Question

The number of defective items produced by a machine (Y) is known to be linearly related to the speed setting of the machine (X). Data is provided below.

a) (3) Fit a linear regression function by ordinary least squares; obtain the residuals and plot the residuals against X. What does the residual plot suggest?

b) (3) Plot the absolute value of the residuals and the squared residuals vs. X. Which plot has a better line?

c) (4) Perform a weighted least square using the squared residuals to compute the weights. Obtain the weighted least squares estimates for the estimated parameters and their standard errors. Are these values similar to the ones produced in a)? Which results are better, the ones generated in a) or c)? Please explain your answer.

y	x
28	200
75	400
37	300
53	400
22	200
58	300
40	300
96	400
46	200
52	400
30	200
69	300

Answer 1

SOLUTION

a)

We shall use R for all numeric computation

model10<-lm(Y ~ X, data = data10)

> model10

Call:

lm(formula = Y ~ X, data = data10)

Coefficients:

(Intercept) X  

-5.7500 0.1875  

> summary(model10)

Call:

lm(formula = Y ~ X, data = data10)

Residuals:

Min 1Q Median 3Q Max

-17.250 -11.250 -2.750 9.188 26.750

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -5.75000 16.73052 -0.344 0.73820

X 0.18750 0.05381 3.484 0.00588 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 15.22 on 10 degrees of freedom

Multiple R-squared: 0.5484, Adjusted R-squared: 0.5032

F-statistic: 12.14 on 1 and 10 DF, p-value: 0.005878

As the independent variable seems to be categorical, the model too much deviates from actual values. Residuals are large!

b)

c. plot(model10$residuals, model10$model$X,xlab = "Residuals", ylab = "X", type = "p")

Call:

lm(formula = model10$residuals ~ data10$X)

Residuals:

Min 1Q Median 3Q Max

-17.250 -11.250 -2.750 9.188 26.750

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 4.467e-17 1.673e+01 0 1

data10$X -3.140e-18 5.381e-02 0 1

c)

Line of Regression Y on X i.e Y = bo + b1 X
					X	Y	(Xi - Mean)^2	(Yi - Mean)^2	(Xi-Mean)*(Yi-Mean)
28	200	506.25	10000	2250
75	400	600.25	10000	2450
37	300	182.25	0	0
53	400	6.25	10000	250
22	200	812.25	10000	2850
58	300	56.25	0	0
40	300	110.25	0	0
96	400	2070.25	10000	4550
46	200	20.25	10000	450
52	400	2.25	10000	150
30	200	420.25	10000	2050
69	300	342.25	0	0

calculation procedure for regression

mean of X = sum ( X / n ) = 50.5

mean of Y = sum ( Y / n ) = 300

sum ( (Xi - Mean)^2 ) = 5129

sum ( (Yi - Mean)^2 ) = 80000

sum ( (Xi-Mean)*(Yi-Mean) ) = 15000

b1 = sum ( (Xi-Mean)*(Yi-Mean) ) / sum ( (Xi - Mean)^2 )

= 15000 / 5129

= 2.925

bo = sum ( Y / n ) - b1 * sum ( X / n )

bo = 300 - 2.925*50.5 = 152.31

value of regression equation is, Y = bo + b1 X

Y'=152.31+2.925* X

bo =152.31

b1 =2.925

Standard Error of Y on X i.e Y = bo + b1 X
					Xi	Yi	Y'=152.31+2.92*X	Y-Y'	(Y-Yi)^2
28	200	234.21	-34.21	1170.324
75	400	371.685	28.315	801.739
37	300	260.535	39.465	1557.486
53	400	307.335	92.665	8586.802
22	200	216.66	-16.66	277.556
58	300	321.96	-21.96	482.242
40	300	269.31	30.69	941.876
96	400	433.11	-33.11	1096.272
46	200	286.86	-86.86	7544.66
52	400	304.41	95.59	9137.448
30	200	240.06	-40.06	1604.804
69	300	354.135	-54.135	2930.598

Standard error = Sqrt( ( sum ( Y -Yi )^2/ n-2 )

sum ( Y -Yi )^2 = 36131.807

Standard Error = 36131.807