why doubling the number in a contingency table changes the p-value? I am doing a statistics problem, which is testing if the evaluation of a person is independent of the person's sex.
I am given a contingency table, I calculated the expected value for each entry and calculated the chi-square value then I got a p-value.
Then the question asked me to do the same thing after doubling all entries in the contingency table, I got a p-value smaller than the one I got before. Why does this happen?
Can anyone give me an explanation for the difference? Thx!
 A: You can verify simply by writing out the equations that if you scale all entries by $c>0$ then the $\chi^2$ is also scaled by $c>0$. Therefore the $\chi^2$ is proportional to sample size for any given strength of relationship.
A larger sample size will allow you to detect a smaller relationship at a set significance level. Conversely, a small sample size will only detect larger effects at the same level. 
So, note that statistical significance under a null of 0 effect does not mean a "significant" effect in everyday use of the word "significant" as "important", but merely a statistically noticeable one. 
See also power analysis.
A: I'll give an intuitive explanation: the test is try to help us to confirm a "guess" of a conclusion over the population, based on the samples (some subset of the population) we have. 
The lower the p-value, the more confidence we have to reject the null-hypothesis and confirm the "guess" we made.
Now we double our sample data, and those extra data follow the same pattern/relationship, since we just copied the entries.
As a result, we observed the same pattern, but now we observed this pattern in more samples - we are more confident that our "guess" is true, since the "guess" lives true for wider samples. And to the extreme extent, if we could exhaust all the population instead of just relying on samples, we do not even need to "guess", because we know it for sure (of course we cannot do that).
A: $$
\begin{array}{ccc}
\begin{array}{cc}
1 & 2 \\ 3 & 4 \\ 5 & 5
\end{array} & \text{Now multiply all entries by 10 } \longrightarrow &
\begin{array}{cc}
10 & 20 \\ 30 & 40 \\ 50 & 50
\end{array}
\end{array}
$$
In the first table, the chi-square test statistic for the null hypothesis of independence of rows and columns is
\begin{align}
& \sum \frac{(\text{observed} - \text{expected})^2}{\text{expected}} \\[10pt]
= {} & \frac{(1-1.35)^2}{1.35} + \frac{(2-1.65)^2}{1.65} + \frac{(3-3.15)^2}{3.15} + \frac{(4 - 3.85)^2}{3.85} + \frac{(5-4.5)^2}{4.5} + \frac{(5-5.5)^2}{5.5}
\end{align}
In the second table, it is
$$
\frac{(10-13.5)^2}{13.5} + \frac{(20-16.5)^2}{16.5} + \frac{(30-31.5)^2}{31.5} + \frac{(40 - 38.5)^2}{38.5} + \frac{(50-45)^2}{45} + \frac{(50-55)^2}{55}
$$
Observe two things:


*

*We multiplied each numerator by $10^2$ and each denominator by $10$, thereby multiplying the entire expression by $10^2/10 = 10.$ Thus the value of the chi-square test statistic is $10$ times as big.

*If the pattern of deviation from independence of rows and columns persists as sampling continues until we have $10$ times as many observations as we had before, then we have $10$ times as much evidence of the non-independent distribution, so it makes sense that the evidence against independence is stronger.

