= Solution
For a <generalized likelihood-ratio test>, maximize the <likelihood> separately over the null parameter space and the unrestricted space. Define $\Lambda=\sup_{H_0}L/\sup L\le1$, and reject for small $\Lambda$, equivalently large $T=-2\log\Lambda$. Choose the rejection threshold to achieve the desired null significance level, using an exact distribution when available or a justified asymptotic calibration; nuisance parameters are fitted under the respective restrictions.
For independent <Poisson distributions>, the likelihood is $L=\prod_i e^{-\lambda_i}\lambda_i^{X_i}/X_i!$. The unrestricted maximizers are $\widehat\lambda_i=X_i$, allowing zero as a boundary value. Under the common-mean null, maximizing $e^{-n\lambda}\lambda^{\sum_iX_i}$ gives $\widehat\lambda=\bar X$. The exponential and factorial factors cancel in the ratio, yielding
$$
\boxed{\Lambda=\prod_{i:X_i>0}\left(\frac{\bar X}{X_i}\right)^{X_i},\qquad T=-2\log\Lambda=2\sum_{i:X_i>0}X_i\log\frac{X_i}{\bar X}.}
$$
If all counts are zero, set $\Lambda=1$ and $T=0$; no division by $\bar X$ is then needed.
For positive $\bar X$, put $X_i=\bar X+\delta_i$, so $\sum_i\delta_i=0$. Expanding the logarithm gives
$$
(\bar X+\delta_i)\log(1+\delta_i/\bar X)=\delta_i+\frac{\delta_i^2}{2\bar X}+O(|\delta_i|^3/\bar X^2).
$$
Consequently
$$
\boxed{T=\frac1{\bar X}\sum_i(X_i-\bar X)^2+O\!\left(\frac{\sum_i|X_i-\bar X|^3}{\bar X^2}\right).}
$$
It is the <quadratic Pearson approximation to the Poisson deviance>. The approximated statistic is the logarithmic ratio $T$, not the bounded raw ratio $\Lambda$.
Under the null in the large-common-count regime, this statistic has an approximate <chi-squared distribution> with $n-1$ degrees of freedom: fitting the common mean removes one independent residual direction. For $n=7$, use six degrees of freedom. At $T=27.3$, the upper-tail probability in that approximation is
$$
P(\chi_6^2\ge27.3)=e^{-13.65}\left(1+13.65+\frac{13.65^2}{2}\right)\simeq1.27\times10^{-4}.
$$
Thus \b[reject the equal-mean null at conventional 5% or 1% significance levels], provided the expected counts justify the chi-square approximation. No particular significance level or common count was supplied, so the numerical decision is stated with those assumptions.
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