Is testing members of a group together a sample size of 1? Say you are measuring the decisions of 50 people in a room. They must either choose to go to the blue corner or the red corner. You count the number of people in each corner. Is this a sample size of 50 or 1?
I can see that the choices are not necessarily independent of each other, but also that this isn't a given.
 A: Suppose you can assume that each individual makes their own independent
decision, whether to go the Red or Blue corner. Then you could
call 'Blue' a 'Success', count the number $x$ of them.
You might test the null hypothesis $H_0: p= 1/2$ against
the alternative $H_a: p \ne 1/2,$ where $p$ is the probability of choosing Blue.
In these circumstances it is reasonable to assume
you have a sample size of $n = 50$ subjects (instead
of a sample size of $n = 1$ group of fifty).
If you observe $x = 35$ in the Blue corner out of
$n = 50,$ then an exact 2-sided binomial test in R would look like this.
binom.test(35, 50, p=.5)

        Exact binomial test

data:  35 and 50
number of successes = 35, number of trials = 50,
 p-value = 0.0066
alternative hypothesis: 
 true probability of success is not equal to 0.5
95 percent confidence interval:
 0.5539177 0.8213822
sample estimates:
 probability of success 
                    0.7 

The P-value $0.0066 < 0.01 = 1\%$ indicates that the
estimated proportion $\hat p = x/n = 0.70$ is significantly
different from $p_0 = 0.5$ at the 1% level of significance.
[If individual choices are independent and unbiased, then
it would be unlikely to have as many as $35$ individuals (or as few as $15)$ in the Blue corner.]
A 95% CI for $p$ based on your data is
$(0.555,\, 0.821).$
The P-value can be computed in R, using a binomial PDF,
as shown below. Alternatively, in this example, a normal approximation to the binomial distribution could be used to get very nearly the
same value.
sum(dbinom(c(0:15, 35:50), 50, 0.5)) 
[1] 0.006600448

In the plot below, the P-value is the sum of the heights
of the blue bars outside the two orange dotted lines.

R code for figure:
x = 0:50;  pdf = dbinom(x, 50, .5)
hdr = "PDF of BINOM(50, .5) Showing P-value"
plot(x, pdf, type="h", lwd=2, col="blue", main=hdr)
 abline(v = c(15.5, 34.6), lwd=2, lty="dotted", col="orange")
 abline(v=0, col="green2")
 abline(h=0, col="green2")


Note: An approximate normal test (with continuity
correction) can be done by finding
$Z = \frac{x - np_0}{\sqrt{np_0(1-p_0)}}
= \frac{34.5 - 25}{\sqrt{12.5}} = 2.687,$ and
rejecting $H_0$ at the 5% level, because $|Z| > 2.69 > 1.96.$
z = (34.5 - 25)/sqrt(12.5);  z
[1] 2.687006

The P-value $0.0072$ can be found in R as below
or approximated using printed tables of the
CDF of the standard normal distribution. This leads
to rejection at the 1% level.
2*(1 - pnorm(z))
[1] 0.007209571

A: I mean, it depends on if this is a controlled experiment or not. Normally you would want to try to manipulate the experimental condition in a way that is able to be modeled on other populations, so in this case a best case scenario is to split up the paradigm between 25 people in one condition (blue or red) and 25 in another (whatever color doesnt already have 25 people). A 49-to-1 paradigm would be a bit odd to draw valid statistical inference from.
