Indu6331 - Study guides, Class notes & Summaries

4.1 The inside diameters of bearings used in an aircraft landing gear assembly are known to have a standard deviation of σ= 0.002cm. A random sample of 15 bearings has an average inside diameter of 8.2535cm a. Test the hypothesis that the mean inside bearing diameter is 8.25cm Use a two-sided alternative and α= 0.05 σ= 0.002 n= 15 Samp. aveg.= 8.2535 α= 0.05 H0: μ0= 8.25 H1: μ0≠ 8.25 Zα /2=Z0.025=1.96 Zo= x´−μ0 σ /√n = 8.2535−8.25 0.002/√15 =6.77 If /Z0/ &g...

9.1. The data in table represent individual observations on molecular weight taken hourly from a chemical process. The target value of molecular weight is 1050 and the process standard deviation is thought to be about σ= 25. a) Set up a tabular cusum for the mean of this process. Design the cusum to quickly detect a shift of about 1.0σ in the process mean. σ= 25 μ0= 1.050 Magnitude of shift= 1,0(25)= 25 μ1= 1.050 + 1,0(25)= 1.075 K = /μଵ − μ଴/ 2 = /1.075 − 1.050/ 2 ...

6.25 Samples of n= 5 units are taken from a process every hour. The x and R values for a particular quality characteristic are determined. After 25 samples have been collected, we calculate x avg.= 20 and R= 4.56 a. What are the three sigma control limits for x and R? n= 5 Three sigma control limits for X UCL=´x+ A2 R´ =20+0.577∗4.56=22.63 Centerline=x´ =20 LCL=´x−A2R´ =20−0.577∗4.56=17.37 Three sigma control limits for R UCL=D4R´ =2.114∗4.56=9.64 Centerline=R´ =4.56 ...

INDU 6331 Assignment 7 Advanced Quality Control As an example considering p= 0.02, the value of Pa is Pa=P{d ≤ c }=∑ d=0 c n! d !(n−d)! p d (1−p) n−d=∑ d=0 c nCd∗p d (1−p) n−d Pa=P{d ≤2}=100C 0∗0.020 (1−0.02) 100−0 +100C1∗0.021 (1−0.02) 100−1 +100C 2∗0.022 (1−0.02) 100−2 Pa, p=0.02=0.677