Question

The following table presents shear strengths (in kN/mm) and weld diameters (mm) for a sample of spot welds.

Diameter x	Strength, y	xiyi	xi2	yi2	yi	(yi –yi)2	(yi- y)2
4.2	51	214.20	17.64	2601.00	53.85	8.15	835.21
4.4	54	237.60	19.36	2916.00	59.64	31.84	670.81
4.6	69	317.40	21.16	4761.00	65.43	12.74	118.81
4.8	81	388.80	23.04	6561.00	71.22	95.68	1.21
5.0	75	375.00	25.00	5625.00	77.01	4.02	24.01
5.2	79	410.80	27.04	6241.00	82.79	14.39	0.81
5.4	89	480.60	29.16	7921.00	88.58	0.17	82.81
5.6	101	565.60	31.36	10201.00	94.37	43.96	445.21
5.8	98	568.40	33.64	9604.00	100.16	4.66	327.61
6.0	102	612.00	36.00	10404.00	105.95	15.57	488.41
Σ = 51.00	Σ = 799.00	Σ = 4170.00	Σ = 263.40	Σ = 66,835.00	∑ = 799.00	∑ = 231.19	∑ = 2,994.90

a. (15 pts) Calculate the regression coefficients for the least-squares-line (linear regression model).

b. (15 pts) Determine the percentage of the variation in strength that is explained by weld diameter (r²)?

c. (15 pts) Can the least squares line be used to predict a strength for a diameter of 8 mm? if so, predict the strength, if not explain why not.

Answer 1

Solution: (a) The linear regression equation of y (strength) on x (weld diameter) is given by y_hat = b0 + b1 x where y_hat is the predicted value of the response variable, x is the given value of the explanatory variable, b0 is the y-intercept and b1 is the slope.

We have and

Here b0 =  -67.691 and b1 = 28.939

Thus the estimated least squares line is y_hat = -67.691 + 28.939x

(b) The coefficient of determination denotes the percentage of the variation in strength that is explained by weld diameter. It is given by R2 = 0.9228. Thus, 92.28% of the variation in strength is explained by weld diameter.

(c) Since 92.28% of variation in strength is explained by the least squares line, it can be used to predict a strength for a diameter of 8 mm. It is obtained by putting x = 8 in the obtained equation in part (a) as -67.691 + (28.939*8)

=  163.821 kn/mm(Ans.)