ssrs 2012 barcode font By Theorem 11-1, the posterior density of u is beta with a 25 and b 17. in Software

Printer QR Code in Software By Theorem 11-1, the posterior density of u is beta with a 25 and b 17.

By Theorem 11-1, the posterior density of u is beta with a 25 and b 17.
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11.11. A poll to predict the fate of a forthcoming referendum found that 480 out of 1000 people surveyed were in favor of the referendum. What are the chances that the referendum would be lost
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CHAPTER 11 Bayesian Methods
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Assume a vague prior distribution (uniform on [0, 1]) for the proportion u of people in the population who favor the referendum. The posterior distribution of u, given the poll result, is beta with parameters 481, 521. We need the probability that u 0.5. Computer software gives 0.90 for this probability, so we can be 90% sure that the referendum would lose.
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11.12. In the previous problem, suppose an additional 1000 people were surveyed and 530 were found to be in favor of the referendum. What can we conclude now
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We take the prior now to be beta with parameters 481 and 521. The posterior becomes beta with parameters 1011 and 991. The probability for u 0.5 is 0.33. This means there is only a 33% chance now of the referendum losing.
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Sampling from a Poisson population 11.13. The number of accidents during a six-month period at an intersection has a Poisson distribution with mean l. It is believed that l has a gamma prior density with parameters a 2 and b 5. If eight accidents were observed during the first six months of the year, find the (a) posterior density, (b) posterior mean, and (c) posterior variance.
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(a) We know from Theorem 11-2 that the posterior density is gamma with parameters nx # b>(1 nb) 5>6. (b) From (32), 4, the posterior mean 50>6 < 8.33 and (c) the posterior variance a 10 and
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250>36 < 6.94.
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11.14. The number of defects in a 1000-foot spool of yarn manufactured by a machine has a Poisson distribution with unknown mean l. The prior distribution of l is gamma with parameters a 2 and b 1. A total of 23 defects were found in a sample of 10 spools that were examined. Determine the posterior density of l.
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By Theorem 11-2, the posterior density is gamma with parameters nx # b>(1 nb) 1>11 < 0.091. a 23 2 25 and
Sampling from a normal population 11.15. A sample of 100 measurements of the diameter of a sphere gave a mean x 4.38 inch. Based on prior # experience, we know that the diameter is normally distributed with unknown mean u and variance 0.36. Determine the posterior density of u assuming a normal prior density with mean 4.5 inch and variance 0.4.
From Theorem 11-3, we see that the posterior density is normal with mean 4.381 and variance 0.004.
11.16. The reaction time of an individual to certain stimuli is known to be normally distributed with unknown mean u but a known standard deviation of 0.35 sec. A sample of 20 observations yielded a mean reaction time of 1.18 sec. Assume that the prior density of u is normal with mean m 1 sec and variance y2 0.13. Find the posterior density of u.
By Theorem 11-3, the posterior density is normal with mean 1.17 and variance 0.006.
11.17. A random sample of 25 observations is taken from a normal population with unknown mean u and variance 16. The prior distribution of u is standard normal. Find (a) the posterior mean and (b) its precision. (c) Find the precision of the maximum likelihood estimator.
(a) By Theorem 11-3, the posterior mean of u is 25 25 x 16 # 25x # . 41 25 16 41 25
25x # (b) The precision of the estimate is the reciprocal of its variance. The variance of 41 the precision is roughly 4.2.
0.24, so
(c) The maximum likelihood estimate of u is x. Its variance is 16 > 25, so the precision is about 1.6. #
11.17. X is normal with mean 0 and unknown precision j, which has prior gamma distribution with parameters a and b. Find the posterior distribution of j based on a random sample x (x1, x2, c, xn) from X.
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