Question

Question 8 )A science student investigated how far a spring would stretch when various weights are attached to it. The stretch was measured in millimetres and the weights in grams. The results are as follows

Mass in grams (g)	10	12	14	16	18	20	22	24
Extension in millimetres (mm)	15	18	20	25	28	30	33	40

(i) Which of the two variables is the dependent variable?

(ii) Find the correlation coefficient r correct to 2 decimal places.

(iii) Describe the relationship between the extension and the weight of the mass used.

(iv) Find the equation of the regression line.

(v) On the graph paper draw a scattergram showing the data.

(vi) Use the regression equation to draw the line of regression on your scatter diagram.

(vii) Use your diagram to estimate the mass needed to cause an extension of 15mm

Answer 1

The calculations are given below,

extension (x)	mass(y)	s1=(x-xbar)^2	s2=(y-ybar)^2	s3=(x-xbar)*(y-ybar)
15	10	123.7656	49	77.875
18	12	66.0156	25	40.625
20	14	37.5156	9	18.375
25	16	1.2656	1	1.125
28	18	3.5156	1	1.875
30	20	15.0156	9	11.625
33	22	47.2656	25	34.375
40	24	192.5156	49	97.125
xbar=sum(x)/8	ybar=sum(y)/8	sxx=sum(s1)	syy=sum(s1)	sxy=sum(s3)
26.125	17	486.8748	168	283
		b1=sxy/sxx	b0=ybar-b1*xbar	correlation=r= sxy/sqrt(sxx*syy)
		0.5813	1.8135	0.9895

i) The Mass is dependent variable because as extension increases mass also incereases.

ii) The correlation between mass and extension is given by,

r= sxy/ sqrt(sxx*syy) = 0.9895 =0.99

iii) There is positive linear relationship between extension and weight of the mass used.

iv) The equation of regression line is given by

mass = bo + b1 * extension

mass= 1.8135 + 0.5813 * extension.

v) and vi)

vii) The estimated mass from graph when extension is 15 is given by 10.5 (g)