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CHAPTER 7 Tests of Hypotheses and Significance
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Let n denote the required sample size and x the number of heads in n tosses above which we reject the hypothesis p 0.5. From Fig. 7-11, (1) Area under normal curve for p (2) Area under normal curve for p 0.5 to right of 0.6 to left of x np x !npq x np 0.5n 0.5 !n is 0.025. is 0.05.
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!npq Actually, we should have equated the area between (n x) 0.6n 0.49 !n and
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0.6n x 0.49 !n
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0.6n x 0.49 !n
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to 0.05; however (2) is a close approximation. Notice that by making the acceptance probability 0.05 in the worst case, p 0.6, we automatically make it 0.05 or less when p has any other value outside the range 0.4 to 0.6. Hence, a weighted average of all these probabilities, which represents the probability of a Type II error, will also be 0.05 or less. From (1), From (2), x 0.5n 0.5 !n 1.96 1.645 or or (3) x (4) x 0.5n 0.6n 0.980 !n. 0.806 !n.
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x 0.6n 0.49 !n
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Then from (3) and (4), n 318.98. It follows that the sample size must be at least 319, i.e., we must toss the coin at least 319 times. Putting n 319 in (3) or (4), x 177. For p 0.5, x np 177 159.5 17.5. Therefore, we adopt the following decision rule: (a) Accept the hypothesis p between 142 and 177. 0.5 if the number of heads in 319 tosses is in the range 159.5 17.5, i.e.,
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(b) Reject the hypothesis otherwise.
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Quality control charts 7.29. A machine is constructed to produce ball bearings having a mean diameter of 0.574 inch and a standard deviation of 0.008 inch. To determine whether the machine is in proper working order, a sample of 6 ball bearings is taken every 2 hours and the mean diameter is computed from this sample, (a) Design a decision rule whereby one can be fairly certain that the quality of the products is conforming to required standards, (b) Show how to represent graphically the decision rule in (a).
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# (a) With 99.73% confidence we can say that the sample mean X must lie in the range (mX 3sX) to (mX 3sX) or (m 3s> !n) to (m 3s> !n). Since m 0.574, s 0.008 and n 6, it follows that with 99.73% confidence the sample mean should lie between (0.574 0.024> !6) and (0.574 0.024> !6) or between 0.564 and 0.584 inches. Hence, our decision rule is as follows: (1) If a sample mean falls inside the range 0.564 to 0.584 inches, assume the machine is in proper working order. (2) Otherwise conclude that the machine is not in proper working order and seek to determine the reason. (b) A record of the sample means can be kept by means of a chart such as shown in Fig. 7-12, called a quality control chart. Each time a sample mean is computed, it is represented by a point. As long as the points lie between the lower limit 0.564 inch and upper limit 0.584 inch, the process is under control. When a point goes outside of these control limits (such as in the third sample taken on Thursday), there is a possibility that something is wrong and investigation is warranted. The control limits specified above are called the 99.73% confidence limits, or briefly the 3s limits. However, other confidence limits, such as 99% or 95% limits, can be determined as well. The choice in each case depends on particular circumstances.
CHAPTER 7 Tests of Hypotheses and Significance
Fig. 7-12
Fitting of data by theoretical distributions 7.30. Fit a binomial distribution to the data of Problem 5.30, page 176.
We have P(x heads in a toss of 5 pennies) f (x) 5Cx pxq5 x, where p and q are the respective probabilities of a head and tail on a single toss of a penny. The mean or expected number of heads is m np 5p. For the actual or observed frequency distribution, the mean number of heads is a fx af (38)(0) (144)(1) (342)(2) (287)(3) 1000 (164)(4) (25)(5) 2470 1000
Equating the theoretical and actual means, 5p 2.47 or p 0.494. Therefore, the fitted binomial distribution is given by f (x) 5Cx (0.494)x(0.506)5 x. In Table 7-5 these probabilities have been listed as well as the expected (theoretical) and actual frequencies. The fit is seen to be fair. The goodness of fit is investigated in Problem 7.43.
Table 7-5 Number of Heads (x) 0 1 2 3 4 5 P(x heads) 0.0332 0.1619 0.3162 0.3087 0.1507 0.0294 Expected Frequency 33.2 or 33 161.9 or 162 316.2 or 316 308.7 or 309 150.7 or 151 29.4 or 29 Observed Frequency 38 144 342 287 164 25
7.31. Use probability graph paper to determine whether the frequency distribution of Table 5-2, page 161, can be closely approximated by a normal distribution.
First the given frequency distribution is converted into a cumulative relative frequency distribution, as shown in Table 7-6. Then the cumulative relative frequencies expressed as percentages are plotted against upper class boundaries on special probability graph paper as shown in Fig. 7-13. The degree to which all plotted points lie on a straight line determines the closeness of fit of the given distribution to a normal distribution. It is seen that there is a normal distribution which fits the data closely. See Problem 7.32.
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