Em que nível um teste

JUSTIFICATIVA: Pule com segurança - está aqui para referência e para legitimar a pergunta.

A abertura deste artigo diz:

"O famoso teste de contingência qui-quadrado de Karl Pearson é derivado de outra estatística, chamada estatística z, com base na distribuição Normal. As versões mais simples do $\chi^2$ podem ser matematicamente idênticas aos testes z equivalentes. Os testes produzem o mesmo resultado em todas as circunstâncias. Para todos os efeitos, "qui-quadrado" pode ser chamado de "z-quadrado". Os valores críticos de $\chi^2$ para um grau de liberdade são o quadrado dos valores críticos correspondentes de z ".

Isso foi afirmado várias vezes no currículo ( aqui , aqui , aqui e outros).

E, de fato, podemos provar que $\chi^2_{1\,df}$ é equivalente a$X^2$com$X\sim N(0,1)$:

Digamos que $X \sim N(0,1)$ e que $Y=X^2$ e encontre a densidade de $Y$ usando o método $cdf$ :

$p(Y \leq y) = p(X^2 \leq y)= p(-\sqrt{y} \leq x \leq \sqrt{y})$. O problema é que não podemos integrar de forma estreita a densidade da distribuição normal. Mas podemos expressá-lo:

Como os valores do normal são simétricos:$pdf$

. Igualando isso aopdfdo normal (agora oxnopdfserá$f_X(y)= F_X'(\sqrt{y})\,\frac{1}{\sqrt{y}}$$pdf$$x$$pdf$ a ser conectado aoe - x $\sqrt{y}$ partes do normalpdf); e lembrando-se de incluir$e^{-\frac{x^2}{2}}$$pdf$ no final:$\frac{1}{\sqrt{y}}$

Compare com o pdf do quadrado chi:

Desde , para1df, derivamos exatamente opdfdo quadrado de chi.$\Gamma(1/2)=\sqrt{\pi}$$1$$pdf$

Além disso, se chamamos a função prop.test()em R, estamos invocando o mesmo teste do como se decidíssemos .$\chi^2$chisq.test()

A QUESTÃO:

Então, eu entendo todos esses pontos, mas ainda não sei como eles se aplicam à implementação real desses dois testes por dois motivos:

1. Um teste z não é quadrado.

2. As estatísticas de teste reais são completamente diferentes:

O valor da estatística de teste para um $\chi^2$ é:

onde$\chi^2 = \sum_{i=1}^{n} \frac{(O_i - E_i)^2}{E_i} = N \sum_{i=1}^n p_i \left(\frac{O_i/N - p_i}{p_i}\right)^2$

$\chi^2$ = Pearson's cumulative test statistic, which asymptotically approaches a $\chi^2$ distribution. $O_i$ = the number of observations of type $i$; $N$ = total number of observations; $E_i$ = $N p_i$ = the expected (theoretical) frequency of type $i$, asserted by the null hypothesis that the fraction of type $i$ in the population is $p_i$; $n$ = the number of cells in the table.

On the other hand, the test statistic for a $z$-test is:

$\displaystyle Z = \frac{\frac{x_1}{n_1}-\frac{x_2}{n_2}}{\sqrt{p\,(1-p)(1/n_1+1/n_2)}}$ with $\displaystyle p = \frac{x_1\,+\,x_2}{n_1\,+\,n_2}$, where $x_1$ and $x_2$ are the number of "successes", over the number of subjects in each one of the levels of the categorical variables, i.e. $n_1$ and $n_2$.

This formula seems to rely on the binomial distribution.

These two tests statistics are clearly different, and result in different results for the actual test statistics, as well as for the p-values: 5.8481 for the $\chi^2$ and 2.4183 for the z-test, where $\small 2.4183^2=5.84817$ (thank you, @mark999). The p-value for the $\chi^2$ test is 0.01559, while for the z-test is 0.0077. The difference explained by two-tailed versus one-tailed: $\small 0.01559/2=0.007795$ (thank you @amoeba).

So at what level do we say that they are one and the same?

Let us have a 2x2 frequency table where columns are two groups of respondents and rows are the two responses "Yes" and "No". And we've turned the frequencies into the proportions within group, i.e. into the vertical profiles:

      Gr1   Gr2  Total
Yes   p1    p2     p
No    q1    q2     q
      --------------
     100%  100%   100%
      n1    n2     N

The usual (not Yates corrected) $\chi^2$ of this table, after you substitute proportions instead of frequencies in its formula, looks like this:

Remember that $p= \frac{n_1p_1+n_2p_2}{n_1+n_2}$, the element of the weighted average profile of the two profiles (p1,q1) and (p2,q2), and plug it in the formula, to obtain

Divide both numerator and denominator by the $(n_1^2n_2+n_1n_2^2)$ and get

the squared z-statistic of the z-test of proportions for "Yes" response.

Thus, the 2x2 homogeneity Chi-square statistic (and test) is equivalent to the z-test of two proportions. The so called expected frequencies computed in the chi-square test in a given column is the weighted (by the group n) average vertical profile (i.e. the profile of the "average group") multiplied by that group's n. Thus, it comes out that chi-square tests the deviation of each of the two groups profiles from this average group profile, - which is equivalent to testing the groups' profiles difference from each other, which is the z-test of proportions.

This is one demonstration of a link between a variables association measure (chi-square) and a group difference measure (z-test statistic). Attribute associations and group differences are (often) the two facets of the same thing.

(Showing the expansion in the first line above, By @Antoni's request):

$n_1[\frac{(p_1-p)^2}{p}+\frac{(q_1-q)^2}{q}]+n_2[\frac{(p_2-p)^2}{p}+\frac{(q_2-q)^2}{q}] = \frac{n_1(p_1-p)^2q}{pq}+\frac{n_1(q_1-q)^2p}{pq}+\frac{n_2(p_2-p)^2q}{pq}+\frac{n_2(q_2-q)^2p}{pq} = \frac{n_1(p_1-p)^2(1-p)+n_1(1-p_1-1+p)^2p+n_2(p_2-p)^2(1-p)+n_2(1-p_2-1+p)^2p}{pq} = \frac{n_1(p_1-p)^2(1-p)+n_1(p-p_1)^2p+n_2(p_2-p)^2(1-p)+n_2(p-p_2)^2p}{pq} = \frac{[n_1(p_1-p)^2][(1-p)+p]+[n_2(p_2-p)^2][(1-p)+p]}{pq} = \frac{n_1(p_1-p)^2+n_2(p_2-p)^2}{pq}.$