The table below shows the shipments (in millions of dollars) of consumer durable

Question

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. Year 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 intermediate values. Round your answer to 6 decimal 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. 0.998 The r value denotes a strong positive correlation. 148037.3757*+ 1.25854sx (199,000)- 102412.08432 101570to 103254 Confidence Interval. 53.16ss The decision is to reject the null hypothesis atisti

Explanation / Answer

The statistical software output for this problem is:

Simple linear regression results:
Dependent Variable: Durables
Independent Variable: Nondurables
Durables = -148037.38 + 1.2585416 Nondurables
Sample size: 16
R (correlation coefficient) = 0.99752401
R-sq = 0.99505415
Estimate of error standard deviation: 1555.5957

Parameter estimates:

Analysis of variance table for regression model:

Predicted values:

Hence,

Confidence interval lower limit = 101569

Parameter Estimate Std. Err. Alternative DF T-Stat P-value Intercept -148037.38 4677.705 0 14 -31.647437 <0.0001 Slope 1.2585416 0.02371377 0 14 53.072186 <0.0001

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