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: y | 12521784 3067 4071 50627062 | 7916 7825 8093 8728 8758 8934 9236 * 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 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). B (c) Estimate the number of patients in the list in 2016. (d) Calculate the fitted value of y corresponding to x= 2008. Find the corresponding residual. Round your answer to the nearest integer (e.g. 9876). V= Round your answer to the nearest integer (e.g. 9876). e =

Answer 1

	y	x	x^2	y^2	xy		Y	res=y-Y	res sqr
	1252	2000	4000000	1567504	2504000		2097.516	-845.516	714898.1
	1784	2001	4004001	3182656	3569784		2796.495	-1012.49	1025145
	3067	2002	4008004	9406489	6140134		3495.473	-428.473	183588.7
	4071	2003	4012009	16573041	8154213		4194.451	-123.451	15240.04
	5062	2004	4016016	25623844	10144248		4893.429	168.5714	28416.33
	7062	2005	4020025	49871844	14159310		5592.407	1469.593	2159705
	7916	2006	4024036	62663056	15879496		6291.385	1624.615	2639375
	7825	2007	4028049	61230625	15704775		6990.363	834.6374	696619.5
	8093	2008	4032064	65496649	16250744		7689.341	403.6593	162940.9
	8728	2009	4036081	76177984	17534552		8388.319	339.6813	115383.4
	8758	2010	4040100	76702564	17603580		9087.297	-329.297	108436.3
	8934	2011	4044121	79816356	17966274		9786.275	-852.275	726372.2
	9236	2012	4048144	85303696	18582832		10485.25	-1249.25	1560632
average	6291.385	2006	4024050	47201254	12630303			sum	10136753

1.

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 = average(cy) – average(2) * averagely) [average(22) - (average(2))2]

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

Substituting the values, we get , b=698.978 and a=-1395859

Thus, the regression line is Y=-1395859 + 698.978 X

1. we calcukate the residuals and then the estimate os variance is given by

σ?, Σy - Y)? η – 2

= 10136753/11

Est var=921523

2. $\widehat{\beta ^{_{1 }}}$ = slope = 698.978 = 699

3. when X = 2016, Y=-1395859 + 698.978*2016 = 13280.65 = 13281

4. When X = 2008,estimated y, Y=-1395859 + 698.978*2008 = 7688.824 =7689

The observed y = 8093

residual e = y-Y = 8093-7689 =404