Answer to Question #322410 in Statistics and Probability for Mari bless

Question

12. To determine the weights of w1, w2 and w3 of three objects A1, A2 and A3, the following weighing

design was adopted. The objects were first weighed separately (A1 then A2 then A3) and then all

three (A1, A2 , A3) using ordinary two-parts balance. If the balancing weights in these four weighing

are denoted by x1, x2, x3 and x4 respectively with the expected values and variances given by

E(x1) = w1, E(x2) = w2, E(x3) = w3

E(x4) = w1 + w2 + w3

and

V ar(xi) = iσ2,

i = 1, 2, 3.

(a) By an appropriate method of estimating w1, w2, and w3, generate their normal equations and

present in a matrix form.

(b) Hence or otherwise, find the least square estimators of w1, w2, and w3 if V ar(xi) = iσ2,

i =

1, 2, 3.

(c) Given that x1 = 2.2, x2 = 1.7, x3 = 6.0, x4 = 9.5 and σ = 1.5, find the estimates for w1 and

w3.

Answer 1

$a:\\x=\left[ \begin{matrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 1& 1& 1\\\end{matrix} \right] \left[ \begin{array}{c} w_1\\ w_2\\ w_3\\\end{array} \right] +\varepsilon ,\\\varepsilon =\left[ \begin{array}{c} \varepsilon _1\\ \varepsilon _2\\ \varepsilon _3\\ \varepsilon _4\\\end{array} \right] \\E\varepsilon =0\\cov\varepsilon =\sigma ^2\left[ \begin{matrix} 1& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 4\\\end{matrix} \right] \\Weigthed\,\,least\,\,squares.\\Matrix\,\,form\,\,of\,\,normal\,\,equations:\\X^T\left( cov\varepsilon \right) ^{-1}X\hat{\omega}=X^T\left( cov\varepsilon \right) ^{-1}x\\\left( cov\varepsilon \right) ^{-1}=\frac{1}{\sigma ^2}\left[ \begin{matrix} 1& 0& 0& 0\\ 0& 1/2& 0& 0\\ 0& 0& 1/3& 0\\ 0& 0& 0& 1/4\\\end{matrix} \right] \\\left[ \begin{matrix} 1& 0& 0& 1\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\\end{matrix} \right] \left[ \begin{matrix} 1& 0& 0& 0\\ 0& 1/2& 0& 0\\ 0& 0& 1/3& 0\\ 0& 0& 0& 1/4\\\end{matrix} \right] \left[ \begin{matrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\\ 1& 1& 1\\\end{matrix} \right] \hat{\omega}=\left[ \begin{matrix} 1& 0& 0& 1\\ 0& 1& 0& 1\\ 0& 0& 1& 1\\\end{matrix} \right] \left[ \begin{matrix} 1& 0& 0& 0\\ 0& 1/2& 0& 0\\ 0& 0& 1/3& 0\\ 0& 0& 0& 1/4\\\end{matrix} \right] x\\\left[ \begin{matrix} 5/4& 1/4& 1/4\\ 1/4& 3/4& 1/4\\ 1/4& 1/4& 7/12\\\end{matrix} \right] \left[ \begin{array}{c} \omega _1\\ \omega _2\\ \omega _3\\\end{array} \right] =\left[ \begin{matrix} 1& 0& 0& 1/4\\ 0& 1/2& 0& 1/4\\ 0& 0& 1/3& 1/4\\\end{matrix} \right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\\end{array} \right] \\\left\{ \begin{array}{c} \frac{5}{4}\omega _1+\frac{1}{4}\omega _2+\frac{1}{4}\omega _3=x_1+\frac{1}{4}x_4\\ \frac{1}{4}\omega _1+\frac{3}{4}\omega _2+\frac{1}{4}\omega _3=\frac{1}{2}x_2+\frac{1}{4}x_4\\ \frac{1}{4}\omega _1+\frac{1}{4}\omega _2+\frac{7}{12}\omega _3=\frac{1}{3}x_3+\frac{1}{4}x_4\\\end{array} \right. \\b:\\The\,\,least\,\,squares\,\,estimators\,\,are\\\left[ \begin{array}{c} \omega _1\\ \omega _2\\ \omega _3\\\end{array} \right] =\left[ \begin{matrix} 5/4& 1/4& 1/4\\ 1/4& 3/4& 1/4\\ 1/4& 1/4& 7/12\\\end{matrix} \right] ^{-1}\left[ \begin{matrix} 1& 0& 0& 1/4\\ 0& 1/2& 0& 1/4\\ 0& 0& 1/3& 1/4\\\end{matrix} \right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\\end{array} \right] =\\=\left[ \begin{matrix} 0.9& -0.2& -0.3\\ -0.2& 1.6& -0.6\\ -0.3& -0.6& 2.1\\\end{matrix} \right] ^{-1}\left[ \begin{matrix} 1& 0& 0& 1/4\\ 0& 1/2& 0& 1/4\\ 0& 0& 1/3& 1/4\\\end{matrix} \right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\\end{array} \right] =\\=\left[ \begin{matrix} 0.9& -0.1& -0.1& 0.1\\ -0.2& 0.8& -0.2& 0.2\\ -0.3& -0.3& 0.7& 0.3\\\end{matrix} \right] \left[ \begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\\end{array} \right] \\c:\\\left[ \begin{array}{c} \omega _1\\ \omega _2\\ \omega _3\\\end{array} \right] =\left[ \begin{matrix} 0.9& -0.1& -0.1& 0.1\\ -0.2& 0.8& -0.2& 0.2\\ -0.3& -0.3& 0.7& 0.3\\\end{matrix} \right] \left[ \begin{array}{c} 2.2\\ 1.7\\ 6\\ 9.5\\\end{array} \right] =\left[ \begin{array}{c} 2.16\\ 1.62\\ 5.88\\\end{array} \right] \\\hat{\omega}_1=2.16,\hat{\omega}_3=5.88 .$