How can the chi square replicate so closely a Z-test despite its heavier tails? The chi square is used often as a test of proportions, goodness of fit, contingency tables, etc. But the point I want to focus on is that it reliably replaces the Z-test under minimal conditions to test proportions and 2 x 2 contingency tables, based on the idea that its behavior for a statistic where the deviations are squared is akin to the behavior of the deviations of the sample means under the CLT from a theoretical population proportion, since the chi square with 1 df (as used in a 2 x 2 contingency table) is a squared standard Gaussian.
The chi square is exponential. However, I was surprised to see how much heavier its tail is upon plotting:

and I want to ask if this is encapsulated in its relative higher kurtosis (12 versus 0), or there are other factors, as well as how the distribution is able to replicate the results of a Z-test so well despite this heavier tail.
 A: If $Z \sim \mathsf{Norm}(0. 1).$ than $Q = Z^2 \sim \mathsf{Chisq}(\nu=1),$ a chi-squared distribution with 1 degree of freedom. A z-test would reject the null hypothesis in a two-sided test at the 5% level if test statistic $Z$ has $|Z| \ge 1.96.$ A corresponding chi-squared test (now one-sided) would reject at the 5% level
if $Q \ge 3.841.$
Computations in R:
qnorm(.975)
[1] 1.959964
1.959964^2
[1] 3.841459
qchisq(.95, 1)
[1] 3.841459

Example:
In City A, $x_1 = 438$ randomly chosen subjects out of $n_1 = 818$ preferred Candidate x for governor. In City B there were $x-2 = 501$ out of $n_2 = 1003$ subjects in favor of Candidate X.
Is there a significant difference in the polling popularity of Candidate X between the two cities?
Chi-squared test statistic. In R, prop.pest uses a chi-squared test statistic.
prop.test(c(438, 501), c(818, 1004))

        2-sample test for equality of proportions 
        with continuity correction

data:  c(438, 501) out of c(818, 1004)
X-squared = 2.2538, df = 1, p-value = 0.1333
...
sample estimates:
   prop 1    prop 2 
0.5354523 0.4990040 

Normal test statistic. Essentially, the same test in Minitab uses a z statistic: and gets the same P-value to two decimal places.
Test and CI for Two Proportions 

Sample    X     N  Sample p
1       438   818  0.535452
2       501  1003  0.499501

Difference = p (1) - p (2)
Estimate for difference:  0.0359508
...
Test for difference = 0 (vs ≠ 0):  
  Z = 1.53  P-Value = 0.126
...

The two software programs use slightly different formulas and round-off conventions, but $Z^2 = 1.53^2 = 2.3409 \approx 2.25 = Q$ X-squred in R printout.
