Why does bigger $n$ mean higher chance to reject the null hypothesis?

Question

I'm new to statistics and to statistical tests, so please bare with me.
The attached sketches are rough estimations, and are meant merely for visualization and clarity.

Suppose the following is the distribution of the income salary in some country:

Denote $X$ as a sample from that population. According to the Central limit theorem, we have $\bar{X}\sim N(\mu, \frac{\sigma}{\sqrt{n}})$, where $n$ is the size of the sample $X$, and $\sigma$ is the standard deviation of the population (the "population" here are salaries, of-course). So roughly, $\bar{X}$ might looks like this:

To my best understanding, the bigger the $n$, the "better", in the sense that the sample will more likely represent the original distribution. This can be seen in the $\frac{\sigma}{\sqrt{n}}$ going to $0$ as $n$ grows, thus the Gaussian bell will be further and further "squashed". (Sorry for the informality, I'm working on sharpening my intuition for now)

My question is this: Why does all of this mean that the bigger $n$ is, the more easier it will be to reject the null hypothesis?

Consider a right-side z-test: both the critical value and my sampled $\bar{X}$ will shift left together as $n$ grows:
The critical value is: $\bar{X}_c=\mu_0+Z_{(1-\alpha)}\cdot\frac{\sigma}{\sqrt{n}}$, where $\mu_0$ taken from the null hypothesis, $Z_{(1-\alpha)}$ is the $Z$ value for significance level of $\alpha$, and indeed, again we see that bigger $n$ means that the critical value will get closer to $\mu_0$, but as $n$ gets bigger, our sample $X$ is bigger and thus $\bar{X}$ is more "accurate" - meaning closer to $\mu_0$!

Answer 1

Why does all of this mean that the bigger n is, the more easier it will be to reject the null hypothesis?

That's true only when the null hypothesis is false (where $\mu_0$ is not the true mean).

When the null hypothesis is true, you are describing the probability of rejecting a null. The test controls this to always be $\alpha$ irrespective of the value of $n$ (once the central limit theorem has kicked in and assuming all the test's assumptions are valid).

You can confirm this by checking that

$$Pr(\bar{X}>\mu_0+Z_{(1-\alpha)}\cdot\frac{\sigma}{\sqrt{n}})=\alpha$$

where

$$\bar{X}\sim N(\mu,\frac{\sigma}{\sqrt{n}})$$

Answer 2

$\mu_0$ is the hypothesized mean, and $\bar{X}$ only converges to the true mean $\mu$. If $\mu \neq \mu_0$, then eventually $\bar{X}$ will be bounded away from $\mu_0$ since $\bar{X} \to \mu$, and since its variance is vanishing as $n \to \infty$, values that are even a very small distance from the observed value of $\bar{X}$ will eventually become "statistically" far, as measured in standard deviations, which means these values are likely not the mean of the distribution from which the $X_i$'s came.