2.830 Homework #6. April 2, 2009

Transcription

1 2.830 Homework #6 Dayán Páez April 2, ANOVA The data for four different lithography processes, along with mean and standard deviations are shown in Table 1. Assume a null hypothesis of equality. The full ANOVA table for this process is shown in Table 2 on the next page. The level of significance for rejecting the null hypothesis is 0.204, which indicates that we cannot reject the notion that the processes are different. Note that we have assumed similar variances for each of the processes, which a cursory check reveal to be reasonable. 1.1 Residual diagnostic A residual histogram is shown in Figure 1 on the following page. There is no strongenough evidence to conclude that the residuals do not obey a normal distribution. 2 Two-way ANOVA: Injection molding 2.1 ANOVA table The parameters used in calculating the ANOVA table are shown in Table 3 on the next page. Using these parameters, the associated sum of squares and the ANOVA table is shown in Table 4. The calculations were done according to Montgomery using Octave. Using the reported F-distribution values shown, we would have to reject the hypotheses that each of the sources of variation did not cause a change in the output. Moreover, I II III IV A B C D Table 1: Measurements of four photolithography processes (A, B, C, and D) 1

2 Source of Variation Sum of Squares D.o.f. Mean Square F Ratio Average S A = ν A = 1 s 2 A = s2 T /s2 R = Between treatments S T = ν T = 3 s 2 T = Within treatments S R = ν R = 10 s 2 R = Total S = N = 14 Table 2: Full ANOVA table for lithography process Figure 1: Overall dot diagram for residuals of photolithography process Parameter Value y y y y y y y y y Table 3: Parameters for two-way ANOVA 2

3 Source Sum of Squares D.o.f. Mean Square F 0 Hold time SS A = MS A = F 0 = 37.6 Velocity SS B = MS B = F 0 = 7.15 Interaction SS AB = MS AB = F 0 = Error SS E = MS E = Total SS T = Table 4: Two-way ANOVA table for injection molding there is not enough indication to reject the notion that the changes in parameters had an effect on their interaction. All of these tests were carried out with α = A different statistic To use a different statistic than ANOVA, we could look at each parameter in turn while maintaining the others fixed, and carry out a two-sample t-test. This means that for the hold time parameter, we choose samples at each of the levels with the injection velocity at the low level. We choose the low level for the constant parameter arbitrarily. Under these conditions, the different hypotheses tests are H A 0 : µ A low = µa high (1) H B 1 : µ B low = µb high (2) where the superscripts A, B represent each of the parameters discussed above. We assume normal distributions, and similar variances in each case. 2.3 Applying new statistic Hold time For the hold time, the pooled estimator is and the t-statistic is s 2 p,a = (3) t = 6.81 ν = 28 (4) At α = 0.05, the critical t = 2.05 < t which is enough to reject the null hypothesis Injection velocity We proceed similarly to Section on the previous page for injection velocity. The pertinent parameters are s 2 p,b = (5) t = 3.21 (6) 3

4 Run I II Total (1) a b ab c ac bc abc Table 5: Data for cutting-tool life experiments with similar degrees of freedom and critical t values. Here, too, we reject the null hypothesis that the parameter had no effect on the output. 2.4 Comparison to ANOVA While the t-tests are not exhaustive since they do not also consider the effect of one parameter at the other extreme of the fixed parameter(s), the results are in accord with those obtained by ANOVA. The t-tests, however, provide no information about the interaction between the two variables, although these could also be independently obtained by examining the diagonal elements of the parameter space. 3 Design of Experiment 3.1 Cutting tool life The data for cutting tool life based on two replicates of a 2 3 experiment are shown in Table 5. To determine if any of three factors affects tool life, we average the effects of each. In the case of parameter A, the cutting speed, A = 1 [ ] = (7) 16 Similarly, we determine the effects of the other parameters B = (8) C = (9) These values indicate that all three factors affect the tool life, although B an C significantly more so than factor A. 4

5 3.1.1 Longest tool life Because factors B and C are positively correlated with improved tool life while A is relatively uncorrelated, the combination of factors that should produce the longest tool life is given by bc, which corresponds with the data Best parameter combination We ponder whethere there is a combination of cutting speed A and cutting angle C that always gives good results regardless of metal hardness B. This interaction is given by AC = 1 [ ] = 10.8 (10) 16 This indicates that the combination of high A and C, or high speed and large angle give better tool life regardless of metal hardness. 3.2 Tool life residuals To find the residuals, use the model y = µ + A + B +C + AB + AC + BC + ABC + ε, (11) where ε is the random error. or the residual. We have determined A, B, C, and AC in Section 3.1 on the previous page. We now determine the other three interaction terms: AB = (12) BC = (13) ABC = (14) with µ = Looking at relative magnitudes, we see that the interaction terms have an absolute effect that is about one-fifth that of the single-factor effects. We reduce our regression model to include only those important ones: y = µ + B 2 + C 2 + ε (15) ŷ = x x 3 (16) The residuals are shown in Table 6 on the preceding page and a QQ-plot is shown Figure 2 on the previous page which demonstrates normality in the residuals. In addition, a plot of residuals versus predicted values is shown in Figure Single-replicate tool life Consider the problem in Section 3.1 on page 4 with only run I and with four center points (as shown in Table 7) added. 5

6 Point Expected Residual (1) a b ab c ac bc abc Table 6: Residuals for cutting-tool life Std. Frequency Figure 2: QQ-plot for tool life residuals Center points Table 7: Center points for single-replicate tool-life 6

7 Residuals Expected values Figure 3: Predicted vs. residuals for tool-life Curvature sum of squares The average of the center points is ȳ C = 420 while the average of the eight factorial points is ȳ F = 412. The curvature sum of squares is then given by Experimental error estimate An estimate of experimental error is given by Lack of fit SS pure quad = n Fn C (ȳ F ȳ C ) 2 n F + n C = (17) ˆσ 2 = C (y i ŷ C ) 2 n C = 250 (18) We compute the lack of fit by performing an F-test comparing the variances of the sum of squares of all the interactions to the sum squares calculated in (17), according to the equation F 0 = MS lack of fit = SS E/ν E MS pure error 250 This computation requires knowledge of SS E which is best obtained for problems of this complexity with the aid of a statistical software package capable of ANOVA, and is beyond the scope of manual computation as exercised in this paper. 7 (19)

8 C. Point Temperature Conc. Catalyst Yield (1) a b ab c ac bc abc (a) Data Factor(s) Value A B C 1.50 AB 1.50 AC BC 0.00 ABC 5.50 (b) Interactions Table 8: Single-replicate 2 3 factorial design of nitride etch process Factor(s) Expected Residual (1) a b ab c ac bc abc Table 9: Residuals for nitide etch process 3.4 Nitride etch process Data for a single-replicate 2 3 factorial design experiment for a nitride-etch process are shown in Table 8a. We analyze the single-factor and interactions as in Section 3.1 on page 4. These results are shown in the companion Table 8b on the previous page. We note that only parameter A and interaction AC have significant effect on output. We model the output, then, using the form ŷ = x 1 = 5x 1 x 3, (20) where µ = The residuals are as shown in Table 9 with a QQ-plot shown in Figure 4 on the next page. The plot shows residuals which are reasonably normally distributed. In addition, the model with just those two factors has impressive accuracy with the results for most control points. In order to get the highest yield, then, I would recommend based on the data a higher temperature and if possible the second catalyst. 8

9 1.5 1 Std. Frequency Figure 4: QQ-plot of residuals for nitride etch process 9