Thread: statistics - stuck on the final question

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	X2	Y2	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	∑X2= 1722	∑Y2= 780.25	∑XY= 1157

Normal equations for Y on X are
∑Y=a∑X+nb & ∑XY=a∑x²+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	x2	y2
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	∑ x2=249.34	∑ y2=86.86
X=15.67	Y=10.75

r=∑xy / √((∑ x²)( ∑ y²))


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.

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.

Thanks for that but how would you comment on your prediction from answer b in relation to what you got for answer c?

D = 16 is an interpolation of the data set and as r is close to one you can say the prediction is very reliable

Hi, try this may be it helps.

Method Of Least Square