Question

In: Statistics and Probability

data file: "STATE" "MALE" "BIRTH" "DIVO" "BEDS" "EDUC" "INCO" "LIFE" AK 119.1 24.8 5.6 603.3 14.1...

data file:

"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

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. Answer the following questions using least square estimates in term of matrix formulas.

compute the following using matrix formulas,

  1. (a) Compute and report the least-squares estimates. Write down the least-squares regression equation.

  2. (b) Explain in context what the coefficients corresponding to MALE and BIRTH mean.

  3. (c) Compute the biased and the unbiased estimates of the error variance σ2.

  4. (d) Using the unbiased estimate of error variance, Compute the standard errors of the estimators of the regression coefficients.

  5. (e) Compute the coefficient of determination. Give a practical interpretation of your result.

Solutions

Expert Solution

R commands and outputs:

> d=read.table("data.txt",header=TRUE)

> dim(d

) [1] 51 8

> 51*8

[1] 408

> head(d)

head(d) STATE MALE BIRTH DIVO BEDS EDUC INCO LIFE

1 AK 119.1 24.8 5.6 603.3 14.1 4638 69.31

2 AL 93.3 19.4 4.4 840.9 7.8 2892 69.05

3 AR 94.1 18.5 4.8 569.6 6.7 2791 70.66

4 AZ 96.8 21.2 7.2 536.0 12.6 3614 70.55

5 CA 96.8 18.2 5.7 649.5 13.4 4423 71.71

6 CO 97.5 18.8 4.7 717.7 14.9 3838 72.06

> y=d$LIFE

> x1=d$MALE

> x2=d$BIRTH

> x3=d$DIVO

> x4=d$BEDS

> x5=d$EDUC

> x6=d$INCO

> ###Matrix notation

> ###betahats

> n=dim(d)[1]

> n=51

> I=rep(1,n)

> X=cbind(I,x1,x2,x3,x4,x5,x6)

> betahat=solve(t(X)%*%X)%*%t(X)%*%y

> ## Coefficients corresponding to MALE and BIRTH

> ## MALE: x1= 0.1261018758. As value of variable MALE increases, value of y (LIFE) increases. If MALE decreases decreases, y decreases

. > ## BIRTH: x2= -0.5160557876. As value of variable MALE increases, value of y (LIFE) decreases and vice-versa

. > k=6

> p=k+1

> SSRes=t(y)%*%y-t(betahat)%*%t(X)%*%y

> SSRes=as.numeric(SSRes)

> SSRes

[1] 60.80295

> MSRes=SSRes/(n-p)

> MSRes

[1] 1.381885

> sigsqhat=as.numeric(MSRes) #Unbiased estimate of error variance

> sigsqhat

[1] 1.381885

> Cov_bhat=sigsqhat*solve(t(X)%*%X)

> Cov_bhat

I x1 x2 x3 x4 x5 x6

I 18.4019304518 -1.633385e-01

-6.622173e-02 2.057874e-02

-2.046046e-03 9.566568e-02

-2.334589e-04 x1 -0.1633385203 2.230839e-03

-2.356535e-03 -2.153210e-04

1.563106e-05 3.974575e-04 -6.254433e-06

x2 -0.0662217348 -2.356535e-03

1.375400e-02 -1.089876e-03 6.602875e-06

-6.480967e-03 2.774321e-05

x3 0.0205787372 -2.153210e-04

-1.089876e-03 5.469090e-03 2.367058e-05

4.033179e-04 -6.281835e-06

x4 -0.0020460458 1.563106e-05

6.602875e-06 2.367058e-05 9.594796e-07

-1.537143e-05 -7.391819e-08

x5 0.0956656811 3.974575e-04

-6.480967e-03 4.033179e-04

-1.537143e-05 1.232599e-02

-3.600185e-05

x6 -0.0002334589 -6.254433e-06

2.774321e-05 -6.281835e-06

-7.391819e-08 -3.600185e-05

2.114108e-07

> v_bhat=c()

> for(j in 1:p)

+ {

+ v_bhat[j]=Cov_bhat[j,j]

+ }

> round(v_bhat,6)

[1] 18.401930 0.002231 0.013754 0.005469 0.000001 0.012326

[7] 0.000000

> SE=sqrt(v_bhat)

> SE

[1] 4.2897471315 0.0472317551

0.1172774622 0.0739532971

0.0009795303 0.1110224836

0.0004597943

> std_error=round(SE,6)

> std_error

[1] 4.289747 0.047232 0.117277 0.073953

0.000980 0.111022 0.000460

> b=c("beta0","beta1","beta2","beta3","beta4","beta5","beta6")

> data.frame(b,std_error)

b std_error

1 beta0 4.289747

2 beta1 0.047232

3 beta2 0.117277

4 beta3 0.073953

5 beta4 0.000980

6 beta5 0.111022

7 beta6 0.000460

> ##Coefficient of Determination

> SSTotal=t(y)%*%y-(sum(y))^2/n

> SSTotal=as.numeric(SSTotal)

> SSTotal

[1] 114.3972

> Rsq=1-(SSRes/SSTotal)

> Rsq

[1] 0.4684927

> ##In statistics,in regression analysis, "R squared", denoted as R² ,also known as the coefficient of determination, is the proportion of the variance in the dependent (response) variable that is predictable from the independent (regressor) variable.

>##This means that 46.84927 % of variation in the model is explained by regressors.

