Breeze Toothpaste Company makes tubes of toothpaste. The product is produced and
ID: 450022 • Letter: B
Question
Breeze Toothpaste Company makes tubes of toothpaste. The product is produced and then pumped into tubes and capped. The production manager is concerned whether the filling process for the tubes of toothpaste is in statistical control. The process should be centered on 6 ounces per tube. Six samples of five tubes were taken and each tube was weighed. The weights are:
Develop a control chart for the mean and range for the available toothpaste data. (Round answers to 3 decimal places, e.g. 15.250.)
x-bar chart:
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Choose the appropriate control chart based on your observations.
Breeze Toothpaste Company makes tubes of toothpaste. The product is produced and then pumped into tubes and capped. The production manager is concerned whether the filling process for the tubes of toothpaste is in statistical control. The process should be centered on 6 ounces per tube. Six samples of five tubes were taken and each tube was weighed. The weights are:
Ouches of Toothpaste per Tube Sample 1 2 3 4 5 1 5.78 6.34 6.24 5.23 6.12 2 5.89 5.87 6.12 6.21 5.99 3 6.22 5.78 5.76 6.02 6.10 4 6.02 5.56 6.21 6.23 6.00 5 5.77 5.76 5.87 5.78 6.03 6 6.00 5.89 6.02 5.98 5.78Explanation / Answer
Ouches of Toothpaste per Tube Sample 1 2 3 4 5 X - Bar Range 1 5.780 6.340 6.240 5.230 6.120 5.942 1.110 2 5.890 5.870 6.120 6.210 5.990 6.016 0.340 3 6.220 5.780 5.760 6.020 6.100 5.976 0.460 4 6.020 5.560 6.210 6.230 6.000 6.004 0.670 5 5.770 5.760 5.870 5.780 6.030 5.842 0.270 6 6.000 5.890 6.020 5.980 5.780 5.934 0.240 35.714 3.090 X-Bar is average of observations of that sample Eg - Sample 1 = (5.78 + 6.34 + 6.24 + 5.23 + 6.12) / 5 = 29.71 / 5 = 5.942 Range is difference between maximum value and minimum value of the observations Eg Sample 1 = 6.34 - 5.23 = 1.110 X - Double Bar = X - Bar / No of samples = 35.714/6 = 5.952 UCL = X-Double Bar + A2 * Range Bar LCL = X-Double Bar - A2 * Range Bar A2 value for 6 observations = 0.48 UCL = 5.952 + 0.48*0.515 = 5.952 + 0.247 = 6.199 LCL = 5.952 - 0.48*0.515 = 5.952 - 0.247 = 5.705 Control Line = 5.952 UCL = 6.199 LCL = 5.705 R - Bar = Range / No of samples = 3.09 / 6 = 0.515 UCL = D4 * Range Bar LCL = D3 * Range Bar D3 value for 6 observations = 0 D4 value for 6 observations = 2.00 UCL = 0.515 * 2 = 1.03 LCL = 0.515 * 0 = 0 Control Line = 0.515 UCL = 1.030 LCL = 0