Solution to Tutorial 2 - Solutions to Tutorial 2 1. (a) Y =...

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Solutions to Tutorial 2 1. (a) ˆ Y =1 0 . 2+ 4 . 00 X ( SE )( 0 . 6633) (0 . 4690) MSE =2 . 199289 ,R 2 =0 . 9009 ,F =72 . 73 see code (R code) for the plot. Yes, the linear regression function Fts the data well. (b) ˆ Y =10 . 2+4 . 00 1=14 . 2 (c) b 1 ± t (0 . 975 , 8) s ( b 1 )=4 ± 2 . 306 0 . 469 = [2 . 918486 , 5 . 081514] The interval does not include 0, indicating that we have (big) conFdence that β 1 is di±erent from 0, i.e. β 1 is not zero, and thus the linear association is signiFcant. (d) We need to test H 0 : β 1 v.s.H 1 : β 1 6 Note that | t | = | b 1 0 s ( b 1 ) | =8 . 528 >t (1 α/ 2 ,n 2) = 2 . 306 We reject H 0 . In other words, there is signiFcant linear association. (e) b 0 ± t (0 . 975 , 8) s ( b 0 )=10 . 2 ± 2 . 306 0 . 6633 = [8 . 67043 , 11 . 72957] The interval does not include 0, indicating we have (big) conFdence that β 0 is di±erent from 0. Even if no shipment, there are still broken ampules (due to the other reasons) (f) We need to test H 0 : β 1 v.s. H 1 : β 1 6 Note that | t | = | b 1 2 s ( b 1 ) | =4 . 264392 (1 α/ 2 2) = 2 . 9 We reject H 0 . In other words, the past experience is not applicable to the data 1
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2. The absolute value of the coefficient 1 .
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Solution to Tutorial 2 - Solutions to Tutorial 2 1. (a) Y =...

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