Homework04Solns - Stat 512-2 Solutions to Homework #4 Dr....

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Stat 512-2 Solutions to Homework #4 Dr. Simonsen Due Wednesday, September 21, 2005. 1. Consider the following SAS output giving 5 confidence intervals for the mean of Y. If you wanted to guarantee that the joint coverage of the five confidence intervals was at least 95%, what level would you use when forming each interval, using the Bonferroni correction? Compute this adjusted confidence interval for X = 5. (Note that some observations have been omitted from the output.) Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model 1 16183 805.62 <.0001 Error 16 321.39597 20.08725 Corrected Total 17 16504 Root MSE 4.48188 R-Square 0.9805 Dependent Mean 64.00000 Adj R-Sq 0.9793 Coeff Var 7.00294 Parameter Estimates Parameter Standard Variable Estimate t Value Pr > |t| Intercept -2.32215 2.56435 -0.91 0.3786 x 14.73826 0.51926 28.38 Output Statistics Dep Var Predicted Std Error Obs y Value Mean Predict 95% CL Mean Residual 3 5 78.0000 71.3691 1.0878 69.0630 73.6752 6.6309 4 10.0000 12.4161 2.1021 7.9598 16.8724 -2.4161 6 62.0000 56.6309 54.3248 58.9370 5.3691 8 39.0000 41.8926 1.3125 39.1103 44.6750 -2.8926 10 2 33.0000 27.1544 1.6737 23.6064 30.7024 5.8456 To obtain a joint coverage probability of at least 95% for g = 5 intervals, we use the Bonferroni correction and construct individual confidence intervals with coverage probability 1- α /g = 1-0.05/5 = 1 – 0.01 = 0.99. Thus we would construct 99% confidence intervals for each value of X. [Note to grader: They can either say 99% confidence level or α = 0.01 in answer to “what level”.] From the SAS output we see that df E = n – 2 = 16, and { } ˆˆ 1.0878, 71.3691 hh sY Y == when X h = 5. To obtain the 99% CI we use t c = t(1- α /2g, n-2) = t(0.995, 16) = 2.921. Thus the 99% CI for the mean when X = 5 is 71.3691 ± 2.921×1.0878 = 71.3691 ± 3.1775 = [ 68.1916, 74.5466 ] 2. Based on the following small data set, construct the design matrix, X , its transpose X’ , and the matrices X’X , ( X’X ) -1 , X’Y , and b = ( X’X ) -1 X’Y . (If you have trouble with matrix multiplication, see pages 4-5 of Topic 3.) X Y 2 1 4 2 6 5 8 7 10 9
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12 14 16 18 11 0 ⎡⎤ ⎢⎥ = ⎣⎦ X , , 1111 1 24681 0 ′ = X 5 3 0 2 4 6 8 10 30 220 0 ⎤⎡ == ⎥⎢ ⎦⎣ XX () 1 2 11 3 220 30 1.1 0.15 1 10 20 30 5 3 1 0.15 0.025 5 220 30 20 40 −− = ×− = 1 2 12579 2 4 5 2 4 6 8 10 2 8 30 56 90 186 7 9 ++++ + +++ XY 1 1.1 0.15 24 26.4 27.9 1.5 0.15 0.025 186 3.6 4.65 1.05 ′′ = = + bX Y For the following 5 problems use the Commercial properties data described in KNNL with problem 6.18 on page 251 — 252. (Note: this dataset is not in NKNW. It replaces the Mathematicians salary dataset.) 3.
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This note was uploaded on 02/02/2012 for the course STAT stat512 taught by Professor Libo during the Spring '11 term at Purdue North Central.

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Homework04Solns - Stat 512-2 Solutions to Homework #4 Dr....

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