W u Y v in .NET Attach QR-Code in .NET W u Y v W u Y vQuick Response Code barcode library for .netuse .net framework qr-codes generating tobuild qr code jis x 0510 for .net(10-26).net Vs 2010 qrcode recognizer in .netUsing Barcode scanner for .net framework Control to read, scan read, scan image in .net framework applications.f 1x2 Visual .net Crystal barcode integration with .netusing barcode maker for .net framework crystal control to generate, create bar code image in .net framework crystal applications.2 v u a b c avb x 2 Barcode maker for .netusing barcode generating for .net control to generate, create barcode image in .net applications.(10-27)Visual Studio .NET qrcode implementation for visual c#.netusing barcode implement for .net framework control to generate, create qrcode image in .net framework applications.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.QR Code barcode library in .netusing asp.net web service toget qr code iso/iec18004 with asp.net web,windows applicationThe mean and variance of the F distribution are Control qr barcode image in vb.netusing visual .net toassign qr bidimensional barcode in asp.net web,windows applicationv (v v 4 Linear barcode library on .netuse .net framework linear barcode generator tocreate 1d barcode for .net2) for v Visual .net qr code jis x 0510 creation on .netusing barcode integrating for vs .net control to generate, create qr code 2d barcode image in vs .net applications.2, and Create barcode in .netusing .net framework crystal toincoporate bar code for asp.net web,windows application2v 2 1u v u1v 22 2 1v UPC A printer on .netuse .net vs 2010 ucc - 12 printing toproduce gs1 - 12 with .net22 , 42Create planet on .netusing barcode integration for .net crystal control to generate, create planet image in .net crystal applications.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 fControl gtin - 128 data for .net ean 128 barcode data with .net, u, v 2 Control ansi/aim code 128 data for .net code 128 barcode data in .net,u,v None ean / ucc - 14 reader with noneUsing Barcode Control SDK for None Control to generate, create, read, scan barcode image in None applications.f 1x2 dx 5 and v 10, we nd from Table V of the Control barcode code39 data with microsoft excelto integrate 3 of 9 barcode and ansi/aim code 39 data, size, image with excel spreadsheets barcode sdkThis is illustrated in Fig. 10-5. For example, if u Appendix that P1F f0.05,5,10 2Word linear 1d barcode creator for wordgenerate, create linear barcode none in word documents projectsP1F5,10QR Code JIS X 0510 barcode library for .netusing barcode printer for sql server reporting service control to generate, create denso qr bar code image in sql server reporting service applications.10-5 INFERENCES ON THE VARIANCES OF TWO NORMAL POPULATIONS Control uss code 39 size in office excel ansi/aim code 39 size in microsoft excelf (x) u = 5, v = 5 recognizing code 39 in .netUsing Barcode reader for .net framework Control to read, scan read, scan image in .net framework applications.u = 5, v = 15 f ( x)f1 , u, v f , u, v 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.,u,v (10-28),v,u 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 0.21110-5.2 10-5.3Development 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-CHAPTER 10 STATISTICAL INFERENCE FOR TWO SAMPLES 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 2EXAMPLE 10-11Oxide 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 20.05 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 2f0 8.
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