Question

1) 8 teachers recieved training (1,385, S51:100) and 8 teachers did not (M2: 80.5522 lay). (Test using a:0.5). (S(M-12)] c) 4 d)8 2) Data SP=100 SSX-20. My las, Mx = 10 a) slope (b) b) y-intercept c) predict y given X= 4 2 1 3) ANOVA result: [F (3, 16) = +2.24 p>0.05] Which option is correct? df = 3, n. 17; result not significant dfun ko 17; result not significant c) df an 3 do wn 16 kou; result not significant not significant. 123, dtn 16, 4:40 result a) af bim b) df,

Answer 1

(x) training (Y) NO (Y) No training 8.1. Solution ni=8 n2 = 8 Ž = M1 = 85 y=M2=80 SS, =100 SS2 = 124 where SS, = E(x-7)? SS2= Ely-y)? S- ssi + SS2 nithz-2 " 100 + 124 8 +B-2 J 224 √ 16 = 4 Tu SCM, - M2) = 4

8.2. Data given SP = (x-)(y-Y) = 100 SS x E (X-70)2 20 My Y = 125 Mx = X = 10 Past (m): Slope (b) b = (x-7)(y-7). {(x-2 100 = 5 20 Slobe (b) = 5 a Part (b): Y-Intercept Y- Intercept is denoted by a = Ý - b whese bis slope

a= 125 – 5x10 a= 125 – 50 a = 75 Y- Intercept = 75 Partic): Predict y given X=y Regrepsion line is denoted by Y = a + bx where y Dependent Variable X = Independent Variable a = Y-In tercept b = slope fitted line û = 75+ 5 X At x=4, Predicted value of y is given by 75 +5X4 75 + 20 Y =95 < Y = = 95

0.3.(Solution) ANOVA result F13,16) =+2.24, p >0.05 correct option is (6) of between = 3, dfwithin 16, K=4 Result not significant. explanation : þ value is greater than significan level 0.05 this shows that result is not significant 2

F(3,16) = 2-24 = where degree of freedom between = 3 K-1 = 3 R=1+3 K=4 k=u means there are four groups. degree of freedom within 16 N-K = 16 N-4 = 16 N = 4 +16 N= 20 N=20 means total 20 observations, 5 observations in each group.