Question

Supp Problems, Question 12.62 The table below shows the shipments (in millions of dollars) of consumer durables and nondurables in Canada. Is there a linear relationship between the shipments of durables and nondurables? In other words, if we know the value of nondurables shipped in any one year, can we predict the value of durables during that year? (Hint: Make the value of nondurables the independent variable.) According to the model, if at any given year the nondurables shipment is $201,000 million, what would the predicted amount for durables shipment be for the same year? Construct a confidence interval for the average y value for $ 201,000 million. Use the t statistic to test to determine whether the slope is significantly different from zero. Use a = 0.05. Year Nondurables Durables 1997 $168,619 $65,619 1998 172,197 68,623 1999 176,917 74,234 2000 181,437 79,798 2001 183,404 82,857 2002 188,191 89,937 2003 191,910 92,945 2004 195,773 95,414 2005 198,610 100,138 2006 201,837 107,247 2007 207,919 114,703 2008 210,979 120,763 2009 212,628 117,706 2010 215,858 123,783 2011 218,637 126,250 2012 220,268 129,730 Source: Statistics Canada, CANSIM Table 380-0106, Gross domestic product at 2007 constant prices, expenditure-based, annual (dollars).

Answer 1

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
168619	65619	781482025.00	1138400252.54	943206941.56
172197	68623	594238129.00	944713222.04	749256042.69
176917	74234	386397649.00	631275046.91	493885810.69
181437	79798	229128769.00	382640056.41	296097695.19
183404	82857	173448900.00	272322192.29	217333809.38
188191	89937	70274689.00	88777617.29	78986197.81
191910	92945	21752896.00	41141801.29	29915770.50
195773	95414	641601.00	15564504.41	3160095.19
198610	100138	4145296.00	606548.91	1585662.25
201837	107247	27699169.00	62217586.04	41513557.19
207919	114703	128709025.00	235432582.04	174075552.81
210979	120763	207504025.00	458123189.54	308321919.06
212628	117706	257730916.00	336605528.91	294539727.88
215858	123783	371872656.00	596522617.04	470988800.25
218637	126250	486775969.00	723115796.91	593291996.19
220268	129730	561405636.00	922386251.91	719606031.38

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	3145184	1589747	4303207350.0000	6849844794.4	5415765610.0
mean	196574.00	99359.19	SSxx	SSyy	SSxy

correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.9975  

the value denotes the positive correlation
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sample size ,   n =   16          
here, x̅ = Σx / n=   196574.00   ,     ȳ = Σy/n =   99359.19  
                  
SSxx =    Σ(x-x̅)² =    4303207350.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   5415765610.0          
                  
estimated slope , ß1 = SSxy/SSxx =   5415765610.0   /   4303207350.000   =   1.25854
                  
intercept,   ß0 = y̅-ß1* x̄ =   -148037.37568          
                  
so, regression line is   Ŷ =   -148037.375675   +   1.258542   *x
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Predicted Y at X=   201000   is                  
Ŷ =   -148037.376   +   1.259   *   201000   =   104929.493
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X Value=   201000                      
Confidence Level=   95%                      
                          
                          
Sample Size , n=   16                      
Degrees of Freedom,df=n-2 =   14                      
critical t Value=tα/2 =   2.145   [excel function: =t.inv.2t(α/2,df) ]                  
                          
X̅ =    196574.00                      
Σ(x-x̅)² =Sxx   4303207350                      
Standard Error of the Estimate,Se=   1555.596                      
                          
Predicted Y at X=   201000   is                  
Ŷ =   -148037.376   +   1.259   *   201000   =   104929.493
                          
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    402.813                      
margin of error,E=t*Std error=t* S(ŷ) =   2.1448   *   402.8131   =   863.9482      
                          
Confidence Lower Limit=Ŷ +E =    104929.493   -   863.948   =   104065.5446     
Confidence Upper Limit=Ŷ +E =   104929.493   +   863.948   =   105793.4409      
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Ho:   ß1=   0          
H1:   ß1╪   0          
n=   16              
alpha =   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    1555.596   /√   4303207350.0   =   0.0237
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    1.2585   /   0.0237   =   53.0722
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Degree of freedom ,df = n-2=   14              
p-value =    0.0000              
decison :    p-value<α , reject Ho             
Conclusion:   Reject Ho and conclude that slope is significantly different from zero