So I need to decide if the following statement is true or false: Let $X_1,X_2,...,X_n \sim N(\theta,1)$. Then a $95\%$ confidence interval for $\theta$ is all of the values of $\theta$ that satisfies: $$\left\{ \theta:0.025\leq\int_{-\infty}^{\sqrt{n}\left(\bar{X}-\theta\right)}f\left(x\arrowvert\theta\right)dx\leq0.975\right\}$$ where the density function is:$$f\left(x\arrowvert\theta\right)=\frac{1}{\sqrt{2\pi}}e^{\frac{-\left(x-\theta\right)^{2}}{2}},-\infty<x<\infty$$
So my initial thought was that the statement is false because we know about CDF that $\int_{-\infty}^{\sqrt{n}\left(\bar{X}-\theta\right)}f\left(x\arrowvert\theta\right)dx=P\left(X\leq\sqrt{n}\left(\bar{X}-\theta\right)\right)=P\left(Z+\theta\leq\sqrt{n}\left(\bar{X}-\theta\right)\right)=P\left(Z\leq\sqrt{n}\left(\bar{X}-\theta\right)-\theta\right)$ and the inequality is true iff $z_{0.025}\leq\sqrt{n}\left(\bar{X}-\theta\right)-\theta\leq z_{0.975}$ from here we get $\bar{X}-\frac{z_{0.975}}{\sqrt{n}}\leq\theta+\frac{\theta}{\sqrt{n}}\leq\bar{X}+\frac{z_{0.975}}{\sqrt{n}}$. but becuase $X\sim N\left(\theta,1\right)$ we know that $$CI_{95\%}\left(\theta\right)=\left\{ \bar{X}-\frac{z_{0.975}}{\sqrt{n}}\leq\theta\leq\bar{X}+\frac{z_{0.975}}{\sqrt{n}}\right\}$$ and intervals are not equal so we can declare the statement as false but it seems that is not the right answer (it's a question from past exam and the answer which was similar to mine was marked as incorrect so possibly the statement may be true) and I can't think of a new way to tackle this so I will appreciate your help.