ssrs 2012 barcode font In Problem 11.16, construct a 95% credibility interval for u. in Software

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11.63. In Problem 11.16, construct a 95% credibility interval for u.
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From Problem 11.16, we see that the posterior mean and variance for u are 1.17 and 0.006. The 95% credibility interval is [1.17 (1.96 0.077), 1.17 (1.96 0.077)] [1.02, 1.32].
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11.64. In Problem 11.25, what can you say about the HPD Bayesian credibility interval for u compared to the conventional interval shown in (1), 6
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The posterior distribution of u is normal with mean x and variance s2 >n. The HPD credibility intervals we # obtain would be identical to the conventional confidence intervals centered at x. #
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11.65. The number of individuals in a year who will suffer a bad reaction from injection of a given serum has a Poisson distribution with unknown mean l. Assume that l has Jeffreys improper prior density p(l) 1> !l, l 0 (see Problem 11.21). Table 11-8 gives the number of such cases that occurred in each of the past 10 years. (a) Derive a 98% equal tail credibility interval for l. (b) With what degree of credibility can you assert that l does not exceed 3 Table 11-8
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Year Number 1 2 2 4 3 1 4 2 5 2 6 1 7 2 8 3 9 3 10 0
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(a) We know from Problem 11.21 that the posterior distribution for l is gamma with parameters nx 1 and # 2 1 n , which in our case are 20.5 and 0.1. We thus need the 1st and 99th percentiles of the gamma distribution with these parameters. Using computer software, we get x0.01 1.146 and x0.99 3.248. The 98% credibility interval is [1.146, 3.248].
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CHAPTER 11 Bayesian Methods
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(b) We need the posterior probability that l does not exceed 3. This is the area to the left of 3 under the gamma density with parameters 20.5 and 0.1. Since this area is 0.972, we can be about 97% certain that l does not exceed 3.
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11.66. In Problem 11.14, obtain the 95% Bayesian equal tail area credibility interval for l.
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The posterior density was obtained in Problem 11.14 as a gamma with parameters 25 and 0.091. The percentiles of this density relevant for our credibility interval are x0.975 3.25 and x0.025 1.47. The 95% Bayesian credibility interval is [1.47, 3.25].
11.67. Obtain an equal tail 95% credibility interval for u in Problem 11.22 assuming n
The posterior is beta with parameters 3 and 7. The percentiles are x0.025 interval is [0.075, 0.600].
10, x
0.075 and x0.975
0.600. The
11.68. Obtain an equal tail area 95% credibility interval for u in Problem 11.23 assuming n
The posterior is beta with parameters 3.5 and 7.5. The percentiles are x0.025 interval is [0.093, 0.606]. 0.093 and x0.975
10, x
0.606. The
11.69. In Problem 11.48, obtain a 99% equal tail area credibility interval for (a) u and (b) 1>u.
(a) The posterior distribution of u is gamma with parameters 10.16 and 0.04. We obtain the following percentiles of this distribution using computer software: x0.005 0.15 and x0.995 0.81. The credibility interval is [0.15, 0.81]. (b) Since u 0.15 3 1>u 1>0.15 and u [1>0.81, 1>0.15] [1.23, 6.67]. 0.81 3 1>u 1>0.81, the equal tail area interval for 1>u is
Bayesian hypothesis tests 11.70. The mean lifetime (in hours) of fluorescent light bulbs produced by a company is known to be normally distributed with an unknown mean u but a known standard deviation of 120 hours. The prior density of u is normal with m 1580 hours and y2 16900. A mean lifetime of a sample of 100 light bulbs is computed to be 1630 hours. Test the null hypothesis H0:u 1600 against the alternative hypothesis H1:u 1600 using a Bayes (a) 0.05 test and (b) 0.01 test.
(a) By Theorem 11.3, the posterior density is normal with mean 1629.58 and standard deviation 11.95. The 1600 1629.58 posterior probability of H0 is P(u 1600 ux) P Z < 0.007. Since this 11.95 probability is less than 0.05, we can reject H0.
(b) Since the posterior probability of the null hypothesis, obtained in (a), is less than 0.01, we can reject H0. 11.71. Suppose that in Example 11.18 a second sample of 100 observations yielded a mean reaction time of 0.25 sec. Test the null hypothesis H0:u 0.3 against the alternative H1:u 0.3 using a Bayes 0.05 test.
We take the prior distribution of u to be the posterior distribution obtained in Example 11.18: Normal with mean 0.352 and variance 0.004. Applying Theorem 11-3 with this prior and the new data, we get a posterior mean 0.269 and variance 0.0007. The posterior probability of the null hypothesis is 0.88. Since this is not less than 0.05, we cannot reject the null hypothesis.
11.72. In Problem 11.21, suppose a sample of size 10 yielded the values 2, 0, 1, 1, 3, 0, 2, 4, 2, 2. Test H0:l against H1:l 1 using a Bayes 0.05 test.
We need the posterior probability of H0. From Problem 11.21, it is the area from 0 to 1 under a gamma density 1 with parameters nx 1 17.5 and n 0.1. Using computer software, we see this probability is 0.02. Since # 2 this is less than the specified threshold of 0.05, we reject the null hypothesis.
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