are calculated below

LCL = D₃*${\bar R}$

UCL = D₄*${\bar R}$

From the given table for sample size of 5, D₃ = 0 and D₄ = 2.115

Hence we have the control limits as below

LCL = 0*0.0369 = 0 inches

UCL = 2.115*0.0369 = 0.0780 inches

Hence UCL_R = 0.0780 inches and LCL_R = 0 inches

We can see that all the sample ranges lie within the control limits. Hence we have the below

Based on R-chart, is one or more samples beyond the control limits : No

Answer 2

$\bar {\bar x}$ is the mean of all the sample means. Hence we have $\bar {\bar x}$ as below

$\bar {\bar x}$ = (9.402+9.404+9.391+9.408+9.399+9.401+9.403+9.403+9.393+9.401+9.401+9.406)/12 = 112.812/12 = 9.401

Hence $\bar {\bar x}$ = 9.401 inches

Also, ${\bar R}$ is the mean of all the sample ranges. Hence ${\bar R}$ is as below

${\bar R}$ = (0.033+0.041+0.034+0.051+0.031+0.036+0.041+0.034+0.027+0.029+0.039+0.047)/12 = 0.443/12 = 0.0369

Hence ${\bar R}$ is 0.0369 inches

Now for the x-bar chart the control limits are given as per below formula

Control Limits = ($\bar {\bar x}$) $\small \pm$ A₂*(${\bar R}$)

It is given that the sample size is 5. From the given table, for sample size of 5, A₂ is 0.577

Hence we have the below control limits

UCL_x = 9.401 + 0.577*0.0369 = 9.401 + 0.0213 = 9.4223 inches

LCL_x = 9.401 - 0.577*0.0369 = 9.401 - 0.0213 = 9.3797 inches

We can see that all the sample means lie within the control limits. Hence we have the below

Based on x-chart, is one or more samples beyond the control limits : No

For the given data, ${\bar R}$ is 0.0369 as calculated above

The control limits for R-chart are calculated below

LCL = D₃*${\bar R}$

UCL = D₄*${\bar R}$

From the given table for sample size of 5, D₃ = 0 and D₄ = 2.115

Hence we have the control limits as below

LCL = 0*0.0369 = 0 inches

UCL = 2.115*0.0369 = 0.0780 inches

Hence UCL_R = 0.0780 inches and LCL_R = 0 inches

We can see that all the sample ranges lie within the control limits. Hence we have the below

Based on R-chart, is one or more samples beyond the control limits : No

Answer 3

$\bar {\bar x}$ is the mean of all the sample means. Hence we have $\bar {\bar x}$ as below

$\bar {\bar x}$ = (9.402+9.404+9.391+9.408+9.399+9.401+9.403+9.403+9.393+9.401+9.401+9.406)/12 = 112.812/12 = 9.401

Hence $\bar {\bar x}$ = 9.401 inches

Also, ${\bar R}$ is the mean of all the sample ranges. Hence ${\bar R}$ is as below

${\bar R}$ = (0.033+0.041+0.034+0.051+0.031+0.036+0.041+0.034+0.027+0.029+0.039+0.047)/12 = 0.443/12 = 0.0369

Hence ${\bar R}$ is 0.0369 inches

Now for the x-bar chart the control limits are given as per below formula

Control Limits = ($\bar {\bar x}$) $\small \pm$ A₂*(${\bar R}$)

It is given that the sample size is 5. From the given table, for sample size of 5, A₂ is 0.577

Hence we have the below control limits

UCL_x = 9.401 + 0.577*0.0369 = 9.401 + 0.0213 = 9.4223 inches

LCL_x = 9.401 - 0.577*0.0369 = 9.401 - 0.0213 = 9.3797 inches

We can see that all the sample means lie within the control limits. Hence we have the below

Based on x-chart, is one or more samples beyond the control limits : No

For the given data, ${\bar R}$ is 0.0369 as calculated above

The control limits for R-chart are calculated below

LCL = D₃*${\bar R}$

UCL = D₄*${\bar R}$

From the given table for sample size of 5, D₃ = 0 and D₄ = 2.115

Hence we have the control limits as below

LCL = 0*0.0369 = 0 inches

UCL = 2.115*0.0369 = 0.0780 inches

Hence UCL_R = 0.0780 inches and LCL_R = 0 inches

We can see that all the sample ranges lie within the control limits. Hence we have the below

Based on R-chart, is one or more samples beyond the control limits : No