Stat 512 Homework 7 Solutions Usa

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 jointcoverage 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 dfE= n – 2 = 16, and { }ˆˆ1.0878,71.3691hhsYY==when Xh= 5. To obtain the 99% CI we use tc= 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)-1X’Y. (If you have trouble with matrix multiplication, see pages 4-5 of Topic 3.) X Y2 14 26 58 7109

Statistics 512: Solution to Homework#6 For the following 3 problems use the computer science data that we have been dis-cussing in class. You can get a copy of the data set csdata.dat from the class website. The variables are: id , a numerical identifier for each student; GPA , the grade point average after three semesters; HSM ; HSS ; HSE ; SATM ; SATV , which were all explained in class; and GENDER , coded as 1 for men and 2 for women. 1. In a data step, create a new variable GENDERW that has values 1 for women and 0 for men (use arithmetic on the original variable GENDER ). Run a regression to predict GPA using the explanatory variables HSM , HSS , HSE , SATM , SATV , and GENDERW . (Do not include any interaction terms.) Solution: The required SAS output is given below: Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > |t| Intercept 1 0.31101 0.40447 0.77 0.4428 genderw 1 0.03237 0.11148 0.29 0.7718 hsm 1 0.14423 0.03979 3.62 0.0004 hss 1 0.03827 0.03874 0.99 0.3244 hse 1 0.05103 0.04228 1.21 0.2287 satm 1 0.00100 0.00071725 1.40 0.1633 satv 1-0.00041086 0.00059323-0.69 0.4893 (a) Give the equation of the fitted regression line using all six explanatory variables. Solution: The estimated regression equation is gpa = 0 . 31101 + 0 . 003237 genderw + 0 . 11423 hsm + 0 . 03827 hss +0 . 05103 hse + 0 . 001 satm-0. 00041086 satv (b) Give the fitted regression line for women (use part a). Solution:

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