Confused about how to interpret Chi square results For my local charity I have to organize random draws, each week to give a prize. I just bought a random machine generator, which randomly spits one ball out of 10.
Before using the machine  in public I made some tests in private to ensure that the machine is properly random and I am not accused of cheating.
So I made about 1300 random draws and performed a chisquare test.
I got the following chi square value: 18.13
Since there are 10 possible outcomes, there are 9 degrees of freedom.
Here are the values for 9 df:
ν     0.100  0.050   0.025   0.010   0.005   0.001
9   14.6837 16.9190 19.0228 21.6660 23.5893 27.8772

18.13 is greater that 14.6 and 16.9 does it mean that I am 95% confident that the machine is not random and I should change it? Or is 1300 experiments not enough to make a conclusion?
 A: If the machine is random, you can expect $\frac{1300}{10}=130$ draws out of 1300 for any given ball.  (This is well above the recommended minimum expected value of 5 per category, so you should be fine.)
With the observed frequencies for each ball, calculate the squared Pearson residual:
$$R_i^2 = \frac{(O_i-E_i)^2}{E_i}$$
(again, for this problem, $E_i = 130$ for all the categories).
The sum of these (squared) residuals is the chi-squared test statistic $\chi^2=\sum R_i^2$, with 9 degrees of freedom (one less than the number of categories).  Last, you need to calculate the $P$-value.  Using R

pchisq(18.13, 9, lower.tail = TRUE)

(where I used the chi-square test statistic you reported in the question).  I get p=.03.  So, we can assess the null hypothesis.
If the machine is random, the observed frequencies should be very close to the expected frequencies. Thus, the null could be
$$H_0 : \text{the machine is random}$$
or
$$H_0 : p_1 = p_2 = ··· = p_{10} = \frac{1}{10}$$
If you use a conventional significance level of $\alpha=.05$, then you would reject the null (machine is not random).  If you use $\alpha=.02$, then you would fail to reject the null (no reason to think the machine is not random).
The problem here is that your sample size can actually be too large.
