chapter 1

Chapter 16: Quality Control Methodsp , n = 100, so UCL = np + 3 np( 1−p) = 6.08 + 3 6.08 (.9392)42. = . 06086 .08 + 7.17 == 13.25 and LCL = 0. All points are between these limits, as was the casefor the p-chart. The p-chart and np-chart will always give identical results since( 1−p) p( 1−p)pp − 3 < pˆi< p + 3iffnnnp− 3 np 1−p < npˆ= x < np + 3 np 1−p( ) ( )ii43. Σ i= 4 ( 16) + ( 3)( 4) = 76( 1)22Σ ni− sis = =Σ( n − 1)n , Σ n = 32,729. 4 , = 430. 65when n = 3,when n = 4,ix i ix ,27,380.16 −5661.4= 590.027976 − 20( 24.2905) 1−(.886), so s = 24.2905. For variation:3UCL = 24.2905 += 24.29 + 38.14 = 62.43 ,.886( 24.2905) 1 − (.921)3UCL = 24.2905 += 24.29 + 30.82 = 55.11..921For location: when n = 3, 430 .65 ± 47.49 = 383.16,478. 14 , and when n = 4,430 .65 ± 39.56 = 391.09,470.21.2244.a. Provided the E ( X i) = µ for each i,t−1tE( W ) = αE( X ) + α( −α) E( X ) + ... + α( 1−α) E( X ) + ( −α) µtt1t−111t−1[ α + α ( 1 − α ) + ... + α ( 1−α ) + ( 1 α ) ]tt−1t= µ −[ α ( 1+( 1−α ) + ... + ( 1 − α ) ) + ( 1 α ) ]= µ −∞∞⎡iit ⎤= µ ⎢α∑ ( 1−α ) − α∑( 1−α) + ( 1−α) ⎥⎦⎣ i=0i=t⎡ αt 1t ⎤= µ ⎢ − α ( 1−α) ⋅ + 1−α=( α )( α )⎥⎣1− 1−1−1−⎦( ) µ22 22 2b. ( ) ( ) ( ) ( ) ( ) ( t−1V W = α V X + α 1−αV X + ... + α 1−α) V ( X )= αttt−1222[ ( ) ( ) ( t−11 + 1−α + ... + 1−α) ] ⋅V( X1)= α2t−1[ 1+C + ... + C ]1 − C t 2⋅σ1−C nσ⋅n2C = 1 −α.)(where ( ) 22= α, which gives the desired expression.1482