Question

Minitab Steps

13-34. + An article in the Jounal of the Electrochemical Society [1992, Vol. 139(2), pp. 524-532)] describes an experi- ment to investigate the low-pressure vapor deposition of polysili con. The experiment was carried out in a large-capacity reactor at Sematech in Austin, Texas. The reactor has several wafer posi- tions, and four of these positions were selected at random. The response variable is film thickness uniformity. Three replicates of the experiment were run, and the data are as follows: (a) Is there a difference in the wafer positions? Use o 0.05. (b) Estimate the variability due to wafer positions. (c) Estimate the random error component.

Answer 1

An experiment is conducted to investigate the low-pressure vapour deposition of polysilicon in a large-capacity reactor, which has several wafer positions. Four of these positions are selected at random; three replicates of the experiment were run. The response variable is film thickness uniformity; the data are given in the table that follows: Wafer Position Uniformity 2.76 5.67 4.49 2 1.43 1.70 2.19 2.34 1.97 1.47 4 0.94 1.36 1.65 a) We determine whether there is a difference in the wafer positions using a 0.05level of significance. We test the hypothesis that all treatments are identical against the alternative that there is variability between the treatments; symbolically the hypotheses to be tested are:

o oF Here σ2 is the variance of the treatment effects τ. The test statistic to test the null hypothesis is MSts oMS This is an F random variable with a-1 and a(n-1)degrees of freedom when Hois true. We reject the null hypothesis if the computed value of the test statistic a-)

The sums of squares for the ANOVA with equal sample sizes in each treatment are ,2 у Treatments The error sum of squares is obtained by subtraction as -SS Treatments Mean squares are given by: MSE = a(n-1) MSTreatments =-a-l Treatments

Computation table for totals and averages: Totals Treatment Observations 2.76 5.67 4.49 12.92 1.43 1.70 2.19 5.32 2.34 1.97 1.47 5.78 4 0.94 1.36 1.65 3.95 y-27.97 Grand total of all measurements is given by

=27.97

The total sum of squares is given by (27.97) = (2.76). (5.67) + (1.65)2-G12 86.6307-65.1934 21.4373 The sum of squares due to treatments is Treatments 12.92)(5.32)(5.78)(3.95) (27.97 12 81.4132 = 16.21 98 65.1934 Sum of squares due to errors is obtained by subtraction SS-SS -SSTreatments 21.4373-16.2198 5.2175

The ANOVA is summarized in the table below Source of Variation Sum of Squares Degrees of Freedom Mean SquareP-value Wafer Position 16.2198 5.4066 8.28995 0.0077 Error 5.2175 0.65219 Total 21.4373

The F table value for a 0.05and (3,8) degrees of freedom is aa-1a-l) 00,38 4.07 Thus the test statistic value is greater thandicating that the data is sufficient to reject the null hypothesis The P-value for this test statistic is P- P(Fs>8.29) 3,8 0.0077 This is considerably smaller than a 0.05 indicating a strong evidence to reject Ho. We can thus conclude that there is significant difference between wafer positions

b) The estimator of the variance component due to treatments (wafer positions) is 2 MS Treatments In 5.4066-0.65219 =11.5848 That is, the estimate of the variability due to wafer positions is 1.5848.

c) The estimate of the variance due to random error component is a2 σ = Ms. = 0.65219 Thus the random error component of the variance is approximately 0.65

d) We analyse the residuals from the experiment A residual is the difference between an observation yu and its estimated value y,--(for the completely randomized design). Each residual is given by The residuals are shown in the table below Treatment |Residuals 1.54671.36330.1833 2 -0.3433 |-0.0733| 0.4167 0.4133 0.0433 -0.4567 4 0.3767 0.0433 0.3333

The normality assumption can be checked by constructing a normal probability plot of the residuals Probability Plot of Residual value Normal-95% CI -0.00001667 0.6887 12 0.463 0.210 StDev AD P-Value 60---4 50 Residual value

The plot shows not much deviation from normality as all the observations fall within the two bands We now plot the residuals against the factor levels and compare the spread in the residuals to check the assumption of equal variances at each factor level (wafer positions 1, 2, 3, and 4).

Residual value vs Wafer Positions 1.5- 1.0 0.5 0.0 ะ -0.5 1.0 2 Wafer Positions (factor levels)

We observe that the variability is high in respect of wafer position 1, giving evidence against the assumption of equal variances at all factor levels

We plot the residuals against y (fitted values) Residual value vs Fitted Values 15-г 1.0 0.5 0.0 1.32 1.771.93 4.31 -0.5 2.5 Fitted Values 1.0 1.5 2.0 3.0 3.5 4.0 4.5

We can see that the variability in the residuals is high for larger values of y, indicating that the variability in the residuals depends on the value of y Thus the residual analysis indicates modal inadequacy