# HW6Sol - Homework 6 solution 1 a The fitted line equation...

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Homework 6 – solution 1. a) The fitted line equation is y = 27 . 2 - 0 . 298 x . The estimated expected value of the strength at carbonation depth of 25.0 mm is ˆ y 0 = 27 . 2 - 0 . 298 · 25 . 0 = 19 . 7. b) We test H 0 : β 1 = 0 versus H 1 : β 1 6 = 0. We obtain t 0 = ˆ β 1 se ( ˆ β 1 ) = - 0 . 298 0 . 041 = - 7 . 23 < - t 0 . 005 , 16 . So the regression is significant. The P-value is zero up to 3 decimal points. c) We obtain f 0 = MS R MS E = 52 . 25 > f 0 . 01 , 1 , 16 . So the regression is significant according to this test as well. The P-value is zero up to 3 decimal points. d) We test H 0 : β 1 = - 320 versus H 1 : β 1 6 = - 0 . 320. We obtain t 0 = ˆ β 1 +0 . 320 se ( ˆ β 1 ) = 0 . 022 0 . 041 = 0 . 54 < t 0 . 025 , 16 . So we can’t reject H 0 and the model has to be deemed valid. e) This time, we test H 0 : β 0 = 28 . 0 versus H 1 : β 0 > 28 . 0. We obtain t 0 = ˆ β 0 - 28 . 0 se ( ˆ β 0 ) = - 0 . 8 1 . 65 < t 0 . 05 , 16 . So we can’t reject H 0 and, therefore, conclude that the claim is false at α = 0 . 05. f) The 95% PI is equal to [13 . 4 , 26 . 1]. Since 12.2 is not in this interval, we conclude that the new observation is not consistent with our model (at 95% confidence).

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