Binomial Distribution: Likelihood Ratio Test for Equality of Several Proportions $\newcommand{\szdp}[1]{\!\left(#1\right)}
\newcommand{\szdb}[1]{\!\left[#1\right]}$
Problem Statement: A survey of voter sentiment was conducted in four midcity
political wards to compare the fraction of voters favoring candidate $A.$
Random samples of $200$ voters were polled in each of the four wards. The numbers of voters favoring $A$ in the four samples can be regarded as four independent binomial random variables. Construct a likelihood ratio test of the hypothesis that the
fractions of voters favoring candidate $A$ are the same in all four wards.
Use $\alpha=0.05.$
Note 1: This is essentially Exercise 10.88 in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Sheaffer.
Note 2: I have looked at several threads asking the same question. This thread has no viable answer. This thread has a solution done mostly in R and is not a theoretical derivation of the needed result. This thread works out exactly zero details: and as you'll see, I'm definitely in the weeds on this one.
Note 3: This is cross-posted here.
My Work So Far: Let $p_i$ be the proportion of voters favoring $A$ in Ward $i.$ So the
null hypothesis is that $p_1=p_2=p_3=p_4,$ while the alternative hypothesis
is that at least one proportion is different from the others. We have
$f$ as the
underlying
distribution:
$$f(y_i)=\binom{n}{y_i}p_i^{y_i}(1-p_i)^{n-{y_i}}.$$
It follows that the likelihood function is
$$L(p_1,p_2,p_3,p_4)
=\prod_{i=1}^4\szdb{\binom{n}{y_i}p_i^{y_i}(1-p_i)^{n-y_i}}.$$
Then we construct $L\big(\hat\Omega_0\big)$ and $L\big(\hat\Omega\big).$ Note
that under the null hypothesis, we will set $p_1=p_2=p_3=p_4=p.$ Hence,
$$L\big(\hat\Omega_0\big)
=\prod_{i=1}^4\szdb{\binom{n}{y_i}p^{y_i}(1-p)^{n-y_i}}.$$
The one remaining parameter $p$ we will replace with its MLE, which we can
confidently say is $\big(\sum y_i\big)/(4n).$ Hence
\begin{align*}
L\big(\hat\Omega_0\big)
&=\prod_{i=1}^4\szdb{\binom{n}{y_i}\szdp{\frac{\sum 
y_i}{4n}}^{\!\!y_i}\szdp{1-\frac{\sum y_i}{4n}}^{\!\!n-y_i}}\\
&=\frac{1}{(4n)^n}\prod_{i=1}^4\szdb{\binom{n}{y_i}\szdp{\sum 
y_i}^{\!y_i}\szdp{4n-\sum y_i}^{\!n-y_i}}.
\end{align*}
Next, we turn our attention to $L\big(\hat\Omega\big):$
\begin{align*}
L\big(\hat\Omega\big)
&=\prod_{i=1}^4\szdb{\binom{n}{y_i}\szdp{\frac{y_i}{n}}^{\!y_i}
        \szdp{1-\frac{y_i}{n}}^{\!n-y_i}}\\
&=\prod_{i=1}^4\szdb{\binom{n}{y_i}\szdp{\frac{y_i}{n}}^{\!y_i}
        \szdp{\frac{n-y_i}{n}}^{\!n-y_i}}\\
&=\frac{1}{n^{4n}}\prod_{i=1}^4\szdb{\binom{n}{y_i}y_i^{y_i}
        \,\szdp{n-y_i}^{n-y_i}}.
\end{align*}
Next we form the likelihood ratio:
\begin{align*}
\lambda
&=\frac{L\big(\hat\Omega_0\big)}{L\big(\hat\Omega\big)}\\
&=\frac{\displaystyle 
        \frac{1}{(4n)^n}\prod_{i=1}^4\szdb{\binom{n}{y_i}
            \szdp{\sum y_i}^{\!y_i}\szdp{4n-\sum y_i}^{\!n-y_i}}}
       {\displaystyle 
        \frac{1}{n^{4n}}\prod_{i=1}^4\szdb{\binom{n}{y_i}y_i^{y_i}
            \,\szdp{n-y_i}^{n-y_i}}}\\
&=\frac{n^{4n}}{4^n\,n^n}\cdot
 \prod_{i=1}^4\szdb{\szdp{\frac{\sum y_j}{y_i}}^{\!y_i}\,
                       \szdp{\frac{4n-\sum y_j}{n-y_i}}^{\!n-y_i}}\\
&=\szdp{\frac{n^3}{4}}^{\!\!n}\cdot
 \prod_{i=1}^4\szdb{\szdp{\frac{\sum y_j}{y_i}}^{\!y_i}\,
                       \szdp{\frac{4n-\sum y_j}{n-y_i}}^{\!n-y_i}}.
\end{align*}
My Questions:

*

*This looks wrong to me, because I'm told (and it totally makes sense) that $0\le\lambda\le 1,$ whereas everything in sight is greater than $1.$

*Supposing this expression can be salvaged, what are the next steps? Should I take logs and try to simplify somehow?

*I'm expecting to be able to obtain a test something along the lines of
$$\frac{(1/(4n))\sum_{j=1}^ny_j-\sum_{j=1}^n(y_j/n)}{\displaystyle\sqrt{\sum_{j=1}^4\dfrac{(y_j/n)(1-y_j/n)}{n}}},$$
although this test doesn't strike me as sensitive enough. We could have $y_1/n$ much too low, and $y_4/n$ much too high, and this test could still mark them down as equal because they "average out" to the right thing. What's the right generalization to the standard difference of proportions test?


In the following, I'm going to use the numbers supplied in the exercise. The null hypothesis is $H_0:p_1=p_2=p_3=p_4=p$. As you wrote, the likelihood function is $$
L(\mathbb{p})=\prod_{i=1}^4{200\choose y_i}p_i^{y_i}(1-p_i)^{200-y_i}
$$
where $y_i$ are the number of voters favoring $A$ in ward $i$. Under $H_0$, the MLE of $p$ is $\hat{p}=\sum_{i=1}^{4}y_i/800$. Otherwise, we have $\hat{p}_i=y_i/200$ for $i=1, 2, 3, 4$. Plugging these into the likelihood function and forming the ratio, we get
$$
\lambda = \dfrac{\left(\dfrac{\sum y_i}{800}\right)^{\sum y_i}\left(1 - \dfrac{\sum y_i}{800}\right)^{800 - \sum y_i}}{\prod_{i=1}^{4}\left(\dfrac{y_i}{200}\right)^{y_i}\left(1 - \dfrac{y_i}{200}\right)^{200 - y_i}}.
$$
According to theorem 10.2 in the book, for large $n$, $-2\log(\lambda)$ has approximately a $\chi^2$ distribution with $\nu$ degrees of freedom when $H_0$ is true, where $\nu$ is the difference between the number of freely varying parameters in $\Omega$ and the number of such parameters in $\Omega_0$. Here, we have $4 - 1 = 3$ degrees of freedom.
Using the data provided in the book, we have $y_1 = 76, y_2 = 53, y_3 = 59, y_4 = 48$. According to the formula above, I get $-2\log(\lambda)=10.54$ and a $p$-value of $0.015$. This is very similar to what Rs prop.test gives:
y <- c(76, 53, 59, 48)
prop.test(y, rep(200, 4), correct = FALSE)

X-squared = 10.722, df = 3, p-value = 0.01333
alternative hypothesis: two.sided
sample estimates:
prop 1 prop 2 prop 3 prop 4 
 0.380  0.265  0.295  0.240 

