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ssrs 2012 barcode font In Problem 11.16, construct a 95% credibility interval for u. in Software
11.63. In Problem 11.16, construct a 95% credibility interval for u. QR Code JIS X 0510 Recognizer In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Making Denso QR Bar Code In None Using Barcode printer for Software Control to generate, create QRCode image in Software applications. 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]. QR Code Recognizer In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Encoding QRCode In Visual C# Using Barcode generator for .NET Control to generate, create QR image in .NET applications. 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 QR Creator In VS .NET Using Barcode generator for ASP.NET Control to generate, create Denso QR Bar Code image in ASP.NET applications. Quick Response Code Creator In .NET Framework Using Barcode encoder for VS .NET Control to generate, create QR Code 2d barcode image in .NET applications. 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. # Making Quick Response Code In VB.NET Using Barcode maker for VS .NET Control to generate, create Denso QR Bar Code image in VS .NET applications. Generating ANSI/AIM Code 128 In None Using Barcode encoder for Software Control to generate, create Code 128A image in Software applications. 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 118 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 118 Draw Code 39 In None Using Barcode creation for Software Control to generate, create Code 39 Extended image in Software applications. Printing Bar Code In None Using Barcode printer for Software Control to generate, create bar code image in Software applications. Year Number 1 2 2 4 3 1 4 2 5 2 6 1 7 2 8 3 9 3 10 0
EAN 13 Creator In None Using Barcode maker for Software Control to generate, create EAN13 image in Software applications. Draw GS1 128 In None Using Barcode generator for Software Control to generate, create GS1128 image in Software applications. (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]. RoyalMail4SCC Creation In None Using Barcode encoder for Software Control to generate, create RM4SCC image in Software applications. Paint Data Matrix ECC200 In None Using Barcode encoder for Online Control to generate, create Data Matrix ECC200 image in Online applications. CHAPTER 11 Bayesian Methods
ANSI/AIM Code 128 Drawer In Java Using Barcode maker for Android Control to generate, create Code 128C image in Android applications. Scanning Code 3 Of 9 In Visual Basic .NET Using Barcode reader for .NET Control to read, scan read, scan image in VS .NET applications. (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. Decode Code128 In .NET Framework Using Barcode recognizer for VS .NET Control to read, scan read, scan image in Visual Studio .NET applications. Bar Code Generator In .NET Using Barcode drawer for Visual Studio .NET Control to generate, create bar code image in .NET applications. 11.66. In Problem 11.14, obtain the 95% Bayesian equal tail area credibility interval for l.
Linear 1D Barcode Generator In VS .NET Using Barcode creator for VS .NET Control to generate, create 1D image in .NET applications. Barcode Creation In None Using Barcode generator for Font Control to generate, create bar code image in Font applications. 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 113 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.

