= Solution
Minimizing the <residual sum of squares> $\|Y-Xb\|^2$ gives the <least-squares normal equations> $X^TX\widehat\beta=X^TY$. Since the <design matrix> has full <column rank>, $X^TX$ has a <matrix inverse>, so
$$
\boxed{\widehat\beta=(X^TX)^{-1}X^TY,\qquad
\widehat\beta\sim N_p\bigl(\beta,\sigma^2(X^TX)^{-1}\bigr).}
$$
The <sampling distribution> follows because a linear transformation of a <multivariate normal distribution> is again a <multivariate normal distribution>. Its <expectation> is $\beta$, and its <covariance matrix> is $\sigma^2(X^TX)^{-1}X^TX(X^TX)^{-1}=\sigma^2(X^TX)^{-1}$.
The <hat matrix> is $H=X(X^TX)^{-1}X^T$. It is the <orthogonal projection matrix> onto the <column space> of $X$: $H^T=H$, $H^2=H$ and $HX=X$. Thus the <regression residuals> are $\widehat\varepsilon=(I-H)Y$, with
$$
\operatorname{Cov}(\widehat\varepsilon,\widehat Y)
=\sigma^2(I-H)H=0,\qquad
\operatorname{Cov}(\widehat\varepsilon,\widehat\beta)
=\sigma^2(I-H)X(X^TX)^{-1}=0.
$$
This is <fitted-residual orthogonality>. The vectors are jointly normal, so <independence of uncorrelated jointly normal variables> strengthens both zero-<covariance> statements to <independence>.
For the quadratic <linear regression>, the columns of the <design matrix> are $1,x,x^2$. The replicated design gives
$$
X^TX=\begin{pmatrix}30&0&20\\0&20&0\\20&0&20\end{pmatrix},\qquad
\boxed{\operatorname{Cov}\begin{pmatrix}\widehat\alpha\\\widehat\beta\\\widehat\gamma\end{pmatrix}
=\sigma^2\begin{pmatrix}1/10&0&-1/10\\0&1/20&0\\-1/10&0&3/20\end{pmatrix}.}
$$
For $v(x)=(1,x,x^2)^T$, the <variance of a fitted regression mean> is $\sigma^2v(x)^T(X^TX)^{-1}v(x)$, hence
$$
\boxed{\operatorname{Var}(\widehat\alpha+\widehat\beta x+\widehat\gamma x^2)
=\frac{\sigma^2}{20}(2-3x^2+3x^4).}
$$
Writing $u=x^2\in[0,1]$, this is the quadratic $(\sigma^2/20)(2-3u+3u^2)$. Its <derivative> vanishes at $u=1/2$; it decreases before that point and increases afterwards. Therefore the \b[maximum] is $\sigma^2/10$ at $x=-1,0,1$, and the \b[minimum] is $\sigma^2/16$ at $x=\pm1/\sqrt2$.
At $100^\circ\mathrm C$, $x=0$ and the expected yield is $\alpha$. The unbiased residual <variance> estimate is $s^2=2.43/(30-3)=0.09$. By <Cochran's theorem>, $27s^2/\sigma^2\sim\chi^2_{27}$ independently of $\widehat\alpha$, giving the <Student t confidence interval>
$$
\boxed{\alpha\in\left[\widehat\alpha-t_{27,0.975}\sqrt{0.009},\;
\widehat\alpha+t_{27,0.975}\sqrt{0.009}\right]
\simeq[\widehat\alpha-0.19465,\widehat\alpha+0.19465].}
$$
The interval estimates the expected yield, so it uses the <variance of a fitted regression mean> without an additional future-observation error term. The supplied summary determines the half-width but does not give a numerical value of $\widehat\alpha$.
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