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barcode font reporting services Under the hypothesis H0, mX1 in Software
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EAN 128 Generation In None Using Barcode generator for Office Excel Control to generate, create EAN128 image in Microsoft Excel applications. Paint USS128 In .NET Framework Using Barcode printer for ASP.NET Control to generate, create EAN / UCC  14 image in ASP.NET applications. CHAPTER 7 Tests of Hypotheses and Significance
(a) The observed difference will be significant at a 0.05 level if 0.7 !n 3.75 1.645 or !n 8.8 or n 78 50 28. Therefore, we must increase the sample size in each group by at least 78 (b) The observed difference will be significant at a 0.01 level if 0.7 !n 3.75 2.33 or !n 12.5 or n 157 50 107. Hence, we must increase the sample size in each group by at least 157
7.13. Two groups, A and B, each consist of 100 people who have a disease. A serum is given to Group A but not to Group B (which is called the control group); otherwise, the two groups are treated identically. It is found that in Groups A and B, 75 and 65 people, respectively, recover from the disease. Test the hypothesis that the serum helps to cure the disease using a level of significance of (a) 0.01, (b) 0.05, (c) 0.10. (d) Find the P value of the test. Let p1 and p2 denote, respectively, the population proportions cured by (1) using the serum, (2) not using the serum. We must decide between the two hypotheses H0: p1 H1: p1 p2, and observed differences are due to chance, i.e., the serum is ineffective p2, and the serum is effective Under the hypothesis H0, mP1 0 sP1 1 pqQ n 1 n2 R (0.70)(0.30)Q 1 100 1 R 100 0.0648
where we have used as an estimate of p the average proportion of cures in the two sample groups, given by (75 65) > 200 0.70, and where q 1 p 0.30. Then Z P1 P2 sP1 P2 0.750 0.650 0.0648 1.54 (a) On the basis of a onetailed test at a 0.01 level of significance, we would reject the hypothesis H0 only if the z score were greater than 2.33. Since the z score is only 1.54, we must conclude that the results are due to chance at this level of significance. (b) On the basis of a onetailed test at a 0.05 level of significance, we would reject H0 only if the z score were greater than 1.645. Hence, we must conclude that the results are due to chance at this level also. (c) If a onetailed test at a 0.10 level of significance were used, we would reject H0 only if the z score were greater than 1.28. Since this condition is satisfied, we would conclude that the serum is effective at a 0.10 level of significance. (d) The P value of the test is P(Z 1.54) 0.0618, which is the probability that a z score of 1.54 or higher in favor of the user group would occur by chance if H0 were true. Note that our conclusions above depended on how much we were willing to risk being wrong. If results are actually due to chance and we conclude that they are due to the serum (Type I error), we might proceed to give the serum to large groups of people, only to find then that it is actually ineffective. This is a risk that we are not always willing to assume. On the other hand, we could conclude that the serum does not help when it actually does help (Type II error). Such a conclusion is very dangerous, especially if human lives are at stake. 7.14. Work Problem 7.13 if each group consists of 300 people and if 225 people in Group A and 195 people in Group B are cured.

