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barcode font reporting services under the hypothesis that the coin is fair. Then Z in Software
under the hypothesis that the coin is fair. Then Z Denso QR Bar Code Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Creating QR Code In None Using Barcode generator for Software Control to generate, create QR Code image in Software applications. CHAPTER 7 Tests of Hypotheses and Significance
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Code 3/9 Drawer In None Using Barcode maker for Software Control to generate, create ANSI/AIM Code 39 image in Software applications. UPC  13 Generation In None Using Barcode generation for Software Control to generate, create EAN / UCC  13 image in Software applications. (1) Accept the hypothesis that the coin is fair if the number of heads is between 24.16 and 39.84, i.e., between 25 and 39 inclusive. (2) Reject the hypothesis otherwise. Second method With probability 0.95, the number of heads will lie between m 1.96s and m 1.96s, i.e., np 1.96 !npq and np 1.96!npq or between 32 1.96(4) 24.16 and 32 1.96(4) 39.84, which leads to the above decision rule. Third method 1.96 Z 1.96 is equivalent to 1.96(4) or 32 1.96(4) X 32 decision rule. Generate Ames Code In None Using Barcode generator for Software Control to generate, create Code27 image in Software applications. Encode EAN13 In Visual C# Using Barcode drawer for Visual Studio .NET Control to generate, create GTIN  13 image in .NET framework applications. 1.96 (X 32)>4 1.96(4), i.e., 24.16 Recognize Code 128 In VB.NET Using Barcode reader for .NET framework Control to read, scan read, scan image in VS .NET applications. Paint GS1  12 In None Using Barcode printer for Microsoft Excel Control to generate, create Universal Product Code version A image in Microsoft Excel applications. 1.96. Consequently 1.96(4) (X 32) X 39.84, which also leads to the above
Code128 Generator In None Using Barcode printer for Font Control to generate, create USS Code 128 image in Font applications. UPC Code Maker In Java Using Barcode drawer for Eclipse BIRT Control to generate, create UPCA Supplement 2 image in BIRT applications. (b) If the level of significance is 0.01, each shaded area in Fig. 73 is 0.005. Then the area between 0 and z1 is 0.5000 0.0050 0.4950, and z1 2.58 (more exactly, 2.575). This can also be read from Table 71. Following the procedure in the second method of part (a), we see that with probability 0.99 the number of heads will lie between m 2.58s and m 2.58s, i.e., 32 2.58(4) 21.68 and 32 2.58(4) 42.32. Therefore, the decision rule becomes: (1) Accept the hypothesis if the number of heads is between 22 and 42 inclusive. (2) Reject the hypothesis otherwise. EAN13 Creator In None Using Barcode creator for Microsoft Word Control to generate, create EAN / UCC  13 image in Word applications. Generate EAN 128 In ObjectiveC Using Barcode encoder for iPad Control to generate, create EAN / UCC  14 image in iPad applications. 7.4. How could you design a decision rule in Problem 7.3 so as to avoid a Type II error
A Type II error is made by accepting a hypothesis when it should be rejected. To avoid this error, instead of accepting the hypothesis we simply do not reject it, which could mean that we are withholding any decision in this case. For example, we could word the decision rule of Problem 7.3(b) as: (1) Do not reject the hypothesis if the number of heads is between 22 and 42 inclusive. (2) Reject the hypothesis otherwise. In many practical instances, however, it is important to decide whether a hypothesis should be accepted or rejected. A complete discussion of such cases requires consideration of Type II errors (see Problems 7.23 to 7.25). 7.5. In an experiment on extrasensory perception (ESP) a subject in one room is asked to state the color (red or blue) of a card chosen from a deck of 50 wellshuffled cards by an individual in another room. It is unknown to the subject how many red or blue cards are in the deck. If the subject identifies 32 cards correctly, determine whether the results are significant at the (a) 0.05, (b) 0.01 level of significance. (c) Find and interpret the P value of the test. If p is the probability of the subject stating the color of a card correctly, then we have to decide between the following two hypotheses: H0: p H1: p 0.5, and the subject is simply guessing, i.e., results are due to chance 0.5, and the subject has powers of ESP. We choose a onetailed test, since we are not interested in ability to obtain extremely low scores but rather in ability to obtain high scores. If the hypothesis H0 is true, the mean and standard deviation of the number of cards identified correctly is given by m np 50(0.5) 25 and s !npq !50(0.5)(0.5) !12.5 3.54 CHAPTER 7 Tests of Hypotheses and Significance
(a) For a onetailed test at a level of significance of 0.05, we must choose z1 in Fig. 74 so that the shaded area in the critical region of high scores is 0.05. Then the area between 0 and z1 is 0.4500, and z1 1.645. This can also be read from Table 71. Fig. 74 Therefore, our decision rule or test of significance is: (1) If the z score observed is greater than 1.645, the results are significant at the 0.05 level and the individual has powers of ESP. (2) If the z score is less than 1.645, the results are due to chance, i.e., not significant at the 0.05 level. Since 32 in standard units is (32 25) > 3.54 1.98, which is greater than 1.645, decision (1) holds, i.e., we conclude at the 0.05 level that the individual has powers of ESP. Note that we should really apply a continuity correction, since 32 on a continuous scale is between 31.5 and 32.5. However, 31.5 has a standard score of (31.5 25) > 3.54 1.84, and so the same conclusion is reached. (b) If the level of significance is 0.01, then the area between 0 and z1 is 0.4900, and z1 2.33. Since 32 (or 31.5) in standard units is 1.98 (or 1.84), which is less than 2.33, we conclude that the results are not significant at the 0.01 level. Some statisticians adopt the terminology that results significant at the 0.01 level are highly significant, results significant at the 0.05 level but not at the 0.01 level are probably significant, while results significant at levels larger than 0.05 are not significant. According to this terminology, we would conclude that the above experimental results are probably significant, so that further investigations of the phenomena are probably warranted. (c) The P value of the test is the probability that the colors of 32 or more cards would, in a random selection, be identified correctly. The standard score of 32, taking into account the continuity correction is z 1.84. Therefore the P value is P(Z 1.84) 0.032. The statistician could say that on the basis of the experiment, the chances of being wrong in concluding that the individual has powers of ESP are about 3 in 100. 7.6. The manufacturer of a patent medicine claimed that it was 90% effective in relieving an allergy for a period of 8 hours. In a sample of 200 people who had the allergy, the medicine provided relief for 160 people. (a) Determine whether the manufacturer s claim is legitimate by using 0.01 as the level of significance. (b) Find the P value of the test. (a) Let p denote the probability of obtaining relief from the allergy by using the medicine. Then we must decide between the two hypotheses: H 0: p H1: p 0.9, and the claim is correct 0.9, and the claim is false We choose a onetailed test, since we are interested in determining whether the proportion of people relieved by the medicine is too low. If the level of significance is taken as 0.01, i.e., if the shaded area in Fig. 75 is 0.01, then z1 2.33 as can be seen from Problem 7.5(b) using the symmetry of the curve, or from Table 71. Fig. 75

