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Tests of means and proportions using normal distributions 7.1. Find the probability of getting between 40 and 60 heads inclusive in 100 tosses of a fair coin.
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According to the binomial distribution the required probability is
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The mean and standard deviation of the number of heads in 100 tosses are given by m np 1 100 2 50 s !npq A
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1 40 2 5
Since np and nq are both greater than 5, the normal approximation to the binomial distribution can be used in evaluating the above sum. On a continuous scale, between 40 and 60 heads inclusive is the same as between 39.5 and 60.5 heads. 39.5 in standard units 39.5 60.5 50 50 60.5 in standard units 2.10 2.10 5 5 area under normal curve between z 2.10 and z 2.10 2(area between z 0 and z 2.10) 2(0.4821) 0.9642
Required probability
7.2. To test the hypothesis that a coin is fair, the following decision rules are adopted: (1) Accept the hypothesis if the number of heads in a single sample of 100 tosses is between 40 and 60 inclusive, (2) reject the hypothesis otherwise. (a) Find the probability of rejecting the hypothesis when it is actually correct. (b) Interpret graphically the decision rule and the result of part (a). (c) What conclusions would you draw if the sample of 100 tosses yielded 53 heads 60 heads (d) Could you be wrong in your conclusions to (c) Explain.
(a) By Problem 7.1, the probability of not getting between 40 and 60 heads inclusive if the coin is fair equals 1 0.9642 0.0358. Then the probability of rejecting the hypothesis when it is correct equals 0.0358.
CHAPTER 7 Tests of Hypotheses and Significance
(b) The decision rule is illustrated by Fig. 7-2, which shows the probability distribution of heads in 100 tosses of a fair coin.
Fig. 7-2
If a single sample of 100 tosses yields a z score between 2.10 and 2.10, we accept the hypothesis; otherwise we reject the hypothesis and decide that the coin is not fair. The error made in rejecting the hypothesis when it should be accepted is the Type I error of the decision rule; and the probability of making this error, equal to 0.0358 from part (a), is represented by the total shaded area of the figure. If a single sample of 100 tosses yields a number of heads whose z score lies in the shaded regions, we could say that this z score differed significantly from what would be expected if the hypothesis were true. For this reason the total shaded area (i.e., probability of a Type I error) is called the level of significance of the decision rule; it equals 0.0358 in this case. We therefore speak of rejecting the hypothesis at a 0.0358, or 3.58%, level of significance. (c) According to the decision rule, we would have to accept the hypothesis that the coin is fair in both cases. One might argue that if only one more head had been obtained, we would have rejected the hypothesis. This is what one must face when any sharp line of division is used in making decisions. (d) Yes. We could accept the hypothesis when it actually should be rejected, as would be the case, for example, when the probability of heads is really 0.7 instead of 0.5. The error made in accepting the hypothesis when it should be rejected is the Type II error of the decision. For further discussion see Problems 7.23 7.25.
7.3. Design a decision rule to test the hypothesis that a coin is fair if a sample of 64 tosses of the coin is taken and if a level of significance of (a) 0.05, (b) 0.01 is used.
(a) First method If the level of significance is 0.05, each shaded area in Fig. 7-3 is 0.025 by symmetry. Then the area between 0 and z1 is 0.5000 0.0250 0.4750, and z1 1.96. Thus a possible decision rule is: (1) Accept the hypothesis that the coin is fair if Z is between (2) Reject the hypothesis otherwise. 1.96 and 1.96.
Fig. 7-3
The critical values 1.96 and 1.96 can also be read from Table 7-1. To express this decision rule in terms of the number of heads to be obtained in 64 tosses of the coin, note that the mean and standard deviation of the exact binomial distribution of heads are given by m np 64(0.5) 32 and s (X !npq m) > s (X !64(0.5)(0.5) 32) > 4. 4
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