Question

An experiment was conducted in which different dosages of a particular pesticide were administered to 5 groups of 5 mice. The 25 mice were females of similar age and condition. One group received no chemical. The basic response y was a measure of activity in the brain. It was postulated that brain activity would decrease with an increase in pesticide dosage. Use the given data to answer parts​ (a) and​ (b).

Animal 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Dose, x (mg/kg body weight) 0.0 0.0 0.0 0.0 0.0 2.3 2.3 2.3 2.3 2.3 4.5 4.5 4.5 4.5 4.5 9.2 9.2 9.2 9.2 9.2 18.5 18.5 18.5 18.5 18.5 Activity, y (moles/liter/min) 10.5 10.6 10.8 9.7 9.0 11.0 11.4 9.8 9.5 10.2 10.9 10.2 8.7 11.4 8.1 9.7 7.6 9.0 8.4 2.4 2.8 2.3 3.3 5.2 8.4

Answer 1

a) n=25, Ei= 174.7 ,Eyis 210.9 E(Xi-) = 1056.306 Elyi- Isei 174.7 6.988 n 25 Eyi 2 10.9 8.436 n 25 V(X): < Ecxi - 5]? - 1056.306 25 42.25 23 ways Zly;-})2 = 201.3776 25 8.0551 COUC4) E cxi-73 (4-9) n 11 25 x - 361-784 25 - 14.4716 a - Coulx,y) Vox) - 14.4716 42.2523 - 0.3425 Bo ý - Bit 8.436 - (-0.3425 X 6.988) 10.4485 CSScanned with CamScanner

The least-square estimate of bois 2:44 88 s] 10.5 The least-square estimate of bi is -0,343 b) To tost the lack of fit here the m=5 m=5, 0:25 m. be the number of distinct lovel. n is number of observation. hypothesis is : Ho: The tentative model is ddequetely descaibes The data CRegression function is linear) VS H. The tentative model is not doe describe the data (Regression function'is quetely not linear) hone our tast statistic is Fo - MSLOF N Fx MSPE Fx , m -2,0-m a= 0.05 The critical region is f> Fa, m-r,nom f) Foros , 3 22 23 f> 10.12 CSScanned with CamScanner

Sou Sum of Square bf 'mean square compeeled f Resgression 124.69 1 124.69 Error 76.69 23 3.33 Lack of fit 50°13 3 1 : 71 0:48 part Error 71.6 20 3.58 Total 201.38 24 calculation def of lack of fit = m-2=5-2 - 3 the dot of pure trouron-m : 25-5-20 mean squote of lack of fit SS Lot = df MSlof xdf SS Lot SS lof = 1.7173 - 5.13 85 pE = DF.7 MS PE 20 Y 3.58 - 71.6 f = Ms Lof 1:71 MSPE 3.58 4 0.48 conclusion: Reject | Ho because the competed value of f in the table is greater than the critical region at oros. essCalled wit Cancemen