95 90 80 70 60 50 40 30 20 10 5 1 0.78 0.83 Coefficient of restitution 0.88 in .NET

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99 95 90 80 70 60 50 40 30 20 10 5 1 0.78 0.83 Coefficient of restitution 0.88
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Figure 9-9. Normal probability plot of the coef cient of restitution data from Example 9-6.
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9-3 TESTS ON THE MEAN OF A NORMAL DISTRIBUTION, VARIANCE UNKNOWN
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Computations: Since x t0
0.83725, s
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0.83725 0.82 0.02456 115
0.02456,
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0.82, and n
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15, we have
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Conclusions: Since t0 2.72 1.761, we reject H0 and conclude at the 0.05 level of signi cance that the mean coef cient of restitution exceeds 0.82.
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Minitab will conduct the one-sample t-test. The output from this software package is in the following display: One-Sample T: COR Test of mu Variable COR Variable COR 0.82 vs mu N 15 0.82 Mean 0.83725 StDev 0.02456 T 2.72 SE Mean 0.00634 P 0.008
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95.0% Lower Bound 0.82608
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Notice that Minitab computes both the test statistic T0 and a 95% lower con dence bound for the coef cient of restitution. Because the 95% lower con dence bound exceeds 0.82, we would reject the hypothesis that H0: 0.82 and conclude that the alternative hypothesis H1: 0.82 is true. Minitab also calculates a P-value for the test statistic T0. In the next section we explain how this is done.
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9-3.2
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P-Value for a t-Test
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The P-value for a t-test is just the smallest level of signi cance at which the null hypothesis would be rejected. That is, it is the tail area beyond the value of the test statistic t0 for a onesided test or twice this area for a two-sided test. Because the t-table in Appendix Table IV contains only 10 critical values for each t distribution, computation of the exact P-value directly from the table is usually impossible. However, it is easy to nd upper and lower bounds on the P-value from this table. To illustrate, consider the t-test based on 14 degrees of freedom in Example 9-6. The relevant critical values from Appendix Table IV are as follows:
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Critical Value: Tail Area: 0.258 0.40 0.692 0.25 1.345 0.10 1.761 0.05 2.145 0.025 2.624 0.01 2.977 0.005 3.326 0.0025 3.787 0.001 4.140 0.0005
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Notice that t0 2.72 in Example 9-6, and that this is between two tabulated values, 2.624 and 2.977. Therefore, the P-value must be between 0.01 and 0.005. These are effectively the upper and lower bounds on the P-value. Example 9-6 is an upper-tailed test. If the test is lower-tailed, just change the sign of t0 and proceed as above. Remember that for a two-tailed test the level of signi cance associated with a particular critical value is twice the corresponding tail area in the column heading. This consideration must be taken into account when we compute the bound on the P-value. For example, suppose that t0 2.72 for a two-tailed alternate based on 14 degrees of freedom. The value t0 2.624 (corresponding to 0.02) and t0 2.977 (corresponding to 0.01), so the lower and upper bounds on the P-value would be 0.01 P 0.02 for this case.
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CHAPTER 9 TESTS OF HYPOTHESES FOR A SINGLE SAMPLE
Finally, most computer programs report P-values along with the computed value of the test statistic. Some hand-held calculators also have this capability. In Example 9-6, Minitab gave the P-value for the value t0 2.72 in Example 9-6 as 0.008.
9-3.3
Choice of Sample Size
The type II error probability for tests on the mean of a normal distribution with unknown variance depends on the distribution of the test statistic in Equation 9-23 when the null hypothesis H0: , the distribution for T0 0 is false. When the true value of the mean is 0 is called the noncentral t distribution with n 1 degrees of freedom and noncentrality parameter 1n . Note that if 0, the noncentral t distribution reduces to the usual central t distribution. Therefore, the type II error of the two-sided alternative (for example) would be P5 t P5 t
2,n 1 2,n 1
T0 T0
2,n 1 0
2,n 1 6
where T denotes the noncentral t random variable. Finding the type II error probability for 0 the t-test involves nding the probability contained between two points of the noncentral t distribution. Because the noncentral t-random variable has a messy density function, this integration must be done numerically. Fortunately, this ugly task has already been done, and the results are summarized in a series of O.C. curves in Appendix Charts VIe, VIf, VIg, and VIh that plot for the t-test against a parameter d for various sample sizes n. Curves are provided for two-sided alternatives on Charts VIe and VIf. The abscissa scale factor d on these charts is de ned as d For the one-sided alternative or 0 0
(9-24)
we use charts VIg and VIh with 0 0 (9-25)
We note that d depends on the unknown parameter 2. We can avoid this dif culty in several ways. In some cases, we may use the results of a previous experiment or prior information to make a rough initial estimate of 2. If we are interested in evaluating test performance after the data have been collected, we could use the sample variance s2 to estimate 2. If there is no previous experience on which to draw in estimating 2, we then de ne the difference in the mean d that we wish to detect relative to . For example, if we wish to detect a small difference in the mean, we might use a value of d 0 0 1 (for example), whereas if we are interested in detecting only moderately large differences in the mean, we might select d 0 0 2 (for example). That is, it is the value of the ratio 0 0 that is important in determining sample size, and if it is possible to specify the relative size of the difference in means that we are interested in detecting, then a proper value of d can usually be selected. EXAMPLE 9-7 Consider the golf club testing problem from Example 9-6. If the mean coef cient 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 0 0.02 0.02456 0.81. By referring to the operating characteristic curves in Appendix Chart VIg (for 0.05) with d 0.81 and n 15, we nd that 0.10,