Testing coefficients separately and simultaneously From the summary output by lm, I have the following from R: 
Call:  
lm(formula = DJIA[2:262] ~ DJIA[1:261])
Residuals:
     Min       1Q   Median       3Q      Max
-176.878  -22.397   -0.641   26.478  125.139
Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) 36.898384  42.989100   0.858    0.392    
DJIA[1:261]  0.994459   0.007477 133.002   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 42.37 on 259 degrees of freedom
Multiple R-squared: 0.9856,     Adjusted R-squared: 0.9855
F-statistic: 1.769e+04 on 1 and 259 DF,  p-value: < 2.2e-16**

How do I test the coefficients separately and simultaneously in this case? What if I assume that the model has an intercept of 0 and slope of 1? Does it translate to B0 = 0 and B1 = 1?
 A: Recall that in the linear regression model
$$
Y_i = \boldsymbol{X}_i'\boldsymbol{\beta} + \varepsilon_i
$$
together with $\mathbb{E}(\varepsilon_i \mid \boldsymbol{X}_i) = 0$, a Wald test of the null hypothesis $\mathbf{r}\boldsymbol{\beta} - \boldsymbol{r}=\boldsymbol{0}$, requires the specification of the non-stochastic matrix $\mathbf{r}$ and  the non-stochastic vector $\boldsymbol{r}$.
In R, you can get away without having to specify the matrix $\mathbf{r}$ (which is typically sparse), and just specify the vector $\boldsymbol{r}$, if, as you slightly paradoxically put it, you want to test restrictions "separately and simultaneously" by  which I presume you mean that your tests do not involve linear combinations of the parameters. 
So, for example, in the following example, you are jointly testing the null hypothesis that the intercept is 0, and the pop75 variable is equal to 1.
    library(aod)
    require(stats)
    lmLCH <- lm(sr ~ pop15 + pop75 + dpi + ddpi, data = LifeCycleSavings)
    vR <- c(0, 0, 1, 0, 0)
    wald.test(b=coef(lmLCH)-vR, Sigma = vcov(lmLCH), Terms=c(1, 3))

The vector vR corresponds to the $\boldsymbol{r}$ above. This gives you the output:
Wald test:
----------

Chi-squared test:
X2 = 16.3, df = 2, P(> X2) = 0.00028

Equivalently, you can take the longer road home, by specifying the matrix $\mathbf{r}$,
$$
\mathbf{r} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0\end{bmatrix}
$$
and $\boldsymbol{r} = [0, 1]'$.
vR <- c(0, 1)
mR <- as.matrix(rbind(c(1, 0, 0, 0, 0), c(0, 0, 1, 0, 0)))
wald.test(b=coef(lmLCH), Sigma = vcov(lmLCH), L = mR, H0 = vR)

Note that the two give identical results.
