I get
a) T = 0.5876D+1.5448
b) T = 10.95
c) r = 0.9954
So you're on the right track. Rounding error is most likely the differnce between your answers and mine.
The torque, T Nm, required to rotate shafts of different diameters, D mm, on a machine has been tested and recorded below.
D (mm) 6 10 14 18 21 25 T (Nm) 5.5 7.0 9.5 12.5 13.5 16.5
(a) Use the method of least squares to determine the linear regression equation that relates the torque to the diameter.
(b) Estimate the value of torque required when the diameter is 16 mm.
(c) Calculate the coefficient of correlation between these variables and comment on your prediction obtained in (b).
What I’ve done so far:
Question a
X Y X^{2} Y^{2} XY 6 5.5 36 30.25 33 10 7 100 49 70 14 9.5 196 90.25 133 18 12.5 324 156.25 225 21 13.5 441 182.25 283.5 25 16.5 625 272.25 412.5 ∑X= ∑Y= 64.5 ∑X^{2}= 1722 ∑Y^{2}= 780.25 ∑XY= 1157
Normal equations for Y on X are
∑Y=a∑X+nb & ∑XY=a∑x^{2}+b∑X
Substitute values from tables gives
64.5=94a=6b & 1157=1722a+94b
Equation 1 gives b= (64.5-94a)/6
Equation 2 gives 1157=1722a+94((64.5-94a)/6)
Solving gives a=0.59 & b=1.507
The regression line equation is Y=0.59X+1.507
Question b
11Nm
Question c
X Y x=X-X y=Y-Y xy x^{2} y^{2} 6 5.5 -9.67 -5.25 50.77 93.51 27.56 10 7 -5.67 -3.75 21.26 32.15 14.06 14 9.5 -1.67 -1.25 2.09 2.79 1.56 18 12.5 2.33 1.75 4.08 5.43 3.06 21 13.5 5.33 2.75 14.66 28.41 7.56 25 16.5 9.33 5.75 53.65 87.05 33.06 ∑X= 94 ∑Y= 64.5 ∑XY= 146.51 ∑ x^{2}=249.34 ∑ y^{2}=86.86 X=15.67 Y=10.75
r=∑xy / √((∑ x^{2})( ∑ y^{2}))
r= 146.51 / √ (249.34 x 86.86)
r= 1
not sure if this is the final answer for c need some help here and guidance if Im completely wrong here on all of the above.
Hi, try this may be it helps.
Method Of Least Square