Q1- A product has three quality characteristics. The nominal values of these quality characteristics andtheir sample covariance matrix have been determined from the analysis of 30 preliminary samples of size3.0n=10 as follows: = [3.5]2.81.4 1.02 1.05 = [1.02 1.35 0.98]. The sample means for each quality characteristic1.05 0.98 1.20for 15 additional samples of size n=10 are show below. Is the process in statistical control? Please providedetails.Sample#123456789101112131415Xbar1Xbar2Xbar33.13.32.62.8343.832.423.23.74.13.83.23.73.9333.34.64.23.332.63.944.743.633.12.42.52.83.532.72.21.8333.22.92.8Answer:Step 1. Definition of the limits of the T2 statistic:-LCL = 0UCL = 16.93 (see below calculation) =( + 1)( 1)3(30 + 1)(10 1)837 ,(,+1) = ,(3,3010303+1) = ,(3,268) + 130 10 30 3 + 1268= 16.93Step 2. Calculation of the difference vector for each sample:DifferenceSamplefor the first#mean1234567891011121314150,10,3-0,4-0,2010,80-0,6-10,20,71,10,80,2DifferenceDifferencefor thefor thesecondthird meanmean0,20,20,40,3-0,5-0,4-0,5-0,3-0,201,10,70,70,2-0,2-0,1-0,5-0,6-0,9-10,40,20,50,21,20,40,50,10,10Step 3. Calculation of the corresponding T2 statistic for each sam ...
To Order an Original Plagiarism Free Paper on the Same Topic Click Here
Other samples, services and questions:
When you use PaperHelp, you save one valuable — TIME
You can spend it for more important things than paper writing.
