Question

We consider a multiple linear regression model with LIFE (y) as the response variable, and MALE (x1), BIRTH (x2), DIVO (x3), BEDS (x4), EDUC (x5), and INCO (x6), as predictors.

"STATE" "MALE" "BIRTH" "DIVO" "BEDS" "EDUC" "INCO" "LIFE"
AK 119.1 24.8 5.6 603.3 14.1 4638 69.31
AL 93.3 19.4 4.4 840.9 7.8 2892 69.05
AR 94.1 18.5 4.8 569.6 6.7 2791 70.66
AZ 96.8 21.2 7.2 536.0 12.6 3614 70.55
CA 96.8 18.2 5.7 649.5 13.4 4423 71.71
CO 97.5 18.8 4.7 717.7 14.9 3838 72.06
CT 94.2 16.7 1.9 791.6 13.7 4871 72.48
DC 86.8 20.1 3.0 1859.4 17.8 4644 65.71
DE 95.2 19.2 3.2 926.8 13.1 4468 70.06
FL 93.2 16.9 5.5 668.2 10.3 3698 70.66
GA 94.6 21.1 4.1 705.4 9.2 3300 68.54
HW 108.1 21.3 3.4 794.3 14.0 4599 73.60
IA 94.6 17.1 2.5 773.9 9.1 3643 72.56
ID 99.7 20.3 5.1 541.5 10.0 3243 71.87
IL 94.2 18.5 3.3 871.0 10.3 4446 70.14
IN 95.1 19.1 2.9 736.1 8.3 3709 70.88
KS 96.2 17.0 3.9 854.6 11.4 3725 72.58
KY 96.3 18.7 3.3 661.9 7.2 3076 70.10
LA 94.7 20.4 1.4 724.0 9.0 3023 68.76
MA 91.6 16.6 1.9 1103.8 12.6 4276 71.83
MD 95.5 17.5 2.4 841.3 13.9 4267 70.22
ME 94.8 17.9 3.9 919.5 8.4 3250 70.93
MI 96.1 19.4 3.4 754.7 9.4 4041 70.63
MN 96.0 18.0 2.2 905.4 11.1 3819 72.96
MO 93.2 17.3 3.8 801.6 9.0 3654 70.69
MS 94.0 22.1 3.7 763.1 8.1 2547 68.09
MT 99.9 18.2 4.4 668.7 11.0 3395 70.56
NC 95.9 19.3 2.7 658.8 8.5 3200 69.21
ND 101.8 17.6 1.6 959.9 8.4 3077 72.79
NE 95.4 17.3 2.5 866.1 9.6 3657 72.60
NH 95.7 17.9 3.3 878.2 10.9 3720 71.23
NJ 93.7 16.8 1.5 713.1 11.8 4684 70.93
NM 97.2 21.7 4.3 560.9 12.7 3045 70.32
NV 102.8 19.6 18.7 560.7 10.8 4583 69.03
NY 91.5 17.4 1.4 1056.2 11.9 4605 70.55
OH 94.1 18.7 3.7 751.0 9.3 3949 70.82
OK 94.9 17.5 6.6 664.6 10.0 3341 71.42
OR 95.9 16.8 4.6 607.1 11.8 3677 72.13
PA 92.4 16.3 1.9 948.9 8.7 3879 70.43
RI 96.2 16.5 1.8 960.5 9.4 3878 71.90
SC 96.5 20.1 2.2 739.9 9.0 2951 67.96
SD 98.4 17.6 2.0 984.7 8.6 3108 72.08
TN 93.7 18.4 4.2 831.6 7.9 3079 70.11
TX 95.9 20.6 4.6 674.0 10.9 3507 70.90
UT 97.6 25.5 3.7 470.5 14.0 3169 72.90
VA 97.7 18.6 2.6 835.8 12.3 3677 70.08
VT 95.6 18.8 2.3 1026.1 11.5 3447 71.64
WA 98.7 17.8 5.2 556.4 12.7 3997 71.72
WI 96.3 17.6 2.0 814.7 9.8 3712 72.48
WV 93.9 17.8 3.2 950.4 6.8 3038 69.48
WY 100.7 19.6 5.4 925.9 11.8 3672 70.29

