1 INFO & DEC SCI 474 SPRING, 1994 QPI STAT METHODS INSTRUCTOR: SCLOVE MINITAB OUTPUT FOR PROBLEM 6: Confounding (Montgomery, Problem 10.1) For Meeting #13 MTB > # Column C50 will contain the data, the results from the 8 treatment combinations in MTB > # in standard order: (1) a b ab c ac bc abc. MTB > # MTB > NAME C50'Data' MTB > SET C50 DATA> 22 32 35 55 44 40 60 39 DATA> # MTB > NAME C1'A' C2'B' C3'C' C12'AB' C13'AC' C23'BC' C123'ABC' MTB > # 'A' will contain the contrast coefficients for Effect A: MTB > SET 'A' DATA> -1 +1 -1 +1 -1 +1 -1 +1 MTB > # Similarly for 'B', 'C', 'AB', etc.: MTB > SET 'B' DATA> -1 -1 +1 +1 -1 -1 +1 +1 MTB > SET 'C' DATA> -1 -1 -1 -1 +1 +1 +1 +1 MTB > LET 'AB'= 'A'*'B' MTB > LET 'AC' = 'A'*'C' MTB > LET 'BC' = 'B'*'C' MTB > LET 'ABC' = 'A'*'B'*'C' Forming contrasts results in differences of sums of four observations. Hence, to compute effects, which are on the scale of means (the same scale as the original observations), a divisor of four is used. MTB > # Effect = Contrast/4 MTB > LET K1 = SUM('A'*'Data')/4 MTB > LET K2 = SUM('B'*'Data')/4 MTB > LET K3 = SUM('C'*'Data')/4 MTB > LET K12 = SUM('AB'*'Data')/4 MTB > LET K13 = SUM('AC'*'Data')/4 MTB > LET K23 = SUM('BC'*'Data')/4 MTB > LET K123 = SUM('ABC'*'Data')/4 MTB > PRINt K1 K2 K3 K12 K13 K23 K123 K1 1.25000 K2 12.7500 K3 9.75000 K12 -1.75000 K13 -13.7500 K23 -5.25000 K123 -6.75000 MTB > NAME C7 'Effects' MTB > COPY K1 K2 K3 K12 K13 K23 K123 into C7 MTB > # These are the effects, in the order MTB > # A B C AB AC BC ABC MTB > PRINT 'Effects' Effects 1.25 12.75 9.75 -1.75 -13.75 -5.25 -6.75 1 CMS DSN= Problem6 IDS 474(QPI Stat Methods): Minitab Output for Problem 6 p.2 -------------------------------------------------------------------- MTB > DESC C7 N MEAN MEDIAN TRMEAN STDEV SEMEAN Effects 7 -0.54 -1.75 -0.54 9.33 3.53 MIN MAX Q1 Q3 Effects -13.75 12.75 -6.75 9.75 MTB > NAME C22 'NScores' MTB > NSCOres of c7, put into c22 MTB > PRINt c7 c22 ROW Efects NScores 1 1.25 0.35147 2 12.75 1.36459 3 9.75 0.75613 4 -1.75 0.00000 5 13.75 -1.36459 6 -5.25 -0.35147 7 -6.75 -0.75613 MTB > SORT c7 c22 put into c11 c42 MTB > NAME C11 'OrdEffs' C42 'OrdNScs' MTB > PRINt c11 c42 ROW OdEffs OrdNScs 1 13.75 -1.36459 2 -6.75 -0.75613 3 -5.25 -0.35147 4 -1.75 0.00000 5 1.25 0.35147 6 9.75 0.75613 7 12.75 1.36459 MTB > PLOT c7 vs c22 - * 0+ * - - - - * Effects0+ - * - - * * - -0+ - - * - --------+---------+---------+---------+---------+-------- -1.00 -0.50 0.00 0.50 1.00 NScores MTB > CENTer C11 put into c21 1 CMS DSN= Problem6 IDS 474(QPI Stat Methods): Minitab Output for Problem 6 p.3 -------------------------------------------------------------------- MTB > NAME C21 'StdEffs' MTB > PRINT 'StdEffs' 'OrdNScs' ROW tdEffs OrdNScs 1 -.41614 -1.36459 2 -.66597 -0.75613 3 -.50522 -0.35147 4 -.13013 0.00000 5 .19137 0.35147 6 .10229 0.75613 7 .42379 1.36459 MTB > PLOT 'StdEffs' vs 