Chapter 9: Problem 12
Fit \(y=a+x\) to the data $$ \begin{array}{l|lll} \mathrm{x} & 0 & 1 & 2 \\ \hline \mathrm{y} & 1 & 3 & 4 \end{array} $$ by the method of least squares.
Chapter 9: Problem 12
Fit \(y=a+x\) to the data $$ \begin{array}{l|lll} \mathrm{x} & 0 & 1 & 2 \\ \hline \mathrm{y} & 1 & 3 & 4 \end{array} $$ by the method of least squares.
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Get started for freeLet \(X_{1 j}, X_{2 j}, \ldots, X_{a_{f} j}\) represent independent random samples of sizes \(a_{j}\) from a normal distribution with means \(\mu_{j}\) and variances \(\sigma^{2}, j=1,2, \ldots, b\). Show that $$ \sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. .}\right)^{2}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}}\left(X_{i j}-\bar{X}_{. j}\right)^{2}+\sum_{j=1}^{b} a_{j}\left(\bar{X}_{. j}-\bar{X}_{. .}\right)^{2} $$ or \(Q^{\prime}=Q_{3}^{\prime}+Q_{4}^{\prime} .\) Here \(\bar{X}_{. .}=\sum_{j=1}^{b} \sum_{i=1}^{a_{j}} X_{i j} / \sum_{j=1}^{b} a_{j}\) and \(\bar{X}_{. j}=\sum_{i=1}^{a_{j}} X_{i j} / a_{j} .\) If \(\mu_{1}=\mu_{2}=\) \(\cdots=\mu_{b}\), show that \(Q^{\prime} / \sigma^{2}\) and \(Q_{3}^{\prime} / \sigma^{2}\) have chi-square distributions. Prove that \(Q_{3}^{\prime}\) and \(Q_{4}^{\prime}\) are independent, and hence \(Q_{4}^{\prime} / \sigma^{2}\) also has a chi-square distribution. If the likelihood ratio \(\Lambda\) is used to test \(H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{b}=\mu, \mu\) unspecified and \(\sigma^{2}\) unknown against all possible alternatives, show that \(\Lambda \leq \lambda_{0}\) is equivalent to the computed \(F \geq c\), where $$ F=\frac{\left(\sum_{j=1}^{b} a_{j}-b\right) Q_{4}^{\prime}}{(b-1) Q_{3}^{\prime}} $$ What is the distribution of \(F\) when \(H_{0}\) is true?
The driver of a diesel-powered automobile decided to test the quality of three types of diesel fuel sold in the area based on mpg. Test the null hypothesis that the three means are equal using the following data. Make the usual assumptions and take \(\alpha=0.05\). $$ \begin{array}{llllll} \text { Brand A: } & 38.7 & 39.2 & 40.1 & 38.9 & \\ \text { Brand B: } & 41.9 & 42.3 & 41.3 & & \\ \text { Brand C: } & 40.8 & 41.2 & 39.5 & 38.9 & 40.3 \end{array} $$
Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be a random sample of size \(n=4\) from the normal distribution \(N(0,1) .\) Show that \(\sum_{i=1}^{4}\left(X_{i}-\bar{X}\right)^{2}\) equals $$ \frac{\left(X_{1}-X_{2}\right)^{2}}{2}+\frac{\left[X_{3}-\left(X_{1}+X_{2}\right) / 2\right]^{2}}{3 / 2}+\frac{\left[X_{4}-\left(X_{1}+X_{2}+X_{3}\right) / 3\right]^{2}}{4 / 3} $$ and argue that these three terms are independent, each with a chi-square distribution with 1 degree of freedom.
Suppose \(\mathbf{A}\) is a real symmetric matrix. If the eigenvalues of \(\mathbf{A}\) are only 0 's and 1 's then prove that \(\mathbf{A}\) is idempotent.
Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a normal distribution \(N\left(\mu, \sigma^{2}\right)\). Show that $$ \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2}+\frac{n-1}{n}\left(X_{1}-\bar{X}^{\prime}\right)^{2}, $$ where \(\bar{X}=\sum_{i=1}^{n} X_{i} / n\) and \(\bar{X}^{\prime}=\sum_{i=2}^{n} X_{i} /(n-1)\). Hint: \(\quad\) Replace \(X_{i}-\bar{X}\) by \(\left(X_{i}-\bar{X}^{\prime}\right)-\left(X_{1}-\bar{X}^{\prime}\right) / n\). Show that \(\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(n-2\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What then is the distribution of $$ \frac{[(n-1) / n]\left(X_{1}-\bar{X}^{\prime}\right)^{2}}{\sigma^{2}} ? $$
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