Applied Statistics and Probability for Engineers

This result is based on the fact that (n 1 1)S^21 /^21 is a chi-square random variable with n 1  1

degrees of freedom, that (n 2 1)S^22 ^22 is a chi-square random variable with n 2 1 degrees

of freedom, and that the two normal populations are independent. Clearly under the null

hypothesis H 0 : ^21 ^22 the ratio has an distribution. This is the basis of

the following test procedure.

F 0 S 12
S 22 Fn 1 1,n 2  1

358 CHAPTER 10 STATISTICAL INFERENCE FOR TWO SAMPLES

Null hypothesis:

Test statistic: (10-29)

Alternative Hypotheses Rejection Criterion

H 1 :  12 ^22 f 0 f 1 , n 1 1,n 2  1

H 1 :  12 

 22 f 0

 f ,n 1 1,n 2  1

H 1 : ^21  22 f 0

 f 2,n 1 1,n 2  1 or f 0 f 1  2,n 1 1,n 2  1

F 0 

S 12

S^22

H 0 : ^21 ^22

EXAMPLE 10-11 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 mixtures

of gases are being studied to determine whether one is superior in reducing the variability of

the oxide thickness. Twenty wafers are etched in each gas. The sample standard deviations of

oxide thickness ares 1 1.96 angstroms ands 2 2.13 angstroms, respectively. Is there any

evidence to indicate that either gas is preferable? Use0.05.

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

1. The parameters of interest are the variances of oxide thickness ^21 and ^22. We will
   assume that oxide thickness is a normal random variable for both gas mixtures.


2. 
   


3. 
   


4. 
   


5. The test statistic is given by Equation 10-29:


6. Since n 1 n 2 20, we will reject
   .


7. Computations: Since s^21 (1.96)^2 3.84 and s^22 (2.13)^2 4.54, the test statistic is


8. Conclusions: Since f0.975,19,190.40f 0 0.85f0.025,19,192.53, we cannot
   reject the null hypothesis H 0 : ^21 ^22 at the 0.05 level of significance. Therefore,
   there is no strong evidence to indicate that either gas results in a smaller variance of
   oxide thickness.

f 0 

s^21

s^22



3.84

4.54

0.85

if f 0 f0.975,19,19 1
f0.025,19,19 1
2.530.40

H 0 : ^21 ^22 if f 0

 f0.025,19,192.53 or

f 0 

s^21

s^22

0.05

H 1 : ^21 ^22

H 0 : ^21 ^22

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