Sampling 4 pieces of precision-cut wire (to be used in computer assembly) every hour for the past 24 hours has produced the following results:Hour Xbar R Hour Xbar R Hour Xbar R Hour Xbar R1 3.15 0.76 7 3.15 0.53 13 3.21 0.80 19 3.31 1.662 3.10 1.13 8 2.65 1.08 14 2.83 1.31 20 2.99 1.143 3.12 1.48 9 2.92 0.76 15 3.12 1.01 21 2.65 1.034 3.49 1.31 10 2.85 1.28 16 2.74 0.55 22 3.28 0.415 3.17 1.22 11 2.93 1.17 17 2.86 1.43 23 2.94 1.636 2.86 0.37 12 3.07 0.35 18 2.84 1.34 24 2.54 0.92Based on the sampling done, the control limits for 3-sigma x overbarx chart are (round all intermediate calculations to three decimal places before proceeding with furthercalculations):Upper Control Limit (UCL Subscript x overbarUCLx) = ?Lower Control Limit (LCL Subscript x overbarLCLx) = ?Based on the x overbarx-chart, the wire cutting process has beenIN CONTROL or OUT OF CONTROL?The control limits for the 3-sigma R-chart are (round all intermediate calculations to three decimal places before proceeding with further calculations):Upper Control Limit (UCL Subscript Upper RUCLR) = ?Lower Control Limit (LCL Subscript Upper RLCLR) = ?

[tex]\text{UCL}_{X} = X + A_{2} \times R = 2.990 + 0.729 \times 1.028 = 3.739\\\\LCL_{X} = X - A_{2} \times R = 2.990 - 0.729 \times 1.028 = 2.240[/tex]

Control Limits, X-bar Chart

LCL UCL Mean

3.739 2.240 2.990

Since, all the points are within the UCLₓ and LCLₓ control Limits, the process is said to be in control.

(B)

R Chart:

Compute the Upper Control Limit and Lower Control Limit for the R-chart as follows:

[tex]UCL_{R} = D_{4} \times R = 2.282 \times 1.028 = 2.345\\\\LCL_{R} = D_{3}\times R = 0 \times 1.028 = 0[/tex]

Control Limits, R Chart

LCL UCL Mean

0 2.345 1.028

Omm2 Omm2 · Answer 2 · 2022-03-14T08:05:55+01:00

So, the required values are,

[tex]UCL_{\bar{x}}=3.7\\LCL_{\bar{x}}=2.23\\CL_{\bar{x}}=2.96[/tex]

And,

[tex]UCL_R=2.32\\LCL_R=0\\CL_R=1.0175[/tex]

Assembling problem:

The problems are studied by the use of statistical methods economic loss is based on the study of collected statistical data.

It is given that,

[tex]n=4\\k=24\\\sum \bar{x_i}=71.27\\\sum R_i=24.42[/tex]

As we know,

[tex]\bar{\bar{a}}=\frac{\sum\bar{x_i}}{k}\\=\frac{71.27}{24}\\=2.9695833\\\bar{R}=\frac{\sum R_i}{k}\\=\frac{24.42}{24}\\=1.0175[/tex]

From the table for [tex]n=4[/tex]

Then,[tex]\bar{x}-chart:[/tex]

[tex]UCL_{\bar{x}}=\bar{\bar{x}}+A_2\bar{R}\\=2.96958333+0.729(1.0175)\\=3.71134083\\\\LCL_{\bar{x}}=\bar{\bar{x}}-A_2\bar{R}\\\\=2.9695833-0.729(1.0175)\\=2.22782583\\CL_{\bar{x}}=\bar{\bar{x}}\\[/tex]

Now, calculating R-chart

[tex]UCL_R=D_4\bar{R}\\=2.282(1.0175)\\=2.321935\\LCL_R=D_3\bar{R}\\=0(1.0175)\\=0\\CL_R=\bar{R}\\=1.0175[/tex]

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