Test hypotheses using statistic value I have an exam today, and the solution to this problem that my teacher provided differs from mine:

A government department estimates that at most the 12% of the people older than 65 years has a health insurance. In a sample of 900 people older than 65 years, 801 has no health insurance. Is true what the government says, given the data? The confidence value is 0.1.
A. The suspicion of the health department is supported by the statistic evidence, with a level of significance of 10%
B. The suspicion of the health department is not supported by the statistic evidence, with a level of significance of 10%

This is meant to be solved using comparing the percent point of 0.1 in a normal distribution, to the statistic value $Zc$ obtained with
\begin{equation}
Zc = \frac{0.11 - 0.12}{\sqrt{\frac{0.12(1-0.12)}{900}}} = -0.9231861823449949
\end{equation}
With Python, the quantile point that leaves an area of 0.1 at its  left is  -1.28. With this, I conclude that the answer to this problem is B. Am I correct?
 A: The Health dept estimates that at most 12% of over 65 has a health insurance.
The data evidence shows that only 11% of the random sample has an insurance. So the suspicious of the Health dept is supported by the statistical evidence, and this without doing any test. The test is necessary only when the statistical evidence is against the null hypothesis....so the correct answer is A
More formally we have the following System to prove ($H_0$ is the suspicios of the Health Dept)
$$
\left\{ 
\begin{array}{c}
H_0: \mu \leq 12\% \\ 
H_0: \mu > 12\%
\end{array}
\right. 
$$
With $n=900$ we can assume a Gaussian model. Under this assumption and applying well known theorems, the Critical region is the following
$C=\{\mathbf{x} \in \mathbf{X} :\bar{X}_n>k\}$
That means, fixed a size $\alpha$ for the test, we reject $H_0$ if and only if
$$\mathbb{P}[\bar{X}_n>k|\mu=0.12]=10 \%$$
To calculate $k$, as usual,
$$\frac{k-0.12}{\sqrt{0.12\times0.88}}\sqrt{900}=1.28 \rightarrow k=13.39\%$$
Concluding, we reject $H_0$ if and only if the sample mean is $>13.39\%$

*

*Our mean is $11\%<13.39\%$ so we do not reject it (A)


*In terms of p.value, for this test we have a pvalule of $\mathbb{P}[\bar{X}_n>0.11|\mu=0.12]=82.20\%$ (to reject null hypothesis the pvalue must be less than 10%)
