Statistical reporting of chi-square and odds ratio I'm looking to see if the process and reporting are correct. (This isn't homework help. I am playing around with a secondary data set trying to learn more about the research process and statistical reporting.)
Research Question: Is there an association between being sexually active in the past year and getting an HIV test?
So I set-up a two by two table, HIV testing on the column and sexual activity on the row.


*

*Not Active, No: 72

*Not Active, Yes: 14

*Active, No: 489

*Active, Yes: 238


Chi-sq = 9.7392, df = 1, p = 0.001804
Because the Chi-square test reveals a significant association, I'd like to report an odds ratio.


*

*The odds of getting HIV tested when not sexually active = 14/72 = 0.194

*The odds of getting HIV tested when sexually active = 238/489 = 0.487

*The odds ratio = 0.487/0.194 = 2.51


Those who are sexually active have 2.5 the odds of getting HIV tested compared to those who are not sexually active.
Is this reporting of the odds ratio correct? If the intention is to apply this finding in practice, would it be better to report a relative risk?
 A: Yes, based on your description of the setup, this appears to be correct. That is, given your Chi Squared value with 1 Degree of Freedom, we can reject the null hypothesis of independence at p < .05 (or p < .01). 
Your caclulated probabilities of getting tested given sexually active status also appear to be correct.
The question of whether to report the probabilities of getting tested as a relative probability depend on many things and the answer is subjective, but as a general suggestion, I would advise that you present both the raw and relative probabilities if I was going to present the relative probability for maximum utility and transparency.
Additionally, however, you should report the upper and lower 95% (or suitable value) confidence interval for the odds ratio:
Given your input frequency table:
$$\begin{matrix} 
    & Sexually Active & Not Sexually Active \\
    Tested &a & b \\
    Not Tested & c & d
\end{matrix}$$
And your odds ratio:
$$OR$$
Upper 95 % CI= $$e^{ln(OR)+1.96\sqrt{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})}}$$
Lower 95% CI= $$e^{ln(OR)-1.96\sqrt{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})}}$$
Which evaluates to:
Upper 95 % CI $\approx$  $4.542$
Lower 95% CI $\approx$ $1.387$
Notably, the interval does not include $1$, so it passes the test for significance.
The method for the above is more fully explored here among other places.
