Question

Reserve Problems Chapter 11 Section 2 Problem 1 The department of health studied the number of patients who need liver transplantation. The following data are the Liver Transplantation Waiting List (LTWL), where y is the size (in number of patients) and x is the corresponding year: 1278 1761 2917 3923 4983 6914 7655 7596 77378524 8376 8639 8946 x 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 Round your intermediate answers to four decimal places (e.g. 98.7654). (a) Fit the simple linear regression model using the method of least squares. Find the estimate of o. Round your answer to the nearest integer (e.g. 9876). (b) How does the LTWL size change in average for a year? Round your answer to the nearest integer (e.g. 9876). Bi = (c) Estimate the number of patients in the list in 2018. y = (d) Calculate the fitted value of y corresponding to x = 2012. Find the corresponding residual. Round your answer to the nearest integer (e.g. 9876). Round your answer to the nearest integer (e.g. 9876).

Answer 1

Solution:

The basic concept of regression models and its parameters and estimates calculation is needed as a prerequisite to under the solution to the given problem. Since a lot of data are there, I have used R program to calculate the necessary requirements, parameter estimation, calculation of model variance & necessary other calculations required. The 4 datasets are named d1, d2, d3 & d4 and all the calculations for each dataset has been done seperately with the models named as 'model1', 'm2', 'm3' & 'm4' respectively. The outputs for each of the dataset are mentioned in the comments under each of the codes done for the seperate datasets. The solution is pasted below:

######Reading 4 seperate datasets######

x1=c(2003:2015)
y1=c(1278,1761,2917,3923,4983,6914,7655,7596,7737,8524,8376,8639,8946)
d1=data.frame(x1,y1)

x2=c(1.91,4.29,3.43,4.01,3.05)
y2=c(7300,15510,13650,19940,13100)
d2=data.frame(x2,y2)

x1_=c(80,80,75,rep(62,5),rep(58,6),rep(50,5),56,70)
x2_=c(rep(27,2),25,24,22,23,24,24,23,18,18,17,18,19,18,18,19,19,rep(20,3))
x3_=c(89,88,90,rep(87,3),rep(93,2),87,80,89,88,82,93,89,86,72,79,80,82,91)
y_=c(42,37,37,28,18,18,19,20,15,14,14,13,11,12,8,7,8,8,9,15,15)
d3=data.frame(x1_,x2_,x3_,y_)

x11=c(1,1,1,3,3,3,5,5,5,7,7,7)
x21=c(20,40,60,20,40,60,20,40,60,20,40,60)
y11=c(1500,1250,180,3100,2600,530,4550,4210,910,6290,5730,1320)
d4=data.frame(x11,x21,y11)

#------------------------------------------------------------------------------------------------------------------------#

#######Basic Regression and estimation of parameters#######


######Start of Data1#######

#Data 1 = d1

#Fit a model

model1=lm(y1~x1,data=d1)
model1
summary(model1)
sigma_sq1=(summary(model1)$sigma)**2
b1=summary(model1)$coef[,1][2]
b1
Y_2018 = -1343102+671.6*2018
Y_2018
Y_2012=-1343102+671.6*2012
res1=8524-Y_2012


#The model is
# Y = -1343102+671.6*X
# The estimate of model variance is 882970.4
# The regression coefficient is 671.6
# The value of Y in 2018 is 12140.23
# The value of Y in 2012 is 8157.2
# The residual in Y in 2012 is 366.8

########End of Data 1######

#----------------------------------------------------------------------------------------------------------------#

######Start of Data2######

#Data 2 = d2

# Fit a model

m2=lm(y2~x2,data=d2)
sigma_sq2=(summary(m2)$sigma)**2
b2=summary(m2)$coef[,1][2]
y_4th=-667.1+4364*4.01
res2_4th_val_x=19940-y_4th
y_est=-667.1+4364*4.5

#The model is
# Y = -667.1+4364*x; units of variables as mentioned in the problem
# The model estimated variance is 5633072
# The regression coefficient is 4364 
# The value of y when x = 4.01 is 16832 kg & the residual is 3107.46 (rounded off to nearest integer)
# The mean mass of crop harvested from 4.5 hectares is 18971 kg (rounded off to nearest integer)

########End of Data 2######

#------------------------------------------------------------------------------------------------------------------#

######Start of Data 3######

#Data 3 = d3

# Fit a model

m3=lm(y_~x1_+x2_+x3_,data=d3)
sigma_sq3=(summary(m3)$sigma)**2
a1=62
a2=27
a3=86
y_cap=-39.92+0.7156*a1+1.30*a2+(-0.16)*a3

#The Regression model is given by for data 3 = d3 (with stackflow as response variable Y) as 
#Y = -39.92+0.7156*x1+1.30*x2-0.16*x3
# The estimated model variance is 10.52 (upto 2 decimal places)
# The fitted value of Y (stackflow) for given x1=62, x2=27 & x3=86 is given by 25.79 (upto 2 decimal places)

######End of Data 3#######

#-------------------------------------------------------------------------------------------------------------------#

######Start of Data 4#######

#Data 4 = d4

# Fit a model

m4=lm(y11~x11+x21,data=d4)
sigma_sq4=(summary(m4)$sigma)**2

#m4-> Y = 3494.5+577.84*x1-78.13*x2

se_b0=as.numeric(summary(m4)$coef[,2][1])
se_b1=as.numeric(summary(m4)$coef[,2][2])
se_b2=as.numeric(summary(m4)$coef[,2][3])

#x2=40, x1=4

x1=4; x2=40

Y_pred = 3494.5+577.84*x1-78.13*x2

#The regression model to predict the number of Internet users is: (all corrected to 2 decimal places)
#Y = 3494.5+577.84*x1-78.13*x2
#The standard error of intercept of the model is 873.72
#The standard error of the regression coefficient of x1 (no. of yrs since research) of the model is 122.35
#The standard error of the regression coefficient of x2 (age) of the model is 16.75
#The predicted value of Y for x1=4 & x2=40 is 2680.66

######End of Data 4######

#-----------------------------------------------------------------------------------------------------------------#

(Answer)