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STATISTICAL PROCESS CONTROL

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\[\hspace{2cm}\ \text{Consider the 4 samples of each size 5 taken}\ \hspace{15cm}\\ \text{from the production lot of a machine}\ \hspace{17cm}\]
Sample
number
Si1Si2Sii3Si4Si5
S16.356.406.326.376.42
S26.346.406.346.366.39
S36.426.646.356.596.72
S46.386.486.446.586.39
\[1.\ \text{To calculate the sample means}\ \bar{S_1}, \bar{S_2}, \bar{S_3}, \bar{S_4}\ and\ \hspace{16cm}\\ \text{the mean of the sample means}\ \bar{S}\ =\ \frac{\bar{S_1}\ +\ \bar{S_2}\ +\ \bar{S_3}\ +\ \bar{S_4}}{4}\ \hspace{10cm}\]
\[2.\ \text{To calculate the sample variances}\ \hspace{19cm}\\ v_1,\ v_2,\ v_3,\ v_4\ and\ \sigma\ =\ \sqrt{\frac{1}{4}\ \Sigma_{i=1}^{4} v_i}\ \hspace{10cm}\]
\[3.\ \text{To determine the central line CL =}\ \bar{S},\ \hspace{20cm}\\ \text{lower control limit LCL =}\ \bar{S}\ -\ \frac{2.58}{\sqrt{5}}\ \sigma\ \hspace{19cm}\\ \text{and Upper control limit UCL}\ =\ \bar{S}\ +\ \frac{2.58}{\sqrt{5}}\ \sigma \hspace{18cm}\]
\[4.\ \text{To draw the}\ \bar{X}\ chart\ and\ determine\ \hspace{18cm}\\ \text{out-of-control signals.}\ \hspace{15cm}\]
\[\color{green}{Step\ 1:}\ \text{Open Geogebra classic 5 (by double clicking on the icon)}\ \hspace{18cm}\]
\[\color{green}{Step\ 2:}\ \text{Open spread sheet view}\ \hspace{19cm}\]
\[\color{green}{Step\ 3:}\ \text{type the data by using the spread sheet cells}\ S_1, S_2, S_3, and S_4\ \hspace{15cm}\\ \text{in columns and A, B, C and D as rows of the spreadsheet.}\ \hspace{10cm}\]
\[\color{green}{Step\ 4:}\ \text{To Calculate the mean of}\ S_1\ by\ using\ the\ \hspace{15cm}\\ \text{input bar to type}\ S_{1m}\ =\ mean(A1:A5)\ \hspace{14cm}\]
\[\color{green}{Step\ 5:}\ \text{Similarly Calculate the mean of}\ S_2\ by\ using\ the\ \hspace{15cm}\\ \text{input bar to type}\ S_{2m}\ =\ mean(B1:B5)\ \hspace{14cm}\]
\[\color{green}{Step\ 6:}\ \text{Similarly Calculate the mean of}\ S_3\ by\ using\ the\ \hspace{15cm}\\ \text{input bar to type}\ S_{3m}\ =\ mean(C1:C5)\ \hspace{14cm}\]
\[\color{green}{Step\ 7:}\ \text{Similarly Calculate the mean of}\ S_4\ by\ using\ the\ \hspace{15cm}\\ \text{input bar to type}\ S_{4m}\ =\ mean(D1:D5)\ \hspace{14cm}\]
\[\color{green}{Step\ 8:}\ \text{To calculate the mean of the sample means}\ \hspace{15cm}\\ \text{by using the input bar to type}\ \hspace{15cm}\\ \text{S_m=mean(S_{1m},S_{2m}, S_{3m}, S_{4m})}\ \hspace{10cm}\]
\[\color{green}{Step\ 9:}\ \text{To Calculate the variance of}\ S_1\ by\ using\ the\ \hspace{15cm}\\ \text{input bar to type}\ V_1\ =\ Variance(A1:A5)\ \hspace{14cm}\]
\[\color{green}{Step\ 10:}\ \text{Similarly calculate the variance of}\ S_2,\ S_3,\ S_4\ by\ using\ the\ \hspace{15cm}\\ \text{input bar to type}\ v_2\ =\ Variance(B1:B5)\ \hspace{5cm}\\ v_3\ =\ Variance(C1:C5)\\ v_4\ =\ Variance(D1:D5)\]
\[\color{green}{Step\ 11:}\ \text{Calculate}\ \sigma\ by\ using\ the\ \hspace{18cm}\\ \text{input bar to type σ=sqrt((1/4(v_1+v_2+v_3+v_4)))}\ \hspace{10cm}\]
\[\color{green}{Step\ 12:}\ \text{To draw the Central Line by using the}\ \hspace{18cm}\\ \text{input bar to type CL:y=S_m}\ \hspace{10cm}\]
\[\color{green}{Step\ 13:}\ \text{To draw the Upper Control Limit (UCL) line by using the}\ \hspace{18cm}\\ \text{input bar to type UCL:y=S_m + (2.58/sqrt(5))σ}\ \hspace{10cm}\]
\[\color{green}{Step\ 14:}\ \text{To draw the Lower Control Limit (LCL) line by using the}\ \hspace{18cm}\\ \text{input bar to type LCL:y=S_m – (2.58/sqrt(5))σ}\ \hspace{10cm}\]
\[\color{green}{Step\ 15:}\ \text{To plot the sample mean against x – values 2,4,6,8}\ \hspace{17cm}\\ \text{by using the input bar to type one by one A:(2,S_{1m}), B:(4,S_{2m}), C:(6,S_{3m}), A:(8,S_{4m})}\ \hspace{5cm}\]
\[\color{green}{Step\ 16:}\ \text{To joint the points A, B, C and D by a poly line}\ \hspace{18cm}\\ \text{using input bar to type Polyline(A,B,C,D)}\ \hspace{10cm}\]
\[\color{green}{Step\ 17:}\ \text{Decision Making Process}\ \hspace{22cm}\]
\[i.\ \hspace{1cm}\ \text{Determine any of the points A,B.C,D lie beyond the control limits}\ \hspace{15cm}\\ \text{LCL and UCL. If any point lies beyond the LCL and UCL, it indicates the}\ \hspace{14cm}\\ \text{out-of-control signal. Record the out-of-control sample(s)}\ \hspace{13cm}\]
\[ii.\ \hspace{1cm}\ \text{If all the points A,B.C,D lie within the control limits LCL and UC,}\ \hspace{15cm}\\ \text{there is no out-of-control signal. The process is control}\ \hspace{12cm}\]

Output for sample mean and sample variance

Sample\[Sample\ mean\ (\bar{S_i})\]\[Sample\ variance\ (\bar{v_i})\]
S1\[\bar{S_1}\]= 6.37v1=0
S2\[\bar{S_2}\]=6.37v2 =0
S3\[\bar{S_3}\]=6.54v3=0.02
S4\[\bar{S_4}\]=6.45v4=0.01

Output for sample mean and sample variance

\[\textbf{Output for}\ \bar{S},\ \sigma\, CL,\ LCL\ and\ UCL\]
Value / Equation
\[\bar{S}\]6.43
σ0.08
CLy = 6.43
LCL\[y\ =\ 6.43\ – \frac{2.58}{\sqrt{5}}\ (0.08)\]
UCL\[y\ =\ 6.43\ + \frac{2.58}{\sqrt{5}}\ (0.08)\]
\[\textbf{Decision making on the process}\]
Is there any out-of-control signalsOne
If “yes”, mention the sample which shows out- of-control signal.C =(6,6.54)
Is the process under control?No
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