Solution verification for hypothesis testing and confidence interval problem

There's a random sample of size $n=10$ observations over a normally distributed population with standard deviation $\sigma =2$:

8.7, 7, 4, 7.6, 3, 8.1, 6.4, 6.1, 9.4, 6.2

- Find 96% confidence interval for the population mean

- Test the hypothesis ${H}_{0}:\mu =7.5$ against Ha: $\mu \ne 7.5$ with significance level $\alpha =0.01$. Find approximation of the observed p value.

My Solution:

We are looking for real numbers l and u s.t.:

$P\left(l<\hat{\theta}<u\right)=0.96$, where $\hat{\theta}=\frac{\sum _{i=1}^{n}{X}_{i}-n\theta}{\sqrt{n{\sigma}^{2}}}\sim N(0,1)$

We are looking for $z}_{0.2$ which from the z-score table is: 2.055. Therefore:

$-2.055<\frac{{\sum}_{i=1}^{n}{X}_{i}-n\theta}{\sqrt{n{\sigma}^{2}}}<2.055\equiv -2.055\sqrt{40}<\sum _{i=1}^{n}{X}_{i}-n\theta <2.055\sqrt{40}\equiv $

$\equiv \xad12.996-66.5<-10\theta <12.996-66.5\equiv 5.350<\theta <7.94.$

For the second part:

We set two z-scores: ${z}_{1}=-2.325\text{}\text{and}\text{}{z}_{2}=2.325$ from which we construct a two-tailed test.

Our test statistic is: $t=\frac{(6.65-7.5)\sqrt{10}}{2}=-1.343\Rightarrow t\ge -2.235$ from where we see the test statistic t is not in the rejection region, therefore we fail to reject the null hypothesis.