Inference Ph in water This is an example from my book and Iam translating from it so there might be some things that are not all clear. Just let me know so i can edit.
Someone is measuring PH-level in a sea and for that she has taken $\textbf{x} =(x_1,...x_n)$ which is a sample taken from $N(\mu, \sigma^2)$ where $\sigma = 0.2$. The confidence interval is done. A 95% upper confidence level for $\mu$ is: $\bar{\mu} = x + \lambda_{0.05}\frac{\sigma}{\sqrt{n}} = \bar{x} + 1.64499\frac{0.2}{\sqrt{10}} \approx 5.8 +0.10= 5.9$ (observe that $\bar{x} = 5.8$ and that $\lambda_{\alpha}$ is defined as the solution $\Phi(\lambda_{\alpha})= 1- \alpha$. $\Phi$ is the Normal distribution function)
A sea is considered acidified if the pH value is less than $\mu_0 = 6.0$ and we want to test if this limit is undershot(i don't know if undershot is a good transalation ,anyhow) So we test the hypothesis $H_0 : \mu = \mu_0$ against $H_1: \mu < \mu_0$ .  
A testvariable should be $\bar{x}$ or equivalently: $T = \dfrac{\bar{x}-\mu_0 }{\frac{\sigma}{ \sqrt{n}}}$.  which is a observation from $N(0,1)$. with $5\% $ error risk we get the test:    $\textbf{what do they mean by $\bar{x}$ is equivalent to $T$ they are just standardizing the variable? }$
"Discard the hypothesis $H_0$" if $T \leq -\lambda_{0.05}$    ($\textbf{i can't understand what they mean here $T\leq -\lambda_{0.05}$ in terms of seeing that $\mu < \mu_0$ }$)
...this occurs when:
$\bar{x}\leq \mu_{0} -\lambda_{0.05}\frac{\sigma}{\sqrt{n}} $ which is equivalent to:
$\mu_{0} \geq \bar{x} + \lambda_{0.05}\frac{\sigma}{\sqrt{n}}$
.....so we discard $H_0$
 A: 
A testvariable should be $\bar{x}$ or equivalently: $T = \dfrac{\bar{x}-\mu_0 }{\frac{\sigma}{ \sqrt{n}}}$.  which is a observation from $N(0,1)$. with $5\% $ error risk we get the test:    $\textbf{what do they mean by $\bar{x}$ is equivalent to $T$ they are just standardizing the variable? }$

Yes exactly they are just standardizing the random variable $\bar{x}$, the sample mean. If you rearrange to get $\bar{x}$ in terms of $T$ you find $\bar{x} = \frac\sigma{\sqrt{n}}T + \mu_0$. As $T$ is normally distributed with mean $0$ and standard deviation $1$ (under the null hypothesis) then this shows you that $\bar{x}$ must be is normally distributed with mean $\mu_0$ and standard deviation $\frac\sigma{\sqrt{n}}$ (under the null hypothesis) -- although of course we start with the distribution of $\bar{x}$ and work out $T$. This distribution for $\bar{x}$ makes sense because under the null hypothesis the true mean is $\mu_0$ and the population standard deviation is scaled by $1/\sqrt{n}$ to get the sampling distribution of the mean. 

"Discard the hypothesis $H_0$" if $T \leq -\lambda_{0.05}$    ($\textbf{i can't understand what they mean here $T\leq -\lambda_{0.05}$ in terms of seeing that $\mu < \mu_0$ }$)

In the earlier part, it seems that they had a 90% confidence interval with the right-hand side given by $\lambda_{0.05} = $ the 95th percentile of the normal distribution $\approx 1.64$. Here they are doing a one-sided test to the left with a significance level of $\alpha=$5% and because of the symmetry of the normal distribution the 5th percentile is exactly as far to the left of the mean (0) as the 95th percentile is to the right. Hence the 5th percentile $=-\lambda_{0.05}$.
Note: usually the point of these examples is to show the equivalence between an $(100\%-\alpha)$ confidence interval and a two-sided hypothesis test with the same $\alpha$. However, here they're showing that a one-sided test of significance $\alpha$ takes its rejection region bound from one side of a $(100\%-\mathbf{2}\alpha)$ confidence interval. Perhaps this is to make the point that using a one-sided test corresponds to a lower-confidence confidence interval than a two-sided test. Usual practice is to use two-sided tests to stop you rigging your test by using a one-sided test and then deciding after collecting the data which side to reject on.
