How do I calculate the probability (%) for chi square test using $X^2$ value and DoF as inputs?
Im trying to create a C++ program to calculate chi square tests with very high DoF, so I cannot use the table to check the probability. I have already written functions to calculate DoF and $X^2$ values.
I'm not a mathematician so an example with the values inserted would be greatly helpful, for example with DoF of $14$ and $X^2$ of $17$.
For very large $n$ the chi-squared distribution is nearly normal with mean equal to df and variance twice the df.
R statistical software available free from r-project.org has density function 'dchisq', cumulative distribution function 'pchisq', quantile function (inverse CDF) 'qchisq', and also 'rchisq' which samples from the distribution.
From what you say I believe you want the area under the density curve of chi-squared with 14 DF to the right of 17 (the P-value for a right-tailed test with test statistic 17). In R that would be
> 1 - pchisq(17, 14)
[1] 0.2561779
Here are other examples, some of which are in your table, and the last of which shows the good agreement with normal for large DF:
> qchisq(.95, 10)
[1] 18.30704 # 95% area below 18.3
> dchisq(2, 15)
[1] 9.829755e-05 # PDF almost 0 at 2
> pchisq(190, 200); pnorm(190, 200, 20) # 3rd arg is SD
[1] 0.3173568 # exact
[1] 0.3085375 # normal appprox
Unless you have considerable experience programming probability functions I do not recommend writing your own C++ code. Surely, there is compiled code you can import that is written by experts, using best procedures for various DFs.
Addendum: If $Q \sim Chisq(df = \nu),$ with large $\nu,$ then $q$ with $P\{Q > q\} = \alpha$ has $$q \approx \nu\left(1 - \frac{2}{9\nu} + z_{\alpha} \sqrt{\frac{2}{9\nu}}\right)^3,$$ where $z_{\alpha}$ cuts probability $\alpha$ from the upper tail of a standard normal curve. For example:
> qchisq(.95, 150)
[1] 179.5806 # cuts 5% from right tail of CHISQ(150)
> 150*(1 - 2/(9*150) + qnorm(.95)*sqrt(2/(9*150)))^3
[1] 179.5788 # good approximation of above
