Indu 6331 - Study guides, Class notes & Summaries

Consider the hole diameter data. Estimate process capability using ´x and R charts. If specifications are at 0±0.01, calculate Cp, Cpk, and Cpkm. Interpret these ratios. n= 5 ´x= ∑ ´x m = (8+0+…+20+18) 20 =10,90 R´ = ∑Ri m = (80+110+…+60+50) 20 =63,50 σ^= R´ d2 = 63,50 2.326 =27,3 * values expressed in ten-thousandths of an inch The following graphics show that the process is in statistical control and that it follows a normal distribution, so process ...

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

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 shif 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 ¿1.075− 1.050/¿ 2 =12.5 ¿ μ1− μ0 /¿ 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 ...

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...

A mechatronic assembly is subjected to a final functional test. Suppose that defects occur at random in these assemblies, and that defects occur according to a Poisson distribution with parameter λ= 0.02 a. What is the probability that an assembly will have exactly one defect? λ=0.02 p ( x )= e −λ λ x x ! p( x=1)= e −0.02 0.021 1! =0.0196=1.96 There is a probability of 1.96% that an assembly will have exactly one defect b. What is the probability that an assembly will ha...

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 ...

What are the three primary technical tools used for quality control and improvement? The three primary technical tools used for quality control and improvement are: 1. Statistical process control (SPC): On-line tool useful for process monitoring and reduction of variability. 2. Acceptance sampling: Connected to inspection and testing of samples selected at random from larger batch. 3. Design of experiments (DOX): Off-line tool useful to discover key variables influencing characteristics o...