Riemann surface
matlab
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Solutions adapted from N.S. Boudreau’s Instructor’s Solution Manual (2011).
MBS, Ex. 12.2. On part (f), do not find or interpret $R_a^2$.
Solution:
(a) $\hat{\beta}0=506.346 ; \hat{\beta}_1=-941.900 ; \hat{\beta}_2=-429.060$ (b) $\hat{y}=506.36-941.90 x_1-429.1 x_2$. (c) $\mathrm{SSE}=151016 ; \mathrm{MSE}=8883 ; \mathrm{s}=94.251$. We expect about $95 \%$ of the $y$ values to be within $\pm 2 \mathrm{~s}= \pm 188.502$ unites of the fitted regression equation. (d) The $\mathrm{p}$-value for $\mathrm{H}_0: \beta_1=0$ against the alternative $\mathrm{H}{\mathrm{a}}: \beta_1 \neq 0$ is $\mathrm{p}=$ .003. Since $p<.05$, we would reject $\mathrm{H}0$; there is sufficient evidence to indicate $\beta_1 \neq 0$ at significance level $\alpha=.05$. (e) The $95 \%$ confidence interval for $\beta_2$ is $$ \begin{aligned} \hat{\beta}_2 \pm t{.025, n-k-1} \operatorname{SE}\left(\hat{\beta}2\right) & =-429.060 \pm 2.110(379.83) \ & =-429.060 \pm 801.4413 \ & =(-1230.5013,372.3813) . \end{aligned} $$ We have used that $n-k-1=20-2-1=17$, so that $t{.025, n-k-1}=$ 2.110 .
(f) $\mathrm{R}^2=45.9 \%$; the fitted regression model explains $45.9 \%$ of the variability in $y$.
$(\mathrm{g}) \mathrm{F}=7.22$
(h) The observed significance level is $p=0.005$. Since the p-value is so small, we would reject $H_0: \beta_1=\beta_2=0$ for most values of the significance level $\alpha$. We have very strong evidence that the model is useful (at least one of the predictor variables is useful for predicting $y$ ).
\begin{prob}
The Gesell dataset concerns a study of whether intelligence can be predicted based on the age at which a child starts to speak. For each of 21 participants in the study, the variable Age represents the age (in months) at which they spoke their first word, and the variable Score represents the Gesell Adaptive Score. (The Gesell test is an adult intelligence test).
(a) Without looking at the data, how would you expect Score to be related to Age? (Positively or negatively?)
Solution: Negatively: children who learn to speak earlier likely have higher intelligence. (Full credit for any answer.)
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