94444 4 solution a the standard deviation of the

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Unformatted text preview: a l m o n e l l a $ d o s e ) #t h e d i s t i n c t v a l u e o f d o s e > n=numeric ( l e n g t h ( d o s e s ) ) > ybar=numeric ( l e n g t h ( d o s e s ) ) > SDsquared=numeric ( l e n g t h ( d o s e s ) ) > for ( i in 1: length ( doses )) { + n [ i ]=sum ( s a l m o n e l l a $ d o s e==d o s e s [ i ] ) + ybar [ i ]=mean ( s a l m o n e l l a $ c o l o n i e s [ s a l m o n e l l a $ d o s e==d o s e s [ i ] ] ) + SDsquared [ i ]= var ( s a l m o n e l l a $ c o l o n i e s [ s a l m o n e l l a $ d o s e==d o s e s [ i ] ] ) + i f ( n [ i ]==1) { SDsquared [ i ]=0 } +} > SSpe <− sum ( ( n − 1) ∗ SDsquared ) > d f p e <− sum ( n − 1) > SSpe / d f p e [ 1 ] 90.94444 4. Solution √ a. The standard deviation of the measurement error is 0.32 + 0.42 = 0.5. b. 2 S1 ˆ, n − 1 = 45 − 1 = 44 E (β1 ) ≈ β1 [1 − n ¯2 i=1 (xi − x) /(n − 3) ∴ ˆ E [β1 ] 0.52 ≈1− ≈ 0.705 β1 44 × 0.92 /42 ˆ c. Here is R output and we have: E (β1 ) ≈ 162.16 > > > > > > n = 100000 x t r u e=rnorm ( n , 0 , s q r t ( 0 . 5 6 ) ) e=rnorm ( n , 0 ,...
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