Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Chapter 10Statistical Inference for Two SamplesApplied Statistics and Probability for EngineersSixth EditionDouglas C. Montgomery George C. RungerCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.10-4: Paired t-Test2Null hypothesis: H0: D 0Test statistic: Alternative Hypothesis P-ValueRejection Criterion for Fixed-Level TestsH1: D≠ 0Probability above t0 and probability belowt0H1: D 0Probability above t0H1: D 0Probability below t0nSDTD/00(10-24)1,201,20nnttortt1,0 ntt1,0 nttSec 10-4 Paired t-TestCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.Example 10-11 Shear Strength of Steel Girder3An article in the Journal of Strain Analysis [1983, Vol. 18(2)] reports a comparison of several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied to nine specific girders, are shown in the table below. Determine whether there is any difference (on the average) for the two methods.GirderKarlsruhe Method Lehigh Method Difference djS1/1 1.186 1.061 0.125S2/1 1.151 0.992 0.159S3/1 1.322 1.063 0.259S4/1 1.339 1.062 0.277S5/1 1.2 1.065 0.135S2/1 1.402 1.178 0.224S2/2 1.365 1.037 0.328S2/3 1.537 1.086 0.451S2/4 1.559 1.052 0.507Sec 10-4 Paired t-TestCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.Example 10-11 Shear Strength of Steel Girder4The seven-step procedure is:1. Parameter of interest: The parameter of interest is the difference in mean shear strength for the two methods.2. Null hypothesis: H0: µD= 03. Alternative hypothesis: H1: µD 04. Test statistic: The test statistic is5. Reject H0if: Reject H0if the P-value is <0.05.6. Computations: The sample average and standard deviation of the differences djare and sd= 0.1350, and so the test statistic is7. Conclusions: Because t0.0005.8= 5.041 and the value of the test statistic t0= 6.15 exceeds this value, the P-value is less than 2(0.0005) = 0.001. Therefore, we conclude that the strength prediction methods yield different results.Interpretation: The data indicate that the Karlsruhe method produces, on the average, higher strength predictions than does the Lehigh method.nsdtd/02769.0d15.69/1350.02769.0/0nsdtdSec 10-4 Paired t-TestCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.A Confidence Interval for D from Paired Samples5If and sDare the sample mean and standard deviation of the difference of n random pairs of normally distributed measurements, a 100(1 - α)% confidence interval on the difference in means 5 µD= µ1- µ2iswhere tα/2,n-1is the upper α /2% point of the t distribution with n - 1 degrees of freedom.dnstdnstdDnDDn//1/2,1,/2 (10-25)Sec 10-4 Paired t-TestCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.Example 10-12 Parallel Park Cars6The journal Human Factors (1962, pp. 375–380) reported a study in which n = 14 subjects were asked to parallel park two cars having very different wheel bases and turning radii. The time in seconds for each subject was recorded and is given in Table 10-4. From the column of observed differences, we calculate d = 1.21 and sD= 12.68. Find the 90% confidence interval for µD= µ1- µ2. 21.779.414/68.12771.121.114/68.12771.121.1//DDDDDnstdnstdNotice that the confidence interval on µD includes zero. This implies that, at the 90% level of confidence, the data do not support the claim that the two cars have different mean parking times µ1 and µ2. That is, the value µD= µ1- µ2 = 0 is not inconsistent with the observed data.Subject1(x1j)2(x2j)(dj)137.017.819.2225.820.25.6316.216.8-0.6424.241.4-17.2522.021.40.6633.438.4-5.0723.816.87.0858.232.226.0933.627.85.81024.423.21.21123.429.6-6.21221.220.60.61336.232.24.01429.853.8-24.0Table 10-4Sec 10-4 Paired t-TestCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.10-5.1 The F Distribution10-5 Inferences on the Variances of Two Normal PopulationsWe wish to test the hypotheses:72221122210::HHLet W and Y be independent chi-square random variables with u and v degrees of freedom respectively. Then the ratio(10-28)has the probability density function(10-29)and is said to follow the distribution with u degrees of freedom in the numerator and v degrees of freedom in the denominator. It is usually abbreviated as Fu,v.vYuWF// xxvuvuxvuvuxfvuuu0,1222)(/21/2)(/2Sec 10-5 Inferences on the Variances of Two Normal PopulationsCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.10-5.2 Hypothesis Tests on the Ratio of Two Variances8Let be a random sample from a normal population with mean µ1and variance , and let be a random sample from a second normal population with mean µ 2and variance . Assume that both normal populations are independent. Let and be the sample variances. Then the ratiohas an F distribution with n1 1 numerator degrees of freedom and n2 1 denominator degrees of freedom.111211,,,nXXX 21222221,,,nXXX 2221S22S22222121//SSFSec 10-5 Inferences on the Variances of Two Normal PopulationsCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.10-5.2 Hypothesis Tests on the Ratio of Two Variances9Null hypothesis: Test statistic: Alternative Hypotheses Rejection Criterion22210: H22210SSF 222112221122211:::HHH1,1,101,1,01,1/2,101,1/2,021212121ornnnnnnnnffffffff(10-31)Sec 10-5 Inferences on the Variances of Two Normal PopulationsCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.Example 10-13 Semiconductor Etch Variability10Oxide 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
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