Applied Statistics and Probability for Engineers

We may also find a P-value for the F-statistic in Example 10-11. Since f0.50, 19, 191.00,

the computed value of the test statistic f 0 3.844.540.85 is nearer the lower

tail of the Fdistribution than the upper tail. The probability that an F-random variable with

19 numerator and denominator degrees of freedom is less than 0.85 is 0.3634. Since it is ar-

bitrary which population is identified as “one,” we could have computed the test statistic as

f 0 4.543.841.18. The probability that an F-random variable with 19 numerator and

denominator degrees of freedom exceeds 1.18 is 0.3610. Therefore, the P-value for the test

statistic f 0 0.85 is the sum of these two probabilities, or P0.3634 0.36100.7244.

Since the P-value exceeds 0.05, the null hypothesis H 0 : ^21 ^22 cannot be rejected. (The

probabilities given above were computed using a hand-held calculator.)

10-5.4 -Error and Choice of Sample Size

Appendix Charts VIo, VIp, VIq, and VIrprovide operating characteristic curves for the F-test

given in Section 10-5.1 for 0.05 and 0.01, assuming that n 1 n 2 n. Charts VIo

and VIpare used with the two-sided alternate hypothesis. They plot against the abscissa

parameter

(10-30)

for various n 1 n 2 n. Charts VIqand VIrare used for the one-sided alternative hypotheses.

EXAMPLE 10-12 For the semiconductor wafer oxide etching problem in Example 10-11, suppose that one gas

resulted in a standard deviation of oxide thickness that is half the standard deviation of oxide

thickness of the other gas. If we wish to detect such a situation with probability at least 0.80,

is the sample size n 1 n 2 20 adequate?

Note that if one standard deviation is half the other,

By referring to Appendix Chart VIowith n 1 n 2 n20 and 2, we find that

Therefore, if 0.20, the power of the test (which is the probability that the difference in

standard deviations will be detected by the test) is 0.80, and we conclude that the sample sizes

n 1 n 2 20 are adequate.

10-5.5 Confidence Interval on the Ratio of Two Variances

To find the confidence interval on recall that the sampling distribution of

is an Fwith n 2 1 and n 1 1 degrees of freedom. Therefore, 

Substitution for Fand manipulation of the inequalities will lead to

the 100(1)% confidence interval for ^21 ^22.

f 2,n 2 1, n 1  12  1 .

P 1 f 1  2, n 2 1, n 1  1 F

F

S^22 ^22

S^21 ^21

^21 ^22 ,

0.20.



 1

 2 ^2



 1

 2

s^21 s^22

10-5 INFERENCES ON THE VARIANCES OF TWO NORMAL POPULATIONS 359

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