Q2. Consider the example that we did in the lecture. The purpose is to relate Y,

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Q2. Consider the example that we did in the lecture. The purpose is to relate Y, the length of an infant to the variables x1 - Age (day), x2 - Length at Birth (cm), x3 - Weight at Birth (kg), and x4- Chest Size at Birth (cm) The data is given in the following Infant Length, y (cm) 57.50 52.80 61.30 67.00 53.50 62.70 56.20 68.50 69.20 Weight at Cheast Size at Birth, X3 (kg) | Birth, X4 (cm) Length at Age, X1 (days) | Birth, X2 (cm) | 78.00 69.00 77.00 88.00 67.00 80.00 74.00 94.00 102.00 48.20 45.50 46.30 49.00 43.00 48.00 48.00 53.00 58.00 2.75 2.15 4.41 5.52 3.21 4.32 2.31 4.30 3.71 29.50 26.30 32.20 36.50 27.20 27.70 28.30 30.30 28.70 The result of the regression of the full model is given in the following

Explanation / Answer

a) Regression d.f

M.S= S.S/D.F

D.F= S.S/M.S= 318.27442/79.5686= 4.0

REGRESSION D.F= 4

M.S ( RESIDUAL)= 2.9655807/4= 0.741395175

F= MS(REGRESSION)/M.S(RESIDUAL)

F= 79.5686/0.741395175

F= 107.32279

TOTAL = 4+4= 8

OBSERVATION= TOTAL+1= 8+1= 9

ADJUSTED R SQUARE= 1-MSE/MST

MST= SST/n-1= 321.24/8= 40.155

= 1-0.741395175/40.155

= 1-0.018463

= 0.981537

Intercept ( tstat)= 7.148/16.460= 0.4343

X1= tstat= coefficient/S.E

= >0.295= 0.1/S.E

S.E= 0.1/0.295= 0.33898

upper 95% confidence interval= 0.726+2.776*0.786= 2.908

2.776 is critical value of t

Lower 95% confidence interval = 3.076-1.059*2.776= 0.136

b) Regression equation

y= 7.148+0.1x1+0.726x2+3.076*x3-0.030x4

c) Only X3 is significant X1, X2 and X4 are not significant.

e) R square is 0.991 which means it explains 99.1% of variability explained by this model.

d) y= 7.148+3.076*1= 7.148+3.076= 10.224cm infant length get increase

I have done the questions from a) to e) Please repost rest of the question along with data.

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