= Solution
\b[The association is strongly <statistically significant> under the reported regression assumptions, but age alone has only moderate predictive ability.] The reported <confidence interval> for the age <regression coefficient> excludes zero by a wide margin. If it is the usual two-sided 95% interval from a <normal linear model> with an intercept and one predictor, there are $206-2=204$ <degrees of freedom>. Taking $t_{0.975,204}\simeq1.97$ gives
$$
\operatorname{SE}(\widehat\beta)\simeq\frac{0.05}{1.97}=0.0254,
\qquad t_{\rm obs}\simeq\frac{0.22}{0.0254}=8.67.
$$
Thus the two-sided <p-value> against zero slope is far below $0.001$. This conclusion concerns a statistical association in the sampled population; it does not establish an individual growth rate. Clustering by practitioner, nonconstant <variance> or nonindependent observations would require an appropriate <standard error> rather than blind reliance on this calculation.
The <scatter plot> has substantial vertical spread at each age. A narrow slope <confidence interval> describes uncertainty about the average trend, whereas a <prediction interval> for a new person's ear length must also include the residual <variance>. The reported interval even permits an approximate numerical assessment: write $S_{xx}=\sum_i(x_i-\bar x)^2$ and let $s_e^2=\mathrm{RSS}/(n-2)$. In <simple linear regression>, $t^2=\widehat\beta^2S_{xx}/s_e^2$, while the fitted and residual sums of squares are $\widehat\beta^2S_{xx}$ and $(n-2)s_e^2$. Hence the <coefficient of determination> is
$$
\boxed{R^2=\frac{t^2}{t^2+n-2}\simeq0.27.}
$$
Under these assumptions, age explains only about 27% of the sample variation; roughly 73% remains unexplained. The calculation is approximate because the printed slope and interval are rounded. For a new individual aged $x_0$, the usual <prediction interval> is based on
$$
\widehat Y(x_0)\ \pm\ t_{0.975,n-2}s_e
\sqrt{1+\frac1n+\frac{(x_0-\bar x)^2}{S_{xx}}}.
$$
The leading one represents individual residual variation. The provided summaries do not determine $s_e$ and $S_{xx}$ separately, so they do not supply a numerical <prediction interval>.
Record sex, height or another measure of body size, ancestry, family characteristics and practitioner or measurement method. Use these <covariates> in a <multiple linear regression>, allowing scientifically justified nonlinear age effects or interactions, and assess <prediction error> on held-out observations. A guessed overall sex proportion cannot replace individual sex measurements or identify sex-adjusted effects.
There is a source inconsistency: the PDF's numerical ear-length summary is ten times the scale of its plotted measurements and fitted line. Indeed, the fitted line at the reported mean age gives about $67.7$ mm, whereas the printed mean is about ten times larger. Missing decimal points in the summary are a plausible explanation. The significance and approximate $R^2$ above use the mutually consistent slope, interval and graph; they do not treat the inconsistent summary as valid data.
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