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and is said to follow the F distribution with u degrees of freedom in the numerator and v degrees of freedom in the denominator. It is usually abbreviated as Fu,v.
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The mean and variance of the F distribution are
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Two F distributions are shown in Fig. 10-4. The F random variable is nonnegative, and the distribution is skewed to the right. The F distribution looks very similar to the chi-square distribution; however, the two parameters u and v provide extra exibility regarding shape. The percentage points of the F distribution are given in Table V of the Appendix. Let f ,u,v be the percentage point of the F distribution, with numerator degrees of freedom u and denominator degrees of freedom v such that the probability that the random variable F exceeds this value is P1F f
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f 1x2 dx 5 and v 10, we nd from Table V of the
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This is illustrated in Fig. 10-5. For example, if u Appendix that P1F f0.05,5,10 2
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f (x) u = 5, v = 5
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Figure 10-4 Probability density functions of two F distributions.
Figure 10-5 Upper and lower percentage points of the F distribution.
That is, the upper 5 percentage point of F5,10 is f0.05,5,10 3.33. Table V contains only upper-tail percentage points (for selected values of f ,u,v for 0.25) of the F distribution. The lower-tail percentage points f1 ,u,v can be found as follows.
For example, to nd the lower-tail percentage point f0.95,5,10, note that f0.95, 5,10 1 f0.05,10, 5 1 4.74 0.211
10-5.2 10-5.3
Development of the F Distribution (CD Only) Hypothesis Tests on the Ratio of Two Variances
A hypothesis-testing procedure for the equality of two variances is based on the following result.
Let X11, X12, p , X1n1 be a random sample from a normal population with mean 1 and variance 2, and let X21, X22, p , X2n2 be a random sample from a second normal pop1 ulation with mean 2 and variance 2. Assume that both normal populations are 2 2 independent. Let S 1 and S 2 be the sample variances. Then the ratio 2 F has an F distribution with n1 inator degrees of freedom.
2 S1 S2 2 2 1 2 2
1 numerator degrees of freedom and n2
1 denom-
This result is based on the fact that (n1 1)S 2/ 2 is a chi-square random variable with n1 1 1 1 degrees of freedom, that (n2 1)S 2 2 is a chi-square random variable with n2 1 degrees 2 2 of freedom, and that the two normal populations are independent. Clearly under the null 2 hypothesis H0: 2 S 2 S 2 has an Fn1 1,n2 1 distribution. This is the basis of 1 2 1 2 the ratio F0 the following test procedure.
Null hypothesis: Test statistic:
H0: F0
2 1 2 S1 S2 2
(10-29) Rejection Criterion f0 f0 f0 f 2,n1 1,n2 1 or f0 f ,n1 1,n2 1 f1 , n1 1,n2 1 f1
2,n1 1,n2 1
Alternative Hypotheses H1: H1: H1:
2 1 2 1 2 1 2 2 2 2 2 2
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 are s1 1.96 angstroms and s2 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. 2. 3. 4. 5. The parameters of interest are the variances of oxide thickness 2 and 2. We will 1 2 assume that oxide thickness is a normal random variable for both gas mixtures. H0: H1:
2 1 2 1 2 2 2 2
0.05 The test statistic is given by Equation 10-29: f0 s2 1 s2 2
2 Since n1 n2 20, we will reject H0: 2 f0.025,19,19 2.53 or 1 2 if f0 if f0 f0.975,19,19 1 f0.025,19,19 1 2.53 0.40. 7. Computations: Since s2 (1.96)2 3.84 and s2 (2.13)2 4.54, the test statistic is 1 2
f0 8.