'OrdNScs' StdEff - - * - * 10+ - - - - * 00+ - * - - * * - -10+ - - * - --------+---------+---------+---------+---------+-------- -1.00 -0.50 0.00 0.50 1.00 OrdNScs MTB > PRINt c42 OrdNScs -1.3645 -0.75613 -0.35147 0.00000 0.35147 0.75613 1.36459 MTB > SET c31 DATA> -1.5 -1.25 -1 -.75 -.5 -.25 0 .25 .5 .75 1.0 1.25 1.5 Column 31 is just some numbers to plot a straight line. MTB > LET K51 = MEAN('Effects') MTB > PRINT K51 K51 -0.535714 MTB > LET K52 = STDEV('Effects') MTB > PRINt K52 K52 9.33120 MTB > Let C32 = c31*K52 + K51 1 CMS DSN= Problem6 IDS 474(QPI Stat Methods): Minitab Output for Problem 6 p.4 -------------------------------------------------------------------- Normal scores: MTB > PRINt c42 OrdNScs -1.3645 -0.75613 -0.35147 0.00000 0.35147 0.75613 1.36459 Loosely speaking, these "normal scores" can be interpreted as follows: the number -1.36459 is the smallest value you would get, on the average, if you took samples samples of size n = 7 from a standard normal distribution. This number is paired with the smallest value in C1. The number -0.75613 is the second smallest value you would get, on the average, in samples of size 7 from a standard normal distribution. It is paired with the second smallest value in C1. This is continued for all 7 values. More precisely, the i-th smallest normal score is usually taken as the value of the inverse cdf of (i - 3/8)/(n + 1/4). If the 7 values are like a random sample from a normal distribution, they will fall in a straight line when plotted against the normal scores. In this example three of the seven are not close to such a straight line (one in the lower left and two at the upper right). Their sizes are appreciably larger than the sizes of the expected values. They correspnd to the effects B = 12.75, C = 9.75 and AC =-13.75. Note there are no d.f. left over for error, so we need to use methods other than ANOVA for assessing the sizes of the effects. MTB > mplot c32 c31 'Effects' 'NScores' - A - A B 0+ B - A - A - A - A B 0+ A - A B - - B A B - A -0+ A - A - A B - --------+---------+---------+---------+---------+-------- -1.20 -0.60 0.00 0.60 1.20 A = C32 vs. C31 B = Effects vs. NScores MTB > # These are the effects, in the order MTB > # A B C AB AC BC ABC 1.25 12.75 9.75 -1.75 -13.75 -5.25 -6.75 1 CMS DSN= Problem6 IDS 474(QPI Stat Methods): Minitab Output for Problem 6 p.5 -------------------------------------------------------------------- MTB > SET C18 DATA> -6.75 -5.25 -1.75 1.25 MTB > LET K41 = MEAN(C18) MTB > LET K42 = STDEV(C18) MTB > Let C32 = c31*K42 + K41 MTB > mplot c32 c31 'Effects' 'NScores' - - - B 0+ B - - - - B A A 0+ A A - B A A - A A A - A 2 B - A A -0+ - - B - --------+---------+---------+---------+---------+-------- -1.20 -0.60 0.00 0.60 1.20 A = C32 vs. C31 B = Effects vs. NScores MTB > stop *** Mintab Release 7.2 *** Minitab, Inc. *** 23-Feb-95 MtoProb6.IDS474