Why does the McNemar's test use $\chi^{2}$ and not the normal distribution? I just noticed how the non exact McNemar's test uses the $\chi^{2}$ asymptotic distribution.  But since the exact test (for the two case table) relies on the binomial distribution, how come it is not common to suggest the normal approximation to the binomial distribution?
 A: Won't the two approaches come to the same thing? The relevant chi-square distribution has one degree of freedom so is simply the distribution of the square of a random variable with a standard normal distribution. I'd have to go through the algebra to check, which I haven't got time to do right now, but I'd be surprised if you don't end up with exactly the same answer both ways.
A: A close-to-intuitive answer:
Take a closer look at the formula for the McNemar test, given the table
      pos | neg
----|-----|-----
pos |  a  |  b
----|-----|-----
neg |  c  |  d

The McNemar statistic M is calculated as:
$$ M = {(b-c)^2 \over b+c} $$
The definition of a $\chi^2$ distribution with k degrees of freedom is that it consists of the sum of squares of k independent standard normal variables. if the 4 numbers are large enough, b and c, and thus b-c and b+c can be approximated by a normal distribution. Given the formula for M, it's easily seen that with large enough values M will indeed follow approximately a $\chi^2$ distribution with 1 degree of freedom.

EDIT :
As onstop rightfully indicated, the normal approximation is in fact completely equivalent. That's rather trivial given the argument using the approximation of b-c by the normal distribution.
The exact binomial version is also equivalent to the sign test, in the sense that in this version the binomial distribution is used to compare b to $Binom(b+c,0.5)$. Or we can say that under the null hypothesis the distribution of b can be approximated by $N(0.5\times(b+c),0.5^2\times(b+c)$. 
Or, equivalently:
$$\frac{b-(\frac{b+c}{2})}{\frac{\sqrt{b+c}}{2}}\sim N(0,1)$$
which simplifies to
$$ \frac{b-c}{\sqrt{b+c}}\sim N(0,1)$$
or, when taken the square on both sides, to $M \sim \chi^2_1$.
Hence, the normal approximation is used. It is the same as the $\chi^2$ approximation.
