# "why doubling the number in a contingency table changes the p-value? i am doing a facts trouble, which is checking out if the evaluation of someone is impartial of the individual's intercourse. i'm given a contingency desk, I calculated the expected fee for each access and calculated the chi-rectangular value then I got a p-fee. Then the query requested me to do the same thing after doubling all entries in the contingency desk, I were given a p-price smaller than the only I got earlier than. Why does this take place? Can all and sundry supply me an cause of the distinction? "

why doubling the number in a contingency table changes the p-value?
i am doing a facts trouble, which is checking out if the evaluation of someone is impartial of the individual's intercourse. i'm given a contingency desk, I calculated the expected fee for each access and calculated the chi-rectangular value then I got a p-fee.
Then the query requested me to do the same thing after doubling all entries in the contingency desk, I were given a p-price smaller than the only I got earlier than. Why does this take place? Can all and sundry supply me an cause of the distinction?
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trivialaxxf
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.
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Kareem Mejia

In the first table, the chi-square test statistic for the null hypothesis of independence of rows and columns is
$\begin{array}{rl}& \sum \frac{\left(\text{observed}-\text{expected}{\right)}^{2}}{\text{expected}}\\ =& \frac{\left(1-1.35{\right)}^{2}}{1.35}+\frac{\left(2-1.65{\right)}^{2}}{1.65}+\frac{\left(3-3.15{\right)}^{2}}{3.15}+\frac{\left(4-3.85{\right)}^{2}}{3.85}+\frac{\left(5-4.5{\right)}^{2}}{4.5}+\frac{\left(5-5.5{\right)}^{2}}{5.5}\end{array}$
In the second table, it is
$\frac{\left(10-13.5{\right)}^{2}}{13.5}+\frac{\left(20-16.5{\right)}^{2}}{16.5}+\frac{\left(30-31.5{\right)}^{2}}{31.5}+\frac{\left(40-38.5{\right)}^{2}}{38.5}+\frac{\left(50-45{\right)}^{2}}{45}+\frac{\left(50-55{\right)}^{2}}{55}$
Observe two things:
We multiplied every numerator by means of ${10}^{2}$ and every denominator through 10, thereby multiplying the complete expression by means of ${10}^{2}/10=10.$. for this reason the fee of the chi-rectangular take a look at statistic is 10 times as massive.
If the pattern of deviation from independence of rows and columns persists as sampling keeps until we have 10 times as many observations as we had before, then we've 10 times as a great deal evidence of the non-impartial distribution, so it makes experience that the evidence against independence is more potent.