<s>[INST] 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!
 [/INST] $$
\begin{array}{ccc}
\begin{array}{cc}
1 & 2 \\ 3 & 4 \\ 5 & 5
\end{array} & \text{Now multiply all entries by 10 } \longrightarrow &
\begin{array}{cc}
10 & 20 \\ 30 & 40 \\ 50 & 50
\end{array}
\end{array}
$$
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.

</s>