How to test if the slopes in the linear model are equal to a fixed value? Suppose we have a simple linear regression model $Z = aX + bY$ and would like to test the null hypothesis $H_0: a=b=\frac{1}{2}$ against the general alternative. 
I think one can use the estimate of $\hat{a}$ and $SE(\hat{a})$ and further apply a $Z$-test to get the confidence interval around $\frac{1}{2}$. Is this ok?
The other question is strongly related to this one. Suppose that we have a sample $\{(x_1,y_1,z_1),\ldots ,(x_n,y_n,z_n) \}$ and we compute $\chi^2$ statistics 
\begin{equation}
\sum_{i=1}^n \frac{(z_i-\frac{x_i+y_i}{2})^2}{\frac{x_i+y_i}{2}}.
\end{equation}
Can these statistics be used to test the same null hypothesis?
 A: You can test this hypothesis with a full versus reduced model test. Here is how you do this. First, fit the model $Z = aX + bY$ and get the residuals from that model. Square the residuals and sum them up. This is the sum of square error for the full model. Let's call this $SSE_f$. Next, calculate $Z - \hat{Z}$ where $\hat{Z} = 1/2*X + 1/2*Y$. These are your residuals under the null hypothesis. Square them and sum them up. This is the sum of square error for the reduced model. Let's call this $SSE_r$.
Now compute:
F = $((SSE_r - SSE_f)/2) / (SSE_f / (n-2))$,
where $n$ is the sample size. Under $H_0$, this F-statistic follows an F-distribution with $2$ and $n-2$ degrees of freedom.
Here is an example using R:
x <- rnorm(n)
y <- rnorm(n)
z <- 1/2*x + 1/2*y + rnorm(n) ### note I am simulating under H0 here

res <- lm(z ~ x + y - 1)
summary(res)
SSE.f <- sum(resid(res)^2)

zhat  <- 1/2*x + 1/2*y
SSE.r <- sum((z-zhat)^2)

F <- ((SSE.r - SSE.f) / 2) / (SSE.f / (n-2))
pf(F, 2, n-2, lower.tail=FALSE) ### this is the p-value

Reject the null if the p-value is below .05 (if your $\alpha$ is indeed .05).
I assume you really meant for your model not to contain an intercept. In other words, I assume you are really working with the model $Z = aX + bY$ and not $Z = c + aX + bY$.
A: In linear regression the assumption is that $X$ and $Y$ are not random variables.  Therefore, the model
$$Z = a X + b Y + \epsilon$$
is algebraically the same as
$$Z - \frac{1}{2} X - \frac{1}{2} Y = (a  - \frac{1}{2})X + (b  - \frac{1}{2})Y + \epsilon = \alpha X + \beta Y + \epsilon.$$
Here, $\alpha = a - \frac{1}{2}$ and $\beta  =b - \frac{1}{2}$.  The error term $\epsilon$ is unaffected.  Fit this model, estimating the coefficients as $\hat{\alpha}$ and $\hat{\beta}$, respectively, and test the hypothesis $\alpha = \beta = 0$ in the usual way.

The statistic written at the end of the question is not a chi-squared statistic, despite its formal similarity to one.  A chi-squared statistic involves counts, not data values, and must have expected values in its denominator, not covariates.  It's possible for one or more of the denominators $\frac{x_i+y_i}{2}$ to be zero (or close to it), showing that something is seriously wrong with this formulation.  If even that isn't convincing, consider that the units of measurement of $Z$, $X$, and $Y$ could be anything (such as drams, parsecs, and pecks), so that a linear combination like $z_i - (x_i+y_i)/2$ is (in general) meaningless.   It doesn't test anything.
