Using a z-test for proportion in place of Chi-Square Here's the study:
Pam knows from prior research that children who are seated in a room with 4 doors are equally likely to pick any door when asked which door they would choose then leaving the room. Pam wants to demonstrate that children labeled by a psychologist as oppositional are more likely to choose the
door behind them.
◦She recruits 80 children with the diagnosis and tests her
hypothesis.
The answer key says to use Chi-Square goodness of fit. My problem is that this will not answer the research question: whether or not oppositional children are more likely to choose the door behind them. It will only tell us if the model of all doors equally likely is a good fit. I was curious if I could just set up a z-test for proportions with two categories: "doors not behind" and "door behind." Would this work?
 A: Eighty is a large enough sample size that you could
use the normal approximation to a binomial distribution. It seems that there are various way to interpret the behavior of "oppositional' children. I will assume that you wish to test $H_0: p \le .25$ against $H_a: p > .25.$ (A traditional chi-squared test would test that all doors are equally likely against the alternative that they are not equally likely.)
If you get 27 children leaving by the rear door, then here is output for a z-test from Minitab statistical software.
Test and CI for One Proportion 

Test of p = 0.25 vs p > 0.25

Sample   X   N  Sample p  95% Lower Bound  Z-Value  P-Value
1       27  80  0.337500         0.250541     1.81    0.035

Because the P-value is less than or equal to $0.05 = 5\%,$ you can reject $H_0$ at the 5% level.
The null distribution (distribution assuming $H_0$ is true) is $\mathsf{Binom}(n=80, p=1/4).$ If $X$ has this distribution, you can standardize to
approximate $P(X \ge 27),$ which is the P-value of this test. After standardizing you can use
printed tables of the standard normal distribution to find the P-value. So it is not necessary to use
software to solve your problem.
Another approach would be to use an exact binomial test. Minitab will also do an exact binomial test, but for variety, I will show results of binom.test from R:
 binom.test(27, 80, p=.25, alt="greater")

        Exact binomial test

data:  27 and 80
number of successes = 27, number of trials = 80, 
 p-value = 0.04989
alternative hypothesis: 
 true probability of success is greater than 0.25
95 percent confidence interval:
 0.2500512 1.0000000
sample estimates:
probability of success 
                0.3375 

In this case a statistical calculator or statistical software would be convenient to
find $P(X \ge 27) = 1 - P(X \le 26) = 0.0499.$
In R, the computation looks like this:
1 - pbinom(26, 80, .25)
[1] 0.04988718

The figure below shows the PDF of $\mathsf{Binom}(80, 0.25)$. The exact binomal P-value is the sum of the heights of the bars to the right of the dotted vertical line. The approximate normal P-value is the area under the approximating normal
density curve to the right of that line.

R code for figure:
x = 0:80;  PDF = dbinom(x, 80, .25)
plot(x, PDF, lwd=2, type="h", main="PDF of BINOM(80,.25)")
 abline(v=0, col="green2")
 abline(h=0, col="green2")
 abline(v=26.5, col="orange", lwd=2, lty="dotted")
mu = 80*.25;  sg = sqrt(80*.25*.75)
 curve(dnorm(x, mu, sg), add=T, col="blue", lwd=2)
                      

