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When $X_1,ldots,X_n sim N(theta, 1)$ with $H_0:theta = theta_0$ versus $H_0:theta neq theta_0$, the the likelihood ratio test is given by

begin{align}
lambda({x}) &= frac{(2pi)^{-n/2}expleft{ frac{-sum_i^n(x_i - theta_0)^2}{2} right}}{(2pi)^{-n/2}expleft{ frac{-sum_i^n(x_i - bar{x})^2}{2} right}}
\&= expleft{frac{-n(bar{x} -theta_0)^2}{2}right}
end{align}

It follows that the rejection region ${xhspace{0.1cm}:lambda(x)hspace{0.1cm}leq c}$ is

begin{align}
left{xhspace{0.1cm}: |bar x - theta_0|geq sqrt{frac{-2log c}{n}}right}tag{1}
end{align}

However, when we have a single trial $X$ with $Xsim text{Binomial}(n, p)$ and with $H_0:p = 0.5$ versus $H_0:p neq 0.5$, I'm having trouble deriving the a similar result as in $(1)$. So to begin,

begin{align}
lambda({x}) &= frac{{n choose x}0.5^n}{{n choose x}hat{p}^x(1-hat{p})^{n-x}} \&= frac{0.5^n}{(frac{x}{n})^x(1-frac{x}{n})^{n-x}} hspace{2cm} text{ since $hat{p} = frac{x}{n}$} hspace{1cm}text{(MLE for $p$)}
end{align}

Now, we reject whenever $lambda(x) < c$ where $c$ is some constant.

begin{align}
frac{0.5^n}{left(frac{x}{n}right)^xleft(1-frac{x}{n}right)^{n-x}} < c
end{align}

However, this is where I get stuck. I'm not sure how to proceed from here? I want to end up with something similar to $(1)$ telling me when to reject $H_0$.

Suppose you have $n = 10$ trials with $x$ successes and you want
to test $H_0: p = 1/2$ vs $H_a: p ne 1/2$ at (somewhere near) the 5% level.
I say 'somewhere near' because the binomial distribution is discrete, so
it is not possible in general to achieve an exact significance level.

Under $H_0,$ (that, is assuming the null hypothesis to be true) the number of successes $X sim mathsf{Binom}(n=10,, p = 1/2).$

In your last inequality the fraction on the left-hand side is smallest when $hat p = X/n$ is far from $1/2.$ So you need to reject when the number of successes $X$ is far from $n/2.$

n = 10; x = 1:(n-1)

p = x/n; frac=.5^n/(p^x*(1-p)^(n-x))

plot(x, frac, pch=19)

Accordingly, we might reject for $X = 0,1,9,10,$ the four values most removed from $10/2 = 5.$ A calculation using the binomial PDF gives $P(X le 2) = P(X ge 9) = 0.0214.$ So that rejection rule leads to a test at about the 2% level.

If we try to reject for $X = 0,1,2,8,9,10,$ then the significance level escalates to $0.109,$ so you would be testing at about the 11% level. If you want to keep the significance level below 5%, then you'll have to use
the rule to reject for $X = 0,1,9,10.$

Here is a graph of the relevant binomial PDF:

Computations using R statistical software:

rej = c(0,1,9,10);  sum(dbinom(rej, 10, .5))

[1] 0.02148438

rej = c(0,1,2,8,9,10);  sum(dbinom(rej, 10, .5))

[1] 0.109375

Note: For larger values of $n,$ one might approximate binomial probabilities using a normal distribution, but $n = 10$ is a bit too small for completely satisfactory normal approximations.