Question
A factory producing dry-cells wanted to test the life of cells produced daily. The cells will be considered satisfactory if their life is 25 hours. For this, a sample of 4 cells was drawn on 7 consecutive days. The results are as follows:
| Day 1 | Day 2 | Day 3 | Day 4 | Day 5 | Day 6 | Day 7 |
| 24 20 25 20 | 16 23 23 20 | 20 25 24 21 | 26 18 28 20 | 20 22 24 22 | 25 23 25 20 | 26 24 20 25 |
i) Estimate the process mean.
ii) Determine the centre line and control limits of the control charts for controlling the process mean and process variability.
iii) Plot the charts on the graph papers.
iv) What managerial decisions should be made on the basis of these charts?
v) If necessary, compute revised control limits.
Answer :
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We will solve the problem step by step: --- ### Step 1: Organizing the Data The given data consists of the life (in hours) of 4 dry cells tested over 7 days: | Day | Readings | |---------|--------------| | 1 | 24, 20, 25, 20 | | 2 | 16, 23, 23, 20 | | 3 | 20, 25, 24, 21 | | 4 | 26, 18, 28, 20 | | 5 | 20, 22, 24, 22 | | 6 | 25, 23, 25, 20 | | 7 | 26, 24, 20, 25 | --- ### Step 2: Compute the Process Mean ( \(\bar{X}\) ) The process mean is calculated using: \[ \bar{X} = \frac{1}{n} \sum X_i \] where \( X_i \) is the sample mean for each day. We compute the daily means: \[ \bar{X}_1 = \frac{24+20+25+20}{4} = 22.25 \] \[ \bar{X}_2 = \frac{16+23+23+20}{4} = 20.5 \] \[ \bar{X}_3 = \frac{20+25+24+21}{4} = 22.5 \] \[ \bar{X}_4 = \frac{26+18+28+20}{4} = 23 \] \[ \bar{X}_5 = \frac{20+22+24+22}{4} = 22 \] \[ \bar{X}_6 = \frac{25+23+25+20}{4} = 23.25 \] \[ \bar{X}_7 = \frac{26+24+20+25}{4} = 23.75 \] Now, compute the overall process mean: \[ \bar{\bar{X}} = \frac{22.25 + 20.5 + 22.5 + 23 + 22 + 23.25 + 23.75}{7} = 22.61 \] Thus, the estimated process mean is 22.61 hours. --- ### Step 3: Compute
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