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CHAPTER 7 Tests of Hypotheses and Significance
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(a) On the basis of a one-tailed test at a 0.01 level of significance, we would reject H0 if T were greater than t0.99, which, for n1 n2 2 12 12 2 22 degrees of freedom, is 2.51. Therefore, we cannot reject H0 at a 0.01 level of significance. (b) On the basis of one-tailed test at a 0.05 level of significance, we would reject H0 if T were greater than t0.95, which for 22 degrees of freedom is 1.72. Therefore, we can reject H0 at a 0.05 level of significance. We conclude that the improvement in yield of wheat by use of the fertilizer is probably significant. However before definite conclusions are drawn concerning the usefulness of the fertilizer, it may be desirable to have some further evidence. (c) The P value is P(T P 0.039. 1.85). The table in Appendix D shows 0.025 P 0.05. By computer software,
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Tests involving the chi-square distribution 7.20. In the past the standard deviation of weights of certain 40.0 oz packages filled by a machine was 0.25 oz. A random sample of 20 packages showed a standard deviation of 0.32 oz. Is the apparent increase in variability significant at the (a) 0.05, (b) 0.01 level of significance (c) What is the P value of the test
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We have to decide between the hypotheses H 0: s H1: s The value of x2 for the sample is x2 0.25 oz and the observed result is due to chance 0.25 oz and the variability has increased ns2 >s2 20(0.32)2 >(0.25)2 32.8.
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(a) Using a one-tailed test, we would reject H0 at a 0.05 level of significance if the sample value of x2 were greater than x2 , which equals 30.1 for n 20 1 19 degrees of freedom. Therefore, we would reject 0.95 H0 at a 0.05 level of significance. (b) Using a one-tailed test, we would reject H0 at a 0.01 level of significance if the sample value of x2 were greater than x2 ,which equals 36.2 for 19 degrees of freedom. Therefore, we would not reject H0 at a 0.01 0.99 level of significance. We conclude that the variability has probably increased. An examination of the machine should be made. (c) The P value is P(x2 P 0.0253. 32.8). The table in Appendix E shows 0.025 P 0.05. By computer software,
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Tests involving the F distribution 7.21. An instructor has two classes, A and B, in a particular subject. Class A has 16 students while class B has 25 students. On the same examination, although there was no significant difference in mean grades, class A had a standard deviation of 9 while class B had a standard deviation of 12. Can we conclude at the (a) 0.01, (b) 0.05 level of significance that the variability of class B is greater than that of A (c) What is the P value of the test
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(a) We have, on using subscripts 1 and 2 for classes A and B, respectively, s1
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We have to decide between the hypotheses H0: s1 H1: s2 s2, and any observed variability is due to chance s1, and the variability of class B is greater than that of A
The decision must therefore be based on a one-tailed test of the F distribution. For the samples in question,
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CHAPTER 7 Tests of Hypotheses and Significance
The number of degrees of freedom associated with the numerator is r2 25 1 24; for the denominator, r1 16 1 15. At the 0.01 level for 24, 15 degrees of freedom we have from Appendix F, F0.99 3.29. Then, since F F0.99, we cannot reject H0 at the 0.01 level.
(b) Since F0.95 2.29 for 24, 15 degrees of freedom (see Appendix F), we see that F reject H0 at the 0.05 level either. (c) The P value of the test is P(F software, P 0.134. 1.74). The tables in Appendix F show that P