Question

1. Four objects are weighed 2 at a time on a spring balance. Denote the 4 unknown weights by B1,...,BA. Six observations are made and are expressed in these forms: Y = B: + 32 + €1. Y = B1 + B3 + €2, Y3 = B1 + B1 + €3, Y4 = 32 +33 + €4, Y = B2 + 4 + €5, Y6 = 33 +34 + €6. Assume that €, N (0,0%), i = 1,...,6. (a) Find expressions for the least squares estimators B1,...,B4 (specify the expressions in terms of Y1,...,Y6). (b) Find an expression for Cov(B) (specify matrix entries, may involve o?). (c) Find expressions for the residuals (specify the expressions in terms of Y1,...,Y). (d) Create a small data set for this study, for (Y.....76) = (5, 8, 6, 7, 10,9). Use Im() func- tion in R to fit the data. Check the results for (a), (b) and (c). Does the output from Im() fit agree with the corresponding calculation results for the data set based on the expressions you derived above? (e) Explain how you will construct a 95% confidence interval for B1 + B2. We can still use the In- distribution. Find the confidence interval for the given data.

Answer 1

1. We can write the model as /B1 Y = BBBBBB 11 1 0 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 Y = X3 +

(a) The normal equation : X'X3 = X'Y = @= (X' or 3111\" Yi + B+ Y37 1131 1| |Yi+X4+X5 113 1| YS + Y4 + Y6 (1113) 【Y3 +5 + Y6 (一直一直一直 = 1孕YPs 空下下 ting以上页 (Yi +8 + Y3} Yi + Y+ F5 + 4 + Y6 {3+} + Y6/ / 2(H+H+})-( +15+}}} 2(X+Y+Ys)-(H+} +Ys) 2(X+Y+Ys)-(Fi +} + Y) 2(X ++Ys)-(Fi + +i)

3 == 2(Y + + )-(H+} + Y) | 2(Y + H+ s)-(YS + 3 + Y6) : = 2(Y +Y+ + 2S +5 + Y)-Y+B+ YA) 6)-(H+ 6 (6) Cour(3) = (XX)-1-2 = | 51平1页 IPang,平卫 1下Pang,

/ 27– Yz-Ys–Y4-Y5+2Y (Y1 – 1:+}+Ys+Ya+Y5–2Y. Y, – Yi+49, +Ys+y-2Y; +Y. -Yi+2Y, – Y; -Y, +2Y; -Y Yg - Yi+Y+4Y3 –2Y+Y;+Y -Y-Y2+2Y3 +2Y4 -Y5-Y (c) Residuals = e = Y-X2 = Yi+Y2-2Y3 +4Y4+Ys+Y6 -Yi-Y2+243 +244 – Ys - Yo Y - 11-2Y, +Y; +YA+4Y+Y 2Y - Y-Y-Y,+2y; -Y Y6 - - 27, +,+Y;+Y, +Ys+4Y. ( 27; – Y-Ys –Y, – Ys+2Y

(d)

R code:

y=c(5,8,6,7,10,9)
x1=c(1,1,1,0,0,0)
x2=c(1,0,0,1,1,0)
x3=c(0,1,0,1,0,1)
x4=c(0,0,1,0,1,1)
x=cbind(x1,x2,x3,x4)
beta=solve(t(x)%*%x)%*%t(x)%*%y
beta
Residual=y-x%*%beta
Residual
lm(y~x1+x2+x3+x4-1)

Output:

> beta
   [,1]
x1 2.0
x2 3.5
x3 4.5
x4 5.0
> Residual=y-x%*%beta
> Residual
     [,1]
[1,] -0.5
[2,] 1.5
[3,] -1.0
[4,] -1.0
[5,] 1.5
[6,] -0.5
> lm(y~x1+x2+x3+x4-1)

Call:
lm(formula = y ~ x1 + x2 + x3 + x4 - 1)

Coefficients:
x1   x2   x3   x4
2.0 3.5 4.5 5.0