Homework 5.docx - Ryne Templeman Module 5 HW 10.20 10.22...

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Ryne Templeman 3/5/18 Module 5 HW 10.20, 10.22, 10.28, 10.30, 10.36 10.20 a. X X-M (X-M) 2 2 0 0 1 -1 1 3 1 1 2 0 0 Mean=2 2/3=.67 S 2 =.67 Y Y-M (Y-M) 2 4 .5 0.25 3 -.5 0.25 3 -.5 0.25 5 1.5 2.25 2 -1.5 2.25 4 .5 0.25 Mean=3.5 5.5/5=1. 1 S 2 =1.1 b. Liberals. D f x =n-1=4-1=3 Conservatives.D f y =n-1=6-1=5 D f total =df x +df y = 3+5=8 c. –2.306 2.306 d. S 2 pooled = ( dfx dftotal ) S 2 x +( dfy dftotal ) S 2 Y S 2 pooled = ( 3 8 ) .67 + ( 5 8 ) 1.1 =0.94 e. S 2 Mx = s 2 pooled N x S 2 My = s 2 pooled N y S 2 Mx = .94 4 = .24
S 2 Mx = .94 6 = 0.16 f. S 2 difference = S 2 MX +S 2 MY = .24+.16=0.4 S difference = S 2 difference = .4 = .63 g. ( m x M y ) ( μ x μ y ) S difference = ( 2 3.5 ) ( 0 ) 0.63 =− 2.38 h. (M X -M Y ) upper = t(S difference ) + (M X -M Y ) sample -2.38(.63) + (2-3.5) = -3.00 (M X -M Y ) lower = -t(S difference ) + (M X -M Y ) sample -(-2.38)(.63) + (2-3.5) = -0.001 i. Cohen’s d = ( M X M Y ) −( μ X μ Y ) S pooled = ( 2 3.5 ) −( 0 ) .94 = ¿ -1.55 10.22 a. t (71)=2.13, p<0.05 We reject the null hypothesis; significant difference b. t (39)=1.77, p>0.05 We retain the null hypothesis; no significant difference c. t (13)=3.02, p>0.05 We retain the null hypothesis; no significant difference 10.28 a. Single sample t test, because we are comparing a sample to a population. The population

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