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#123456789101112131415Xbar1Xbar2Xbar18.104.22.168.8343.832.422.214.171.124.126.96.36.199188.8.131.52.332.63.944.743.6184.108.40.206.83.5220.127.116.1118.104.22.168Answer: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 ...
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