UMP Test of $H_0 : \theta \leq 0.1$ vs. $H_1 : \theta \gt 0.1$ for iid Geometric Random Variables 
Consider $90$ iid geometric random variables having parameter $\theta$
  (and thus mean $1/\theta$), and a UMP test of the null hypothesis that
  $\theta \leq 0.1$ against the alternative that $\theta \gt 0.1$. Give the
  approximate p-value (based on large sample results) which results for
  the case of the sum of the observations equal to $723$.

This is a practice problem (not homework). I'm given that the answer should be $0.025$. 
I have that if $X_i$ are iid random variables from a full one-parameter exponential family distribution, then for a UMP test of
$$H_0:\theta\leq\theta_0 \text{ vs. } H_1:\theta\gt\theta_0$$
we can use 
$$T\left(\vec{X}\right)=\sum_{i=1}^n t(X_i)$$
as a test statistic.
We have that the geometric distribution is a full one-parameter exponential family with
$h(x)=I_{\{1,2,...,\}}(x)$, $c(\theta)=\theta$, $t(x)=x-1$, and $w(\theta)=\text{log}(1-\theta)$
We can use 
$$T\left(\vec{X}\right)=\sum_{i=1}^n t(X_i)$$
as a test statistic and reject $H_0$ for sufficiently small values of $T(\vec{X})$ since $w(\theta)$ is decreasing. 
Applying the CLT to obtain an approximate p-value, we get
$$\mu_{t,0}=E_{H_0}(t(X_1))=E_{H_0}(X_1-1)=\frac{1}{0.1}-1=9$$
and 
$$\sigma_{t,0}=\sqrt{\mathsf{Var}_{H_0}(t(X_1))}=\sqrt{\frac{1-0.1}{0.1^2}}=\sqrt{90}$$
So $$\frac{T\left(\vec{X}\right)-n\cdot\mu_{t,0}}{\sqrt{n}\cdot\sigma_{t,0}}\overset{d}{\longrightarrow} \mathsf N(0,1)$$
But
$$\Phi\left(\frac{723-90\cdot9}{\sqrt{90}\cdot\sqrt{90}}\right)\approx0.167\neq0.025$$
Any ideas on where I went wrong?
 A: Since $t(x) = x-1$, a sufficient statistic for the data is the sample mean $\bar{x}_n$, which can be used as your test statistic.  For the geometric distribution this has moments:
$$\mathbb{E}(\bar{X}_n) = \frac{1}{\theta} \quad \quad \quad \mathbb{V}(\bar{X}_n) = \frac{1-\theta}{n \theta^2}.$$
For large $n$ we can appeal to the central limit theorem to obtain the approximate distribution:
$$\frac{\bar{X}_n - \mathbb{E}(\bar{X}_n)}{\sqrt{\mathbb{V}(\bar{X}_n) }} = \sqrt{n} \cdot\frac{\theta \bar{X}_n - 1}{\sqrt{1-\theta}} \sim \text{N}(0,1).$$
Consider the hypotheses $H_0: \theta \leqslant \theta_0$ and $H_A: \theta > \theta_0$.  For these hypotheses, smaller values of the sample mean are more conducive to the alternative hypothesis, and so you have p-value:
$$\begin{equation} \begin{aligned}
p(x_1,...,x_n) 
&= \sup_{\theta \leqslant \theta_0} \mathbb{P}( \bar{X}_n \leqslant \bar{x}_n | \theta) \\[6pt]
&= \sup_{\theta \leqslant \theta_0} \mathbb{P} \Bigg( \sqrt{n} \cdot\frac{\theta \bar{X}_n - 1}{\sqrt{1-\theta}} \leqslant \sqrt{n} \cdot\frac{\theta \bar{x}_n - 1}{\sqrt{1-\theta}} \Bigg| \theta \Bigg) \\[6pt]
&\approx \sup_{\theta \leqslant \theta_0} \Phi \Bigg( \sqrt{n} \cdot\frac{\theta \bar{x}_n - 1}{\sqrt{1-\theta}}  \Bigg) \\[6pt]
&= \Phi \Bigg( \sqrt{n} \cdot\frac{\theta_0 \bar{x}_n - 1}{\sqrt{1-\theta_0}}  \Bigg) \\[6pt]
&= \Phi \Bigg( \frac{1}{\sqrt{n}} \cdot\frac{\theta_0 n \bar{x}_n - n}{\sqrt{1-\theta_0}}  \Bigg). \\[6pt]
\end{aligned} \end{equation}$$
Your example: In your particular case you have $\theta_0 = 0.1$, $n = 90$ and $\bar{x}_n = 723/90$.  So your approximate p-value should be:
$$\begin{equation} \begin{aligned}
p 
&\approx \Phi \Bigg( \frac{1}{\sqrt{90}} \cdot\frac{0.1 \cdot 723 - 90}{\sqrt{1-0.1}} \Bigg) \\[6pt]
&= \Phi \Bigg( \frac{1}{\sqrt{90}} \cdot\frac{72.3 - 90}{\sqrt{0.9}} \Bigg) \\[6pt]
&= \Phi \Bigg( - \frac{1}{\sqrt{90}} \cdot\frac{17.7}{\sqrt{0.9}} \Bigg) \\[6pt]
&= \Phi \Bigg( - \frac{17.7}{\sqrt{81}} \Bigg) \\[6pt]
&= \Phi \Bigg( - \frac{17.7}{9} \Bigg) \\[6pt]
&= \Phi ( - 1.9 \bar{6} ) \\[6pt]
&= 0.02461083. \\[6pt]
\end{aligned} \end{equation}$$
In your calculation you made an error in the last line where you substituted $T(\vec{X}) = 723$ instead of $T(\vec{X}) = 723-90 = 633$ (consistent with your definition of $T(\vec{X})$).  If you correct this error then you get the same value as in my working.
