How to interpret the p-value for chi-square-test I am fairly new to statistics and programming. I hope my question is neither stupid nor trivial.
I have 8 categories, and every category should have the same count of 156.125. My empirical observation is: [200, 156, 133, 178, 148, 130, 150, 154]. I calculated $\chi^2$ = 23.88390712570056.
Now, I want to calculate the critical value and here I am starting to struggle.

$H_0$ would be my hypothesis that the distribution is uniform. Since $p=0.95$ is around 14, I assume I have to reject $H_0$. But the higher my $p$-values go, the higher the critical value goes. This means that around $p=0.9999$ my calculated $\chi^2$ would be below the critical value and thus, I won't have to reject $H_0$ with a much higher $p$-value. This confuses me because I would have assumed, to have $H_0$ not rejected, the $p$-value has to go down.
What am I doing wrong? Thank you so much in advance.
 A: The $p$ you mentioned in the question is actually 1- significance level ($\alpha$), it is not the $p$-value of the test.
To use the Chi-square test to test the hypothesis:
Null: the data is following a uniform distribution,
Alternative: the data does not follow a uniform distribution.
Under a 5% level of significance,
You should, generally follow this procedure:
#1. find the chi-square test statistic
which you found as  23.88390712570056
#2. Find the critical value (CV),
Where CV is the Chi-square value with (K-1) degrees of freedom and ($1-\alpha$) cumulative probability.
In other words, CV follows a $\chi^2$ distribution with df = $k-1$
and $P(X < CV) = 1-\alpha$
In your case:
you used the graph to plot the chi-square values with df = 7 on the y-axis and the probability $P(X < \text{chi-square values})$ on the x-axis.
and you found $1-\alpha = 1-0.05 = 0.95$.
then from the graph $P(X < 14.07) = 0.95$
so, CV = 14.07 approximately.
#3. Compare the CV with $\chi^2$ test statistic.
If $\chi^2$ test statistic >= CV --> Reject $H_0$

In your case
$\chi^2$ test statistic (23.89) $\geq$ CV (14), so we reject the null hypothesis.
Therefore now, you can conclude with strong evidence under a 5% level of significance that the data are not uniformly distributed.
You mentioned in the question that as higher the $p$ goes, the higher the critical value goes.
$p$ here is $1-\alpha$,
the significance level ($\alpha$) = $P(\text{reject } H_0| H_0 \text{ is True})$ and ($1-\alpha$) is called the confidence level.
α gives you a "space" to make an error.
so, when $\alpha = 0.05$, and confidence level = 1-0.05 = 0.95, you give yourself a 5% chance of doing the error of rejecting a true H0.
$\alpha = 0.05$, you give yourself 5% chance of doing the error of rejecting a true H0" />
The graph shows the $\chi^2$ distribution with df = 7.
The highlighted area is called the rejection region, which is the region of the probability $P(X \geq x)$, any test statistic value in that region will result in rejecting H0.
Now, let's reduce your α to 0.01, for example.

Now, you give yourself less space to make an error, but at the same time, you are more confident in your result because the confidence level increases from 0.95 to (1-0.01 = 0.99)!
But this comes with the cost of increasing the CV, so the rejection region becomes smaller, so now, you are more confident when you reject H0, but it is harder for you to reject H0.
Now, only extreme and rare values of the chi-square test statistic can fall in the rejection region area.
You can take it as a general rule, as you decrease $\alpha$, CV increases, rejection region becomes smaller, so fewer values can fall in it.
You were doing nothing wrong, you just confused $\alpha$ with $1-\alpha$ and the $p$-value (which is something different).
I hope it is clearer now 
A: Comment: What do you mean by "[E]very category should have the same count of 156.125"? Does that mean you're doing a chi-squared test of homogeneity?
In R:
x = c(200, 156, 133, 178, 148, 130, 150, 154)

 chisq.test(x)

        Chi-squared test 
        for given probabilities
data:  x
X-squared = 23.884, df = 7, p-value = 0.001194

The small P-value indicates indicates categories are not
all equally likely.
Note: When specific probabilities are not supplied, the R procedure chicsq.test assumed equal probabilities are intended.
There are $8$ categories, so $7 = 8-1= DF.$
Critical value (5%) for chi-squared statistic is $c = 14.07,$
meaning that $H_0$ is rejected at 5% level if
chi-square statistic exceeds $c.$
qchisq(.95, 7)
[1] 14.06714

In the figure below the vertical dotted line is at
$c = 14.067,$ so that 5% of the area under the density curve lies to the right of $c.$

