= Solution
Write $q=n-p$. In this <normal linear model>, the <ordinary least squares> criterion is $\|Y-Xb\|^2$. Differentiating with respect to $b$ gives the <normal equations for linear least squares>, $X^TXb=X^TY$. Full column rank of the <design matrix> makes $X^TX$ invertible, and completing the square proves that the unique minimum is
$$
\boxed{\widehat\beta=(X^TX)^{-1}X^TY.}
$$
Since $\widehat\beta=\beta+(X^TX)^{-1}X^T\epsilon$, its <sampling distribution> is the <multivariate normal distribution>
$$
\widehat\beta\sim N_p\bigl(\beta,\sigma^2(X^TX)^{-1}\bigr).
$$
Let $H=X(X^TX)^{-1}X^T$ be the <hat matrix>. It is symmetric and idempotent with rank $p$, and $HX=X$. The <regression residual> vector is $(I-H)Y=(I-H)\epsilon$, so
$$
\operatorname{RSS}=\|Y-X\widehat\beta\|^2=Y^T(I-H)Y,\qquad \boxed{s^2=\frac{\operatorname{RSS}}{n-p}.}
$$
An <orthogonal projection of a Gaussian vector> onto the residual space gives $\operatorname{RSS}/\sigma^2\sim\chi^2_q$. The residual projection and fitted projection have zero cross-<covariance>, hence are <independent> because they are jointly <normal>. In particular, $s^2$ is an <unbiased estimator> of $\sigma^2$ and is <independent> of $\widehat\beta$. The <maximum-likelihood estimator> of $\sigma^2$ instead divides by $n$ and is not unbiased.
For the <null hypothesis> $\beta_1=0$, put $v_{11}=((X^TX)^{-1})_{11}$. Under that <null hypothesis>, $\widehat\beta_1/(\sigma\sqrt{v_{11}})$ is standard <normal> and independent of $qs^2/\sigma^2$. Therefore the <Student t-test> uses
$$
T=\frac{\widehat\beta_1}{s\sqrt{v_{11}}}\sim t_q.
$$
\b[A two-sided level-$\alpha$ test rejects when $|T|>t_{q,1-\alpha/2}$.] The denominator is the estimated <standard error> of the <regression coefficient>.
The fitted <linear regression> includes an <intercept> and three <covariates>:
$$
\operatorname{NO2}_i=\beta_0+\beta_w\operatorname{wind}_i+\beta_t\operatorname{maxtemp}_i+\beta_s\operatorname{insol}_i+\epsilon_i,\qquad \epsilon_i\overset{\mathrm{iid}}\sim N(0,\sigma^2).
$$
Thus $n=25$, $p=4$ and the residual <degrees of freedom> are $21$. The fitted <conditional mean> is
$$
\widehat{\operatorname{NO2}}=3.784916-0.527410\operatorname{wind}+0.124991\operatorname{maxtemp}-0.005259\operatorname{insol}.
$$
Each slope measures a conditional association with the response while the other two <covariates> are held fixed. Increasing wind speed by one mile per hour corresponds to a fitted decrease of $0.527410$ in the concentration units used. Its <standard error> is $0.224904$, giving $t=-2.345$ and a two-sided <p-value> of $0.0289$: evidence of a negative conditional association at the 5% level. The temperature slope is $0.124991$ per degree Fahrenheit, with <standard error> $0.075173$, $t=1.663$ and <p-value> $0.1112$. The insolation slope is $-0.005259$ per langley per day, with <standard error> $0.006637$, $t=-0.792$ and <p-value> $0.4370$. Neither of these last two conditional effects is significant at 5%; that does not establish that either true slope is zero. These are associations, not a demonstration of causal effects.
The <intercept> estimates the mean at zero values of all three <covariates>. Its value $3.784916$ has a large <standard error> $7.979647$, giving $t=0.474$ and <p-value> $0.6402$. The zero-covariate combination may be outside the observed range, so this test may have little practical relevance. Every coefficient test uses a <Student t-distribution> with $21$ <degrees of freedom>.
The reported residual <standard deviation> $s=1.844$ measures unexplained variation in concentration units; the subsequent table gives $\operatorname{RSS}=71.428$, consistent with $s=\sqrt{71.428/21}$ after rounding. The five-number residual summary describes the spread, from $-2.3052$ to $3.4033$, with median $-0.4990$. An <intercept> forces the residual sum to zero, not the residual median. These five numbers alone cannot establish <normal> errors, constant <variance> or <independence>; residual-versus-fitted and <quantile-quantile plots> would be more informative, and successive days also warrant checking serial dependence. The <coefficient of determination> $R^2=0.6533$ means that 65.33% of the sample response sum of squares about its mean is explained by this fit. Its adjusted version is
$$
\overline R^2=1-\frac{\operatorname{RSS}/21}{\operatorname{TSS}/24}=0.6037,
$$
which accounts for the number of estimated <regression coefficients>.
The <stepwise selection by the Akaike information criterion> searches between the intercept-only model and the specified model with all three main effects. For these <Gaussian> fits, the printed <Akaike information criterion> is, up to a constant common to the candidates,
$$
\operatorname{AIC}=25\log(\operatorname{RSS}/25)+2p.
$$
The full model has $p=4$ and AIC $34.246$. Removing insolation increases the <residual sum of squares> by $2.136$ but decreases AIC to $32.982$, so that removal is chosen. Removing temperature or wind instead would give AIC $35.337$ or $38.060$ and is worse. In the next step the candidate models include deletions and restoration of insolation. Keeping wind and temperature has the smallest AIC: removing temperature gives $33.389$, restoring insolation gives $34.246$, and removing wind gives $37.960$. Hence the search stops, giving
$$
\boxed{\widehat{\operatorname{NO2}}=4.4368-0.5734\operatorname{wind}+0.1040\operatorname{maxtemp}.}
$$
The accompanying <F-tests> compare a larger model with a one-parameter reduction: $F=\Delta\operatorname{RSS}/s^2$, where $s^2$ is estimated from the larger model. For example, removing insolation gives $2.136/(71.428/21)=0.628$, equal to the corresponding squared <test statistic>. Deletions from the second-stage model use $73.564/22$ in the denominator, whereas restoring insolation uses the full-model denominator. These tests describe the comparisons; they do not determine the AIC choice. In particular, temperature is retained despite its deletion-test <p-value> $0.15016$. AIC balances fit and complexity, and a stepwise optimum need not be a global optimum over a richer model class. Ordinary coefficient inference after choosing a model does not account for the selection itself.
For positive responses, the <Box–Cox transformation> is $T_\lambda(y)=(y^\lambda-1)/\lambda$ for $\lambda\ne0$, and $T_0(y)=\log y$. Its <profile likelihood> refits the <linear regression> and <variance> at each $\lambda$ and includes the transformation Jacobian. The graph peaks at roughly $\lambda=0.3$–$0.4$. The approximate 95% <likelihood-ratio confidence interval> contains values a little below zero through approximately one, using the cutoff $\ell_{\max}-\chi^2_{1,0.95}/2$. Thus a <logarithmic transformation> ($\lambda=0$) and a square-root transformation ($\lambda=1/2$) are plausible, while leaving the response on its original scale ($\lambda=1$) is at or very near the upper confidence boundary. \b[The broad profile supports considering a mild power transformation; it does not determine a sharply estimated power.] The raster does not justify more precise endpoints or a decisive assertion about which side of the boundary $\lambda=1$ lies.
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