13 Semiconductor Etch Variability Oxide layers on semiconductor wafers are etched in a mixture

Example 10-13 Semiconductor Etch Variability Oxide layers on semiconductor wafers are etched in a mixture

  of gases to achieve the proper thickness. The variability in the thickness of these oxide layers is a

  critical characteristic of the wafer, and low variability is desirable for subsequent processing steps. Two different mix- tures of gases are being studied to determine whether one is superior in reducing the variability of the oxide thickness.

  Sixteen wafers are etched in each gas. The sample standard deviations of oxide thickness are s 1 = . angstroms and 1 96

  s 2 = . angstroms, respectively. Is there any evidence to indicate that either gas is preferable? Use a fi xed-level test 2 13 with α = 0.05.

  The seven-step hypothesis-testing procedure may be applied to this problem as follows:

  1. 2 Parameter of interest: The parameters of interest are the variances of oxide thickness σ

  1 and σ 2 . We will assume

  that oxide thickness is a normal random variable for both gas mixtures.

  2. Null hypothesis: H

  0 : σ 1 =

  3. Alternative hypothesis: H 1 : 2 σ≠σ

  4. Test statistic: The test statistic is given by Equation 10-31:

  5. Reject H 0 if: Because n 1 = n

  2 = and α = 0.05, we will reject H 16 0 : σ=σ 1 if f>f 0 . 0 025 15 15 ,, =. 2 86 or if f < f 0 . 0 975 15 15 ,,

  = 1 f . 0 025 15 15 ,,

  1 2 86 .=. 0 35 . Refer to Figure 10-6(a).

  6. Computations: Because s 2 2 1 2 = (. 1 96 ) . 3 84 and s = (. 2 13 )

  . 4 54 , the test statistic is

  7. Conclusion: Because f

  2 . 0 975 15 15 ,, = . 0 35 < 0 85 f . 0 025 15 15 ,, = 2 86 , we cannot reject the null hypothesis H 0 : σ = σ at 1

  the 0.05 level of signifi cance. Practical Interpretation: There is no strong evidence to indicate that either gas results in a smaller variance of oxide

  thickness.

  P-Values for the F -Test The P-value approach can also be used with F-tests. To show how to do this, consider the upper-tailed one-tailed test. The P-value is the area (probability) under the F distribution with

  n 1 − and n 2 − degrees of freedom that lies beyond the computed value of the test statistic 1

  f 0 . Appendix A Table IV can be used to obtain upper and lower bounds on the P-value. For example, consider an F-test with 9 numerator and 14 denominator degrees of freedom for

  which f 0 = . . From Appendix A Table IV, we fi nd that f 3 05 . 0 05 9 14 ,, = . 2 65 and f . 0 025 9 14 ,, = ., 3 21

  Section 10-5Inference on the Variances of Two Normal Distributions

  so because f 0 = . lies between these two values, the P-value is between 0.05 and 0.025; that 3 05

  is, 0 025 . << P 0 05 . The P-value for a lower-tailed test would be found similarly, although Appendix A Table IV contains only upper-tailed points of the F distribution, Equation 10-30 would have to be used to fi nd the necessary lower-tail points. For a two-tailed test, the bounds obtained from a one-tailed test would be doubled to obtain the P-value.

  Finding the

  To illustrate calculating bounds on the P-value for a two-tailed F-test, reconsider

  P-Value for

  Example 10-13. The computed value of the test statistic in this example is f 0 = . . This 0 85

  Example 10-13 value falls in the lower tail of the F , 15 15 distribution. The lower-tailed point that has 0.25

  probability to the left of it is f . 0 75 15 15 ,, = 1 f . 0 25 15 15 ,,

  0 70 , and because 0 70 . < ., 0 85

  the probability that lies to the left of 0.85 exceeds 0.25. Therefore, we would conclude

  that the P-value for f 0 = . is greater than 2 0 25 0 85 (. ) = 05 ., so there is insuffi cient evidence

  to reject the null hypothesis. This is consistent with the original conclusions from Exam- ple 10-13. The actual P-value is 0.7570. This value was obtained from a calculator from which we found that P F ( , 15 15 ≤ . 0 85 ) = . 0 3785 and 2 0 3785 (. ) = . 0 7570 . Computer software can also be used to calculate the required probabilities.

  Some computer packages will perform the F-test on the equality of two variances of inde- pendent normal distributions. The output from the computer package follows.

  Test for Equal Variances

  95 Bonferroni confidence intervals for standard deviations Sample

  F-test (normal distribution) Test statistic = 0.85, P-value = 0.750

  Computer software also gives confi dence intervals on the individual variances. These are the confi dence intervals originally given in Equation 8-19 except that a Bonferroni “adjustment” has been applied to make the confi dence level for both intervals simultaneously equal to at least 95. This consists of using α 2 = 0.05 2 = 0.025 to construct the individual intervals. That is, each individual confi dence interval is a 97.5 CI. In Section 10-5.4, we will show how to construct a CI on the ratio of the two variances.