Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.Chapter 9Tests of Hypotheses for a Single SampleApplied Statistics and Probability for EngineersSixth EditionDouglas C. Montgomery George C. RungerCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.9-3 Tests on the Mean of a Normal Distribution, Variance Unknown9-3.1 Hypothesis Tests on the MeanOne-Sample t-Test2Sec 9-3 Tests on the Mean of a Normal Distribution, Variance UnknownConsider the two-sided hypothesis test H0: 0 and H1: ≠ 0Test statistic:Alternative hypothesis Rejection criteriaH1: 0 t0 t/2,n-1or t0 t/2,n1H1: 0t0 t,n1H1: 0t0 t,n1nSXT/00Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.EXAMPLE 9-6 Golf Club Design3Sec 9-3 Tests on the Mean of a Normal Distribution, Variance UnknownAn experiment was performed in which 15 drivers produced by a particular club maker were selected at random and their coefficients of restitution measured. It is of interest to determine if there is evidence (with 0.05) to support a claim that the mean coefficient of restitution exceeds 0.82. The observations are:0.8411 0.8191 0.8182 0.8125 0.87500.8580 0.8532 0.8483 0.8276 0.79830.8042 0.8730 0.8282 0.8359 0.8660The sample mean and sample standard deviation are and s = 0.02456. The objective of the experimenter is to demonstrate that the mean coefficient of restitution exceeds 0.82, hence a one-sided alternative hypothesis is appropriate.83725.0xThe seven-step procedure for hypothesis testing is as follows:1. Parameter of interest: The parameter of interest is the mean coefficient of restitution, .2. Null hypothesis: H0: 0.823. Alternative hypothesis: H1: 0.82Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.4Sec 9-3 Tests on the Mean of a Normal Distribution, Variance Unknown4. Test Statistic: The test statistic is5. Reject H0if: Reject H0if the P-value is less than 0.05.6. Computations: Since , s = 0.02456, = 0.82, and n 15, we have7. Conclusions: From Appendix A Table II, for a t distribution with 14 degrees of freedom, t0 2.72 falls between two values: 2.624, for which 0.01, and 2.977, for which 0.005. Since, this is a one-tailed test the P-value is between those two values, that is, 0.005 < P < 0.01. Therefore, since P < 0.05, we reject H0and conclude that the mean coefficient of restitution exceeds 0.82.Interpretation: There is strong evidence to conclude that the mean coefficient of restitution exceeds 0.82.nsxt/0083725.0x72.215/02456.082.083725.00tEXAMPLE 9-6 Golf Club Design - ContinuedCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.9-3 Tests on the Mean of a Normal Distribution, Variance Unknown9-3.2 Type II Error and Choice of Sample SizeThe type II error of the two-sided alternative would be5Sec 9-3 Tests on the Mean of a Normal Distribution, Variance Unknown/2, 1 0 /2, 1/2, 1 0 /2, 1( | 0)()nnnnP t T tP t T t Curves are provided for two-sided alternatives on Charts VIIe and VIIf . Theabscissa scale factor d on these charts is defined asFor the one-sided alternative , use charts VIIg and VIIh. Theabscissa scale factor d on these charts is defined as0d 00 or 0d Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.9-3 Tests on the Mean of a Normal Distribution, Variance Unknown6Sec 9-3 Tests on the Mean of a Normal Distribution, Variance UnknownEXAMPLE 9-7 Golf Club Design Sample SizeConsider the golf club testing problem from Example 9-6. If the mean coefficient of restitution exceeds 0.82 by as much as 0.02, is the sample size n 15 adequate to ensure that H0: 0.82 will be rejected with probability at least 0.8?To solve this problem, we will use the sample standard deviation s 0.02456 to estimate . Then d / 0.02/0.02456 0.81. By referring to the operating characteristic curves in Appendix Chart VIIg(for = 0.05) with d = 0.81 and n = 15, we find that = 0.10, approximately. Thus, the probability of rejecting H0: = 0.82 if the true mean exceeds this by 0.02 is approximately 1 = 1 0.10 = 0.90, and we conclude that a sample size of n = 15 is adequate to provide the desired sensitivity.Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution9-4.1 Hypothesis Test on the Variance7Sec 9-4 Tests of the Variance & Standard Deviation of a Normal DistributionSuppose that we wish to test the hypothesis that the variance of a normal population 2equals a specified value, say , or equivalently, that the standard deviation is equal to 0. Let X1, X2,... ,Xnbe a random sample of n observations from this population. To test(9-6)we will use the test statistic:(9-7)20212020::HH20220)1(SnXCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution9-4.1 Hypothesis Test on the Variance8Sec 9-4 Tests of the Variance & Standard Deviation of a Normal DistributionIf the null hypothesis is true, the test statistic defined in Equation 9-7 follows the chi-square distribution with n 1 degrees of freedom. This is the reference distribution for this test procedure. Therefore, we calculate , the value of the test statistic , and the null hypothesis would be rejected ifwhere and are the upper and lower 100 /2 percentage points of the chi-square distribution with n1 degrees of freedom, respectively. Figure 9-17(a) shows the critical region.2020: H20X20X2020: H2121ifornn21n21nCopyright © 2014 John Wiley & Sons, Inc. All rights reserved.9-4 Hypothesis Tests on the Variance and Standard Deviation of a Normal Distribution9-4.1 Hypothesis Test on the Variance9Sec 9-4 Tests of the Variance & Standard Deviation of a Normal DistributionThe same test statistic is used for one-sided alternative hypotheses. For the one-sided hypotheses.(9-8)we would reject H0 if , whereas for the other
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