> AdjRsq=1-((SSRes/(n-p))/(SSTotal/(n-1)))

> AdjRsq

[1] 0.3960144

> ##One can cross-check using following:

> fit=lm(y~x1+x2+x3+x4+x5+x6)

> s=summary(fit)

> s Call:

lm(formula = y ~ x1 + x2 + x3 + x4 + x5 + x6)

Residuals:

Min 1Q Median 3Q Max

-2.5563 -0.6629 0.0755 0.6983 3.3215

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 70.5577813 4.2897471 16.448 < 2e-16 ***

x1 0.1261019 0.0472318 2.670 0.01059 *

x2 -0.5160558 0.1172775 -4.400 6.78e-05 ***

x3 -0.1965375 0.0739533 -2.658 0.01093 *

x4 -0.0033392 0.0009795 -3.409 0.00141 **

x5 0.2368223 0.1110225 2.133 0.03853 *

x6 -0.0003612 0.0004598 -0.786 0.43633

--- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.176 on 44

degrees of freedom

Multiple R-squared: 0.4685, Adjusted R-squared: 0.396

F-statistic: 6.464 on 6 and 44 DF, p-value: 6.112e-05

> aov(fit)

Call:

aov(formula = fit)

Terms:

x1 x2 x3 x4 x5 x6 Residuals

Sum of Squares 4.56322 24.40986 4.67299 12.23977 6.85561 0.85279 60.80295

Deg. of Freedom 1 1 1 1 1 1 44

Residual standard error: 1.175536

Estimated effects may be unbalanced


Related Solutions

data file: "STATE" "MALE" "BIRTH" "DIVO" "BEDS" "EDUC" "INCO" "LIFE" AK 119.1 24.8 5.6 603.3 14.1...
data file: "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...
data: "STATE" "MALE" "BIRTH" "DIVO" "BEDS" "EDUC" "INCO" "LIFE" AK 119.1 24.8 5.6 603.3 14.1 4638...
data: "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...
We consider the multiple linear regression with LIFE (y) as the response variable, and MALE, BIRTH,...
We consider the multiple linear regression with LIFE (y) as the response variable, and MALE, BIRTH, DIVO , BEDS, EDUC, and INCO, as predictors. QUESTION: Plot the standardized residuals against the fitted values. Are there any notable points. In particular look for points with large residuals or that may be influential. # please screenshot the Rcode for the plot. # data information are as follows: "STATE" "MALE" "BIRTH" "DIVO" "BEDS" "EDUC" "INCO" "LIFE" AK 119.1 24.8 5.6 603.3 14.1 4638...
A MLR model have LIFE (y) as the response variable, and MALE (x1), BIRTH (x2), DIVO...
A MLR model have LIFE (y) as the response variable, and MALE (x1), BIRTH (x2), DIVO (x3), BEDS (x4), EDUC (x5), and INCO (x6), as predictors. I know you can use first fit the model using lm(y~x) then use anova(model) to check the SSreg,my question is, what is the difference between  SSreg(β2|β0,β3) and SSreg(β1|β0,β3,β2)? What should you put as the argument of lm() function with respect to (β2|β0,β3) and (β1|β0,β3,β2)
The following data represents a random sample of birth weignts (in kgs) of male babies born...
The following data represents a random sample of birth weignts (in kgs) of male babies born to mothers on a special vitamin supplement. 3.73 3.02 4.37 4.09 3.73 2.47 4.33 4.13 3.39 4.47 3.68 3.22 4.68 3.43 (a) Do the data follow a normal distribution?  ? Yes No Report the P-value of the normality test: (b) Do the data support the claim that the mean birth weight of male babies that have been subjected to the vitamin supplement is at least...
Birth and infant death data for all children born in the state of North Carolina dating...
Birth and infant death data for all children born in the state of North Carolina dating back to 1968. The data set for the births in 2001 contains 120,300 records. The data represents a random sample of 800 of those births and selected variables. My goal is to use the data set to test if there is an association between premature births (PREMIE) and smoking during pregnancy (SMOKE) using α=.05 What would be the null and alternative hypothesis? After this...
Birth and infant death data for all children born in the state of North Carolina dating...
Birth and infant death data for all children born in the state of North Carolina dating back to 1968. The data set for the births in 2001 contains 120,300 records. The data represents a random sample of 800 of those births and selected variables. My goal is to use the data set to test if there is an association between premature births (PREMIE) and smoking during pregnancy (SMOKE) using α=.05 I am going to calculate the critical value of the...
Birth and infant death data for all children born in the state of North Carolina dating...
Birth and infant death data for all children born in the state of North Carolina dating back to 1968. The data set for the births in 2001 contains 120,300 records. The data represents a random sample of 800 of those births and selected variables. My goal is to use the data set to test if there is an association between premature births (PREMIE) and smoking during pregnancy (SMOKE) using α=.05. I wanted to see if someone could check the interpretation...
3. Find the data female and male life expectancy for the 13 richest and 14 poorest...
3. Find the data female and male life expectancy for the 13 richest and 14 poorest countries on earth. Country ID Country Name Female LE Male LE 1 Japan 86.8 80.5 2 Switzerland 85.3 81.3 3 Singapore 86.1 80 4 Australia 84.8 80.9 5 Spain 85.5 80.1 6 Iceland 84.1 81.2 7 Italy 84.8 80.5 8 Israel 84.3 80.6 9 Sweden 84 80.7 10 France 85.4 79.4 11 south Korea 85.5 78.8 12 Canada 84.1 80.2 13 Luxembourg 84 79.8...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT