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0.1153 8. Conclusions: Since P 0.1153 is not less than 0.05, we cannot reject the null hypothesis that the median shear strength is 2000 psi. Another way to say this is that 14 was not large or small enough to indithe observed number of plus signs r cate that median shear strength is different from 2000 psi at the 0.05 level of signi cance.

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It is also possible to construct a table of critical values for the sign test This table is shown as Appendix Table VII The use of this table for the two-sided alternative hypothesis in Equation 15-4 is simple As before, let R denote the number of the differences ( Xi 0) that are positive and let R denote the number of these differences that are negative Let R min (R , R ) Appendix Table VII presents critical values r* for the sign test that ensure that P (type I error) P (reject H0 when H0 is true) for 001, 005 and 010 If the observed value of the test statistic r r*, the null hypothesis H0: 0 should be rejected To illustrate how this table is used, refer to the data in Table 15-1 that was used in Example 15-1.

Now r 14 and r 6; therefore, r min (14, 6) 6 From Appendix Table VII with n 20 and 005, we nd that r*05 5 Since r 6 is not less than or 0 equal to the critical value r*05 5, we cannot reject the null hypothesis that the median shear 0 strength is 2000 psi We can also use Appendix Table VII for the sign test when a one-sided alternative hypothesis is appropriate If the alternative is H1: r* ; 0, reject H0: 0 if r if the alternative is H1: r* The level of significance of 0, reject H0: 0 if r a one-sided test is one-half the value for a two-sided test Appendix Table VII shows the one-sided significance levels in the column headings immediately below the two-sided levels.

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Finally, note that when a test statistic has a discrete distribution such as R does in the sign test, it may be impossible to choose a critical value r* that has a level of signi cance exactly equal to The approach used in Appendix Table VII is to choose r* to yield an that is as close to the advertised signi cance level as possible Ties in the Sign Test Since the underlying population is assumed to be continuous, there is a zero probability that we will nd a tie that is, a value of Xi exactly equal to 0 However, this may sometimes happen in practice because of the way the data are collected When ties occur, they should be set aside and the sign test applied to the remaining data..

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The Normal Approximation When p 0.5, the binomial distribution is well approximated by a normal distribution when n is at least 10. Thus, since the mean of the binomial is np and the variance is np(1 p), the distribution of R is approximately normal with mean 0.5n and variance 0.25n whenever n is moderately large. Therefore, in these cases the null hypothesis H0: 0 can be tested using the statistic

EXAMPLE 37-5

(15-5)

The two-sided alternative would be rejected if the observed value of the test statistic 0 z0 0 z 2, and the critical regions of the one-sided alternative would be chosen to re ect the sense of the alternative. (If the alternative is H1: z , for example.) 0, reject H0 if z0 EXAMPLE 15-2 We will illustrate the normal approximation procedure by applying it to the problem in Example 15-1. Recall that the data for this example are in Table 15-1. The eight-step procedure follows: 1. 2. 3. 4. 5. The parameter of interest is the median of the distribution of propellant shear strength. 2000 psi 2000 psi 0.05 The test statistic is z0 r 0.5n 0.51n z0.025 1.96. H0: H1:

(continued)

6. Since 0.05, we will reject H0 in favor of H1 if 0 z0 0 7. Computations: Since r 14, the test statistic is z0 8. 14 0.51202 1.789

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