Question

Based on the above information, answer the following questions. First, find the Center Line and the Control Limits for the x-bar chart. Second, find the Center Line and the Control Limits for the R-chart. Third, use the equation that sigma =((widehat(R)))/(d_(2)), estimate the process standard deviation ( sigma and calculate the Process Capability Indexes C_( ho ) and C_(pl) Lastly, if the process mean has shifted to 65.07711 lexactly 3sigma _(x) above the original Center Line), what is the associated Type-II Error ( eta ]? Select 4 correct arnuerty] R-Chart C.L. 0.1163 LCL =0.239561, LCL =0.0 R-Chart C.L =0.1163,LCL=0.245958,LCL=0.0 R-Chart CL. =0.1163,LCL=0.251855,LCL=0.0 C_(p)=1.500,C_(pk)=1.500 C_(p)=2.000,C_(pk)=1.933 c_(p)=2.000,c_(pk)=2.000 C_(p)=1.800,C_(pk)=1.333 Type-II Error(mean-65.07711) 0.9 .9730 Type-II Error(mean-65.07711) =0.95720 Type-II Error(mean-65.07711) =0.68268 Type-II Error(mean-65.07711) =0.50000 Piston rings for an automotive engine (4.BL) are produced by a forging process. The specification limits are given as 650.3 (mm). We wish to establish statistical control of the inside diameter of the rings manufactured by this process using the x-bar and R charts. Twenty five (25) samples, each of size five (n5), have been taken when we thick the process is in control. The summary of the inside diameter measurement data from these samples are : 25x=1625.25 1 25R=2.9075 Based on the above information, answer the following questions. First, find the Center Line and the Control Limits for the x-bar chart Second, find the Center Line and the Control Limits for the R-chart. R Third, use the equation that , estimate the process standard deviation ( Lastly, if the process mean has shifted to 65.07711 (exactly 3 above the original Center Line), what is the associated Type-II Error ()? Select 4 correct arawer(s) X-Bar Chart C.L.=65.01 UCL65.07711 LCL=64.94289 X-Bar Chart C.L.=65.01 UCL=65.11711 LCL=64.90289 X-Bar Chart C.L.65.01 UCL=65.17011 LCL=64.84989 X-Bar Chart C.L=65.01 UCL=65.09478 LCL=64.92522 R-Chart C.L.=0.1163 LCL=0.233065 LCL=0.0 R-Chart C.L.=0.1163 LCL=0.239561 LCL=0.0 R-Chart C.L.=0.1163 LCL=0.245858 LCL=0.0 R-Chart C.L.=0.1163 LCL=0.251855 LCL=0.0 C1.500 Cpt = 1.500 2000 C1.933 2000 Cpt2000 1.000 C1.333 Type-II Eror(mean=65.077110.99730 Type-II Error(mean=65.07711) = 0.95720 Type-Il Errer(mean=65.07711) = 0.68268 Type-Il Er(mean=65.07711) 0.50000

          Based on the above information, answer the following questions.
First, find the Center Line and the Control Limits for the x-bar chart.
Second, find the Center Line and the Control Limits for the R-chart.
Third, use the equation that sigma =((widehat(R)))/(d_(2)), estimate the process standard deviation ( sigma 
and calculate the Process Capability Indexes C_(
ho ) and C_(pl)
Lastly, if the process mean has shifted to 65.07711 lexactly 3sigma _(x) above the original
Center Line), what is the associated Type-II Error ( eta  ]?
Select 4 correct arnuerty]
R-Chart C.L. 0.1163 LCL =0.239561, LCL =0.0
R-Chart C.L =0.1163,LCL=0.245958,LCL=0.0
R-Chart CL. =0.1163,LCL=0.251855,LCL=0.0
C_(p)=1.500,C_(pk)=1.500
C_(p)=2.000,C_(pk)=1.933
c_(p)=2.000,c_(pk)=2.000
C_(p)=1.800,C_(pk)=1.333
Type-II Error(mean-65.07711) 0.9 .9730
Type-II Error(mean-65.07711) =0.95720
Type-II Error(mean-65.07711) =0.68268
Type-II Error(mean-65.07711) =0.50000
Piston rings for an automotive engine (4.BL) are produced by a forging process. The specification limits are given as 650.3 (mm). We wish to establish statistical control of the inside diameter of the rings manufactured by this process using the x-bar and R charts. Twenty five (25) samples, each of size five (n5), have been taken when we thick the process is in control. The summary of the inside diameter measurement data from these samples are :
25x=1625.25 1 25R=2.9075 Based on the above information, answer the following questions.
First, find the Center Line and the Control Limits for the x-bar chart
Second, find the Center Line and the Control Limits for the R-chart.
R Third, use the equation that   , estimate the process standard deviation (
Lastly, if the process mean has shifted to 65.07711 (exactly 3 above the original Center Line), what is the associated Type-II Error ()?
Select 4 correct arawer(s)  X-Bar Chart C.L.=65.01 UCL65.07711 LCL=64.94289  X-Bar Chart C.L.=65.01 UCL=65.11711 LCL=64.90289  X-Bar Chart C.L.65.01 UCL=65.17011 LCL=64.84989 X-Bar Chart C.L=65.01 UCL=65.09478 LCL=64.92522 R-Chart C.L.=0.1163 LCL=0.233065 LCL=0.0 R-Chart C.L.=0.1163 LCL=0.239561 LCL=0.0 R-Chart C.L.=0.1163 LCL=0.245858 LCL=0.0 R-Chart C.L.=0.1163 LCL=0.251855 LCL=0.0 C1.500 Cpt = 1.500 2000 C1.933 2000 Cpt2000 1.000 C1.333 Type-II Eror(mean=65.077110.99730  Type-II Error(mean=65.07711) = 0.95720 Type-Il Errer(mean=65.07711) = 0.68268 Type-Il Er(mean=65.07711) 0.50000

based on the above information answer the following questions first find the center line and the control limits for the x bar chart second find the center line and the control limits for the 42032

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