= Solution
\b[The fitted age slope is very small and negative, about $-0.00319$ efficacy points per year, with no conventional evidence against a zero linear slope: $t\approx-0.951$, $p\approx0.342$. The model explains only about 0.0367% of the observed efficacy variation. These results concern a linear association, not proof of no possible age relationship or a <causal effect> of ageing.]
The <descriptive statistics> report 2466 observed ages and efficacy scores, with no missing values for either variable. Age ranges from 14 to 97 years, with <sample mean> 52.689 years and <sample standard deviation> 19.064 years. Efficacy ranges from 0 to 16, with <sample mean> 9.348 and <sample standard deviation> 3.176 scale points. The <standard deviations> use different units and should not be compared as though their numerical ratio measured relative substantive variability. The efficacy mean retains the conflict with Question 1's group summaries; that conflict is not resolved by this separate regression output.
A <simple linear regression> models $\mathbb E(Y\mid x)=\beta_0+\beta_1x$, where $Y$ is efficacy and $x$ is age in years. <Ordinary least squares> chooses the coefficients to minimize the <residual sum of squares>. From the printed coefficients the fitted line is
$$
\boxed{\widehat y=9.516515-0.003192\,x.}
$$
Holding to this model, ten additional years correspond to $-0.03192$ fitted scale points. For example, the fitted mean is about 9.453 at age 20 and 9.261 at age 80, a difference of about $-0.192$ points. These are fitted means, not exact predictions of individual scores. The line does not imply that every person's efficacy falls at that rate, nor that cross-sectional age differences identify within-person changes over time.
The <intercept> is the predicted mean at age zero, about 9.517. Since zero is outside the observed 14–97 age range, its direct interpretation requires extrapolation. The printed <standard error> of 0.188043 and $t\approx50.608$ yield the extremely small displayed <p-value> for a zero <intercept>. That test does not test whether age is a useful predictor. Recentring age at its observed <sample mean> would put the <intercept> near $9.516515-0.003192(52.689)\approx9.34833$, the fitted mean at a substantively relevant reference age. This agrees with the rounded overall efficacy mean.
The age <standard error> is 0.003356 points per year. The coefficient divided by this <standard error> gives $-0.003192/0.003356\approx-0.9511$, agreeing with the printed t value. Estimating an <intercept> and one <regression slope> from 2466 observations leaves $2466-2=2464$ residual <statistical degrees of freedom>. Under a correctly specified <normal linear model>, the <Student t-distribution> provides the stated two-sided <p-value>. At 5%, or 1%, it does not reject $H_0:\beta_1=0$. The <p-value> is not a 34.2% <probability> that age has no effect, and nonrejection does not establish a practically negligible <effect size>.<Summarizing Linear Model Fits in the R documentation>[Value: coefficient estimate, standard error, two-sided t and p columns; residual degrees of freedom]{ref}
A 95% <confidence interval> calculated from the printed estimate and <standard error> is approximately
$$
\boxed{\beta_1\in[-0.009773,\ 0.003389]\text{ scale points per year},}
$$
or about $[-0.09773,0.03389]$ points per decade. It includes zero and gives the scale of slopes compatible with these summaries and assumptions. Assessing a practically negligible effect would require an explicit substantive tolerance and an appropriate <equivalence test>, not simply the fact that $p>0.05$. A nonlinear association could also coexist with a near-zero overall linear slope.
The <coefficient of determination> is $R^2=0.000367$, meaning that the fitted line explains about $100R^2=0.0367\%$ of the sample's centred efficacy <sum of squares> compared with an intercept-only fit. This is not 36.7%, a fraction of people whose efficacy is predicted correctly, or the proportion of efficacy caused by age. The <adjusted coefficient of determination>, approximately $-0.00003873$, is slightly negative because its penalty for estimating the extra coefficient exceeds the tiny improvement in fit. It need not be constrained to be nonnegative and does not mean a negative amount of explained variation.
The overall $F\approx0.9045$ on 1 and 2464 <statistical degrees of freedom>, with $p\approx0.3417$, tests the same zero-slope hypothesis as the age t test because there is only one predictor. The <simple-regression output identities> give $F=t_{\mathrm{slope}}^2$ and $F=(2464)R^2/(1-R^2)$; the printed numbers agree allowing their displayed precision. The signed <Pearson correlation coefficient> is consequently about $-\sqrt{R^2}\approx-0.0192$. These are related summaries of the same weak linear association, not <independent> corroborating tests.
The <residual standard error> of 3.177 points estimates the typical unexplained error scale under the fitted model. It is close to the efficacy <sample standard deviation>, as expected when the line explains almost none of the variation. It is not a <standard error> of the slope or a guarantee that each individual's prediction error lies within 3.177 points. The residual summary spans roughly $-9.46$ to $6.75$, with <sample median> about 0.573 and central <quartiles> about $-1.437$ and $2.579$. An ordinary least-squares fit with an <intercept> has zero mean residual, not necessarily zero median residual. These few <quantiles> cannot establish <normality>, constant conditional spread, absence of outliers or linearity. The apparent last-decimal discrepancies among rounded error scales and other summary fields do not force an additional contradiction: their <rounding> intervals admit compatible regression summaries.
The conventional uncertainty calculations assume <independent>, mean-zero errors conditional on age, an appropriate linear mean, and <homoscedasticity>; normal errors give the usual exact finite-sample t and F calibration. Large samples can relax some distributional requirements but do not remove <confounding>, <selection bias> or dependence. A <residual-versus-fitted plot> could reveal curvature or changing spread, and a <normal probability plot> could assess error shape; neither plot can be reconstructed from the printed residual <quantiles>. A <prediction interval> for an individual includes residual variation and is wider than a <confidence interval> for the mean response. \b[The supplied output supports a very weak, imprecisely signed linear association with little predictive gain from age alone, while leaving nonlinear, causal and design-based questions unresolved.]
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