Question 1 The table below shows the shipments (in millions of dollars) of consu

Question

Question 1 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 $199,000 million, what would the predicted amount for durables shipment be for the same year? Construct a confidence interval for the average y value for 199,000 million. Use the t statistic to test to determine whether the slope is significantly different from zero. Use = 0.05 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 Nondurables $168,619 172,197 176,917 181,437 183,404 188,191 191,910 195,773 198,610 201,837 207,919 210,979 212,628 215,858 218,637 220,268 Durables $65,619 68,623 74,234 79,798 82,857 89,937 92,945 95,414 100,138 107,247 114,703 120,763 117,706 123,783 126,250 129,730 Source: Statistics Canada, CANSIM Table 380-0106, Gross domestic product at 2007 constant prices, expenditure-based, annual (dollars). Round your answers to 4 decimal places. Do not round the intermedlate values. Round your answer to 6 declmal places *Do not round the intermediate values. Round your answer to 2 decimal places. Round the intermediate values to 4 decimal places. Round your answers to 0 decimal place. The r value denotes a ks x y (199,000) = Confidence Interval *to The decision is to

Explanation / Answer

a.

calculation procedure for correlation
sum of (x) = x = 3145184
sum of (y) = y = 1589747
sum of (x^2)= x^2 = 622564606966
sum of (y^2)= y^2 = 164805815045
sum of (x*y)= x*y = 317918692388
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)
covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1
= 317918692388 - [ 16 * (3145184/16) * (1589747/16) ]/16- 1
= 338485350.625
and now to calculate r( x,y) = 338485350.625/ (SQRT(1/16*317918692388-(1/16*3145184)^2) ) * ( SQRT(1/16*317918692388-(1/16*1589747)^2)
=338485350.625 / (16399.709*20690.947)
=0.998
value of correlation is =0.998

b.

coeffcient of determination = r^2 = 0.995

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = 0.9975> 0 ,perfect positive correlation

c.

Line of Regression Y on X i.e Y = bo + b1 X

calculation procedure for regression

mean of X = X / n = 196574

mean of Y = Y / n = 99359.1875

(Xi - Mean)^2 = 4303207350

(Yi - Mean)^2 = 6849844794.44

(Xi-Mean)*(Yi-Mean) = 5415765610

b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2

= 5415765610 / 4303207350

= 1.25854

bo = Y / n - b1 * X / n

bo = 99359.1875 - 1.25854*196574 = -148037.37568

value of regression equation is, Y = bo + b1 X

Y'=-148037.37568+1.25854* X

d.

when value of x = 199000
Y'=-148037.37568+1.25854* ( 199000)
Y'=102412.08432

( X) ( Y) X^2 Y^2 X*Y 168619 65619 28432367161 4305853161 11064610161 172197 68623 29651806809 4709116129 11816674731 176917 74234 31299624889 5510686756 13133256578 181437 79798 32919384969 6367720804 14478309726 183404 82857 33637027216 6865282449 15196305228 188191 89937 35415852481 8088663969 16925333967 191910 92945 36829448100 8638773025 17837074950 195773 95414 38327067529 9103831396 18679485022 198610 100138 39445932100 10027619044 19888408180 201837 107247 40738174569 11501919009 21646412739 207919 114703 43230310561 13156778209 23848933057 210979 120763 44512138441 14583702169 25478456977 218637 126250 47802137769 15939062500 27602921250 220268 129730 48517991824 16829872900 28575367640

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