Suppose that in a sample of size $n = 100$ from an AR(1) process with mean $\mu$
$$X_t - \mu = \phi (X_{t-1}-\mu) + Z_t$$
where ${Z_t} \sim \operatorname{WhiteNoise}(0, \sigma^2), \sigma^2 = 2, \phi = 0.6$, we obtain $\bar{x}_{100} = 0.271$.
Construct an approximate 95% confidence interval for $\mu$. Are the data compatible with the hypothesis that $\mu = 0$?
To solve this problem, I let $\bar{X}=\frac{1}{n}\sum\limits_{t=1}^{n}{X_t}$ and
$$\gamma(h) = Cov(X_t, X_{t+h}) = Cov(X_{t+h}, X_t) = \frac{\phi^h\sigma^2}{1-\phi^2}, \forall h\ge0$$
Then I got a problem with calculating $Var(\bar{X})$:
$\begin{equation} \begin{aligned} \operatorname{Var} \left( {\bar X} \right) &= \operatorname{Cov} \left( {\bar X,\bar X} \right) = \operatorname{Cov} \left( {\frac{1}{n}\sum\limits_{t = 1}^n {{X_t}} ,\frac{1}{n}\sum\limits_{t = 1}^n {{X_t}} } \right)\\ &=\frac{1}{{{n^2}}}\left( {n\gamma (0) + 2(n - 1)\gamma (1) + 2(n - 2)\gamma (2) + \ldots+ 2\gamma (n - 1)} \right)\\ &=\frac{1}{{{n^2}}}\left[ {n + 2(n - 1)\phi + 2(n - 2){\phi ^2} + \ldots +2{\phi ^{n - 1}}} \right]\gamma \left( 0 \right)\\ &=\frac{1}{{{n^2}}}\left[ {n + 2\sum\limits_{k = 1}^{n - 1} {\left( {n - k} \right){\phi ^k}} } \right]\gamma \left( 0 \right) \end{aligned} \end{equation}$
Am I doing it right? If I want to estimate $\sigma_{\bar{X}}$, could I use $\sqrt{Var(\bar{X})}$?
Thank you in advance.