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Let\({d_0} = {e_0} = 0\). Use formulas

\(\begin{aligned}{*{20}{l}}{\left. {{d_l} = \max \left( {0,{d_{l - 1}} + \left( {{{\bar x}_l} - \left( {{\mu _0} + k} \right)} \right)} \right)} \right),}&{l = 1,2, \ldots }\\{\left. {{e_l} = \max \left( {0,{e_{l - 1}} - \left( {{{\bar x}_l} - \left( {{\mu _0} - k} \right)} \right)} \right)} \right),}&{l = 1,2, \ldots }\end{aligned}\)

to compute\({d_1},{d_2}, \ldots \), and\({e_1},{e_2}, \ldots \), where\(k\)is the slope of the lower arm of the\(V\)-mask - its value is taken as\(\Delta /2\)(\(\Delta \)is size of a shift in\(\mu \)on which the attention is). If in a time\(r\), both\({d_r}\)and\({e_r}\)are larger than\(h\), the process is out of control. Inequality\({d_r} > h\)means that the process mean is greater than the target value, and inequality\({e_r} > h\)means that the process mean is smaller than the target value.

As one can notice,\({\mu _0} = 0.75\), value\(k\)is

\(k = \frac{\Delta }{2} = 0.001\)

and\(h = 0.003\). Using the formula above, one can compute\({d_i},e_i^\prime \)s for every\(i\)as

\(\begin{aligned}{*{20}{l}}{\left. {{d_l} = \max \left( {0,{d_{l - 1}} + \left( {{{\bar x}_l} - 0.751} \right)} \right)} \right),}&{l = 1,2, \ldots }\\{\left. {{e_l} = \max \left( {0,{e_{l - 1}} - \left( {{{\bar x}_l} - 0.749} \right)} \right)} \right),\;\;\;l = 1,2, \ldots }&{}\end{aligned}\)

The following tables represents all values from which one can see if the process is in-control or out-of-control.

\(\begin{aligned}{*{20}{c}}i&{{{\bar x}_i} - 16.05}&{{d_i}}\\1&{ - 0.0003}&0\\2&{ - 0.0006}&0\\3&{ - 0.0018}&0\\4&{ - 0.0009}&0\\5&{ - 0.0007}&0\\6&0&0\\7&{ - 0.002}&0\\8&{ - 0.0013}&0\\9&{ - 0.0022}&0\\{10}&{ - 0.0006}&0\\{11}&{0.0006}&{0.0006}\\{12}&{ - 0.0038}&0\\{13}&{ - 0.0021}&0\\{14}&{ - 0.0027}&0\\{15}&{ - 0.0039}&0\\{16}&{ - 0.0012}&0\\{17}&{ - 0.005}&0\\{18}&{ - 0.0028}&0\\{19}&{ - 0.004}&0\\{20}&{ - 0.0017}&0\\{21}&{ - 0.0048}&0\\{22}&{ - 0.0029}&0\\{}&{}&{}\end{aligned}\)

\(\begin{aligned}{*{20}{c}}i&{{{\bar x}_i} - 15.95}&{{e_i}}\\1&{0.0017}&0\\2&{0.0014}&0\\3&{0.0002}&0\\4&{0.0011}&0\\5&{0.0013}&0\\6&{0.0020}&{}\\7&0&0\\8&{0.0007}&0\\9&{ - 0.0002}&{0.0002}\\{10}&{0.0014}&0\\{11}&{0.0026}&0\\{12}&{ - 0.0018}&{0.0018}\\{13}&{ - 0.0001}&{0.0019}\\{14}&{ - 0.0007}&{0.0026}\\{15}&{ - 0.0019}&{0.0045}\\{16}&{0.0008}&{0.0037}\\{17}&{ - 0.003}&{0.0067}\\{18}&{ - 0.0008}&{0.0075}\\{19}&{ - 0.002}&{0.0095}\\{20}&{0.0003}&{0.0092}\\{21}&{ - 0.0028}&{0.0120}\\{22}&{ - 0.0009}&{0.0129}\\{}&{}&{}\end{aligned}\)

For \(r = 15{e_r} = 0.0045 > h = 0.003\); thus, the process mean has shifted to a value which is smaller than the target \(\mu = 0.75\).