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Unformatted text preview: Y will increase 15.8 units. 3) The test statistic is t = 15 . 8 4 . 499 = 3 . 512 > 1 . 746 = t . 05 , 16 Thus, we reject the null ( H : β = 0) 1 4) Following the formula b ± t α 2 se b , we obtain the 99% CI to be 15 . 8 ± 2 . 921 * 4 . 499 = (2 . 658 , 28 . 942). We are 99% conﬁdent that the true value β falls between 2.658 and 28.942. 5) The correlation r = √ R 2 where R 2 = SS Regr SS Total = 67525 87021 = 0 . 776. Thus, r = 0 . 88. See notes for the interpretation. 6) The test statistic is t = . 88 q 1. 776 16 = 7 . 43 > 1 . 746 = t . 05 , 16 Thus, we still reject H which is consistent with our part (3). This is a 1sided test where H a : ρ < 0. 2...
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 Spring '08
 TA
 Statistics, Normal Distribution, Regression Analysis, Errors and residuals in statistics, predicted value

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