# Steffen grønneberg bi lecture 5 gra6036 4th february

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Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 46 / 61

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Multiple linear regression Let Y i be birth weight, X i , 1 age. Suppose Y i = β 0 , B + β 1 X i , 1 + ε i for boys and Y i = β 0 , G + β 1 X i , 1 + ε i for girls with ε i N ( 0 , σ 2 ) . Is β 0 , B 6 = β 0 , G ? Inspired by ANOVA, let’s write Y i = β 0 , G + β 1 X i , 1 + β 2 X i , 2 + ε i where X i , 2 = 1 , if individual i is a boy 0 , if individual i is a girl β 2 = β 0 , B - β 0 , G is the additional weight for all boys, compared to girls. Is β 2 statistically significantly different from zero? Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 47 / 61
Multiple linear regression We now have a model with two covariates. In general, suppose we have d covariates, so Y i = β 0 + d X j = 1 β j X i , j + ε i . We want to approximate β 0 , β 1 , . . . , β d . A candidate line would predict ˆ Y i = ˆ β 0 + d X j = 1 ˆ β j X i , j . with residuals e i = Y i - ˆ Y i . Ordinary least squares: Minimize n i = 1 e 2 i . Formula of solution ˆ β 0 , ˆ β 1 , . . . , ˆ β d is complex unless written in terms of matrices (which we will not do). Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 48 / 61

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Multiple linear regression Assessing the adequacy for the multiple linear regression Y i = β 0 + d X j = 1 β j X i , j + ε i is more complicated than for the simple linear regression. The basic assumption is that Y i is linearly related to each X i , j . For the model at hand, this is simple to assess. For complex models this is more demanding, and we’ll get back to the general case next week. Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 49 / 61
Multiple linear regression Our modeling assumptions are that Y i = β 0 + d X j = 1 β j X i , j + ε i where 1 ε i has zero mean and constant variance. 2 ε i are independent of each other and either 1 ε i N ( 0 , σ 2 ) or 2 the number of observations n is large. Again, we cannot observe ε i , only e i = ˆ Y i - Y i . If these assumptions are fulfilled, ˆ β j - β j SE ( ˆ β j ) is approximately distributed according to the standard Normal distribution N ( 0 , 1 ) . Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 50 / 61

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Multiple linear regression Hence, we test H 0 : β j = 0 versus H A : β j 6 = 0 just as for simple linear regression: Compare the absolute value of t = ˆ β j SE ( ˆ β j ) to quantiles of the standard Normal distribution. The p-value of the test is approximately P ( | N ( 0 , 1 ) | > | t | ) , and is rejected at a 5 % level if | t | > 1 . 96. For our example, the sample size is not very large, and we must assess the approximate Normality of the residuals before we can trust computed p-values. Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 51 / 61
Multiple linear regression Steffen Grønneberg (BI) Lecture 5, GRA6036 4th February 2016 52 / 61

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Multiple linear regression For our data-set, PP-plot of residuals shows slight curvature, but no clear deviation from Normality. We can therefore use the fitted regression model to test H 0 : β 2 = 0 , against H A : β 2 6 = 0 .
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