Question

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:

Reserve Probleme 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 Waking List (LTWL), where y is the size (in number of patients) and is the corresponding year! 1349 1870 3134 40505127 71327928 7870 797687048730 8918 9179 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 Round your intermediate answers to four decimal places (0.9. 98.7654). (*) Rt the simple linear regression model using the method of least squares. Find the estimate of o. Round your answer to the nearest Integer (e-9876). (b) How does the LTWL size change in average for a year? Round your answer to the nearest Integer (0.9. 9876). 2, (c) Estimate the number of patients in the list in 2017. 9 (d) Calculate the fitted value of y corresponding to 2011. Find the corresponding residual. Round your answer to the nearest Integer (0.9. 9876). 9- Round your answer to the nearest integer (e. 9876).

Answer 1

	y	x	x^2	y^2	xy
	1349	2001	4004001	1819801	2699349
	1870	2002	4008004	3496900	3743740
	3134	2003	4012009	9821956	6277402
	4050	2004	4016016	16402500	8116200
	5127	2005	4020025	26286129	10279635
	7132	2006	4024036	50865424	14306792
	7928	2007	4028049	62853184	15911496
	7870	2008	4032064	61936900	15802960
	7976	2009	4036081	63616576	16023784
	8704	2010	4040100	75759616	17495040
	8730	2011	4044121	76212900	17556030
	8918	2012	4048144	79530724	17943016
	9179	2013	4052169	84254041	18477327
average	6305.154	2007	4028063	47142819	12664059

Using the method of least squares, the estimated regression line is given by

Y=a+bX, where a is the intercept and b is the slope estimate.

$slope =b =\frac{average(xy)-average(x)*average(y)}{[average(x^2)-(average(x))^2]]}$

and intercept =a = average(y) - b*average(x)

Substituting the values, we get , b=686.82 and a=-1372151

Thus, the regression line is Y= -1372151 + 686.82X

Now, we calcukate the residuals and then the estimate os variance is given by

$\sigma ^{2} = \frac{\sum (y-Y)^{2}}{n-2}$

y	x	Y	(y-Y)^2
1349	2001	2175.82	683631.3124
1870	2002	2862.64	985334.1696
3134	2003	3549.46	172607.0116
4050	2004	4236.28	34700.2384
5127	2005	4923.1	41575.21
7132	2006	5609.92	2316727.526
7928	2007	6296.74	2661009.188
7870	2008	6983.56	785775.8736
7976	2009	7670.38	93403.5844
8704	2010	8357.2	120270.24
8730	2011	9044.02	98608.5604
8918	2012	9730.84	660708.8656
9179	2013	10417.66	1534278.596
		sum	10188630.38

Thus, estimated variance = 10188630.38/(13-2) = 926239

b. this value is given by the slope estimate , i.e 686.82 = 687

c. When X = 2017, Y= -1372151 + 686.82*2017 = 13165

d. From the table, when X=2011, Y = 9044

Thus, residual = 8730-9044.02 = -314