Let's break down the solution as requested: --- ### Given Data - **Target value of molecular weight (\(\mu_0\))**: 1050 - **Standard deviation (\(\sigma\))**: 25 - **Data (\(x\))**: 20 individual observations (see table) - **n**: 1 (individual observations) --- ## (a) **Tabular CUSUM Chart for the Mean** **Step 1: Select parameters \(k\) and \(h\)** - \(k\) (reference value): Typically \(k = 0.5 \sigma\) for detecting a shift of 1 \(\sigma\) - \(k = 0.5 \times 25 = 12.5\) - \(h\) (decision interval): Typically \(h = 5 \sigma\) - \(h = 5 \times 25 = 125\) (you can adjust this depending on desired ARL) **Step 2: Calculate CUSUM statistics** For each observation \(x_i\): - \(C^+_0 = 0\), \(C^-_0 = 0\) - \(C^+_i = \max[0, C^+_{i-1} + (x_i - \mu_0) - k]\) - \(C^-_i = \max[0, C^-_{i-1} - (x_i - \mu_0) - k]\) **Step 3: Apply the CUSUM formulas** Let's calculate the first few points as an example: - **Observation 1:** \(x_1 = 1045\) - \(C^+_1 = \max[0, 0 + (1045 - 1050) - 12.5] = \max[0, -5 - 12.5] = 0\) - \(C^-_1 = \max[0, 0 - (1045 - 1050) - 12.5] = \max[0, 5 - 12.5] = 0\) - **Observation 2:** \(x_2 = 1055\) - \(C^+_2 = \max[0, 0 + (1055 - 1050) - 12.5] = \max[0, 5 - 12.5] = 0\) - \(C^-_2 = \max[0, 0 - (1055 - 1050) - 12.5] = \max[0, -5 - 12.5] = 0\) Continue this calculation for all 20 observations. - **Alarm:** If either \(C^+\) or \(C^-\) exceeds \(h\), the process is out of control. --- ## (b) **EWMA Control Chart** **Step 1: Choose parameters \(\lambda\) and \(L\)** - \(\lambda\): Typically 0.2 or 0.3 (let’s use 0.2) - \(L\): Usually between 2.7 and 3.0 for 3-sigma limits (let’s use 3) **Step 2: Calculate EWMA statistics** - \(z_0 = \mu_0 = 1050\) - For each \(i\): - \(z_i = \lambda x_i + (1-\lambda)z_{i-1}\) - **Control limits:** - \(UCL = \mu_0 + L \sigma \sqrt{ \frac{\lambda}{2-\lambda} [1 - (1-\lambda)^{2i} ] }\) - \(LCL = \mu_0 - L \sigma \sqrt{ \frac{\lambda}{2-\lambda} [1 - (1-\lambda)^{2i} ] }\) **Step 3: Calculate for first few points** - \(z_1 = 0.2 \times 1045 + 0.8 \times 1050 = 1049\) - \(z_2 = 0.2 \times 1055 + 0.8 \times 1049 = 1050.2\) - \(z_3 = 0.2 \times 1037 + 0.8 \times 1050.2 = 1047.36\) Continue for all 20 points and plot the EWMA. --- ## (c) **Compare Results (Only for IEM 5103 Students)** - You would compare the sensitivity of the CUSUM and EWMA charts to small shifts in the process mean, and comment on which detected a shift earlier, if at all. --- ### **Summary Table (First Few Calculations)** | Obs | \(x\) | \(C^+\) | \(C^-\) | \(z\) | |-----|-------|---------|---------|----------| | 1 | 1045 | 0 | 0 | 1049 | | 2 | 1055 | 0 | 0 | 1050.2 | | 3 | 1037 | 0 | 0 | 1047.36 | | 4 | 1064 | 1.5 | 0 | ... | | ... | ... | ... | ... | ... | --- **You can use Excel or statistical software to automate the calculations for all 20 points and plot both charts.** If you want the full calculation or Excel template, let me know!