Please find the least-square regression using 1)lm()command in R, and 2)the matrix formulas(use R to compute) below. Please provide all the R commands you have used and a screenshot of the result. I just wanna verify what I did was right as I got two really different result referring to these two methods. Thanks

Least Squares Estimates (LSE) The least squares estimates of%,A, ,A, are the values of bo, bi,... , bp for which the sum of squared residuals (RSS) is a minimum bo bxpXip i-1 i-1 i-1 In term of matrix, we have RSS(β) = (Y-X8) (Y-X8) If the matrix X has full rank p the minimum of RSS(B) is taken at The fitted or predicted values are given by The residuals are

Answer 1

The following R code is used to complete the problem and the outputs are attached with it,

***********************************************************************************************************************************************************

#### reading data into R as 'd' #########

d=read.table("tt.txt",sep=" ",header=T)
data=d[,-1]                 ## data without STATE column
attach(data)

Y=LIFE
design=data[,-7]                         #Removing LIFE Column from design matrix
X=cbind(rep(1,51),design)          #Final Design Matrix with first Column as 1
X

rep(1, 51) MALE BIRTH DIVO BEDS EDUC INCO 1 119.1 24.8 5.6 603.3 14.1 4638 1 93.3 19.4 4.4 840.9 7.8 2892 1 94.1 18.5 4.8 569.6 6.7 2791 1 96.8 21.2 7.2 536.0 12.6 3614 1 96.8 18.2 5.7649.5 13.4 4423 1 97.5 18.8 4.7717.7 14.9 3838 1 94.2 16.7 1.9 791.6 13.7 4871 1 86.8 20.1 3.0 1859.4 17.8 4644 1 95.2 19.2 3.2 926.8 13.1 4468 1 93.2 16.9 5.5 668.2 10.3 3698 1 94.6 21.1 4.1 705.4 9.2 3300 1 108.1 21.3 3.4 794.3 14.0 4599 1 94.6 17.1 2.5 773.9 9.1 3643 1 99.7 20.3 5.1 541.5 10.0 3243 1 94.2 18.5 3.3 871.0 10.3 4446 1 95.1 19.1 2.9 736.1 8.3 3709 1 96.2 17.0 3.9 854.6 11.4 3725 1 96.3 18.7 3.3 661.9 7.2 3076 1 94.7 20.4 1.4 724.0 9.0 3023 1 91.6 16.6 1.9 1103.8 12.6 4276 1 95.5 17.5 2.4 841.3 13.9 4267 1 94.8 17.9 3.9 919.5 8.4 3250 1 96.1 19.4 3.4 754.7 9.4 4041 1 96.0 18.0 2.2 905.4 11.1 3819 1 93.2 17.3 3.8 801.6 9.0 3654 1 94.0 22.1 3.7763.1 8.1 2547 1 99.9 18.2 4.4 668.7 11.0 3395 1 95.9 19.3 2.7658.8 8.5 3200 1 101.8 17.6 1.6 959.9 8.4 3077 1 95.4 17.3 2.5 866.1 9.6 3657 1 95.7 17.9 3.3 878.2 10.9 3720 1 93.7 16.8 1.5 713.1 11.8 4684 1 97.2 21.7 4.3 560.9 12.7 3045 1 102819 6 18 7 560 7 108 458 1 4 10 12 13 14 15 16 18 19 20 21 23 24 25 26 27 28 29 30 31 32

inverse=solve(t(X)%*% as.matrix(X),diag(1,7)) ## inverse of X'X
beta_hat=inverse%*%t(X)%*%Y                        ## coefficients using the formula]
beta_hat

> beta hat rep(1, 51) 70.5577812705 BIRTH DIVO BEDS EDUC INCO 0.1261018758 -0.5160557876 -0.1965375074 0.0033392036 0.2368222541 -0.0003612011

#### Now computing using lm() function,

beta_lm =coefficients(lm(LIFE~.,data))|
beta_lm

> beta Im - (Intercept) BIRTH BEDS EDUC INCO 70.5577812704 0.1261018758 -0.5160557876 -0.1965375074 -0.0033392036 0.2368222541 -0.0003612011 MALE DIVO

**************************************************END*********************************************************************************************

Conclusion : the above code is synonymous to the problem and gives correct and similar result through both approaches. Note : the variable 'd' stores the entire data as listed in the question.