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Alexis
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  1. $\text{H}_{0}\text{: }P(\text{Positive}|\text{Group = A}) = P(\text{Positive}|\text{Group = B})$, $\text{H}_{\text{A}}\text{: }P(\text{Positive}|\text{Group = A}) \ne P(\text{Positive}|\text{Group = B})$

    $\text{H}_{0}\text{: }P(\text{Positive}|\text{Group = A}) = P(\text{Positive}|\text{Group = B})$, $\text{H}_{\text{A}}\text{: }P(\text{Positive}|\text{Group = A}) \ne P(\text{Positive}|\text{Group = B})$

    This null hypothesis implies that the best estimate of $P(\text{Positive})$ combines counts from both groups, or (No. Positive in A + No. Positive in B) ÷ Total. When multiplied by the total counts in each group, these create the expected counts, $E_{ij}$ for the top row in a table of expected counts. For example, the expected count of positives for Group A = Total Count in A × (No. Positive in A + No. Positive in B) ÷ Total and (No. Negative in A + No. Negative in B) ÷ Total gives $P(\text{Negative})$ under the null, which does similar for the bottom row in the table of expected counts:

This null hypothesis implies that the best estimate of $P(\text{Positive})$ combines counts from both groups, or (No. Positive in A + No. Positive in B) ÷ Total. When multiplied by the total counts in each group, these create the expected counts, $E_{ij}$ for the top row in a table of expected counts. For example, the expected count of positives for Group A = Total Count in A × (No. Positive in A + No. Positive in B) ÷ Total and (No. Negative in A + No. Negative in B) ÷ Total gives $P(\text{Negative})$ under the null, which does similar for the bottom row in the table of expected counts:

  1. $\text{H}_{0}\text{: }P(\text{Positive}|\text{Group = A}) = P(\text{Positive}|\text{Group = B})$, $\text{H}_{\text{A}}\text{: }P(\text{Positive}|\text{Group = A}) \ne P(\text{Positive}|\text{Group = B})$

This null hypothesis implies that the best estimate of $P(\text{Positive})$ combines counts from both groups, or (No. Positive in A + No. Positive in B) ÷ Total. When multiplied by the total counts in each group, these create the expected counts, $E_{ij}$ for the top row in a table of expected counts. For example, the expected count of positives for Group A = Total Count in A × (No. Positive in A + No. Positive in B) ÷ Total and (No. Negative in A + No. Negative in B) ÷ Total gives $P(\text{Negative})$ under the null, which does similar for the bottom row in the table of expected counts:

  1. $\text{H}_{0}\text{: }P(\text{Positive}|\text{Group = A}) = P(\text{Positive}|\text{Group = B})$, $\text{H}_{\text{A}}\text{: }P(\text{Positive}|\text{Group = A}) \ne P(\text{Positive}|\text{Group = B})$

    This null hypothesis implies that the best estimate of $P(\text{Positive})$ combines counts from both groups, or (No. Positive in A + No. Positive in B) ÷ Total. When multiplied by the total counts in each group, these create the expected counts, $E_{ij}$ for the top row in a table of expected counts. For example, the expected count of positives for Group A = Total Count in A × (No. Positive in A + No. Positive in B) ÷ Total and (No. Negative in A + No. Negative in B) ÷ Total gives $P(\text{Negative})$ under the null, which does similar for the bottom row in the table of expected counts:

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Alexis
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  1. The test statistic $G = 2\sum{ij}O_{ij}\ln\left(\frac{O_{ij}}{E_{ij}}\right)$$G = 2\sum_{ij}O_{ij}\ln\left(\frac{O_{ij}}{E_{ij}}\right)$, where $i,j$ index the cells across rows and columns.

  2. The probability of rejecting $\text{H}_{0}$, assuming it is true, is $p = P(G_{\text{df}=1}\ge G)$, where $\text{df} = \text{No. rows}-1 \times \text{No. columns} - 1$, and the $p$ value is based on the $\chi^{2}$ distribution with $\text{df}$ degrees of freedom.

  3. Reject $H_{0}$ if $p\le \alpha$.

  4. If you rejected $\text{H}_0$, conclude that you found evidence of a difference in probability of positives by group/found that probability of positive is associated with group. If you did not reject $\text{H}_0$, conclude that you failed to find such evidence.

  1. The test statistic $G = 2\sum{ij}O_{ij}\ln\left(\frac{O_{ij}}{E_{ij}}\right)$.

  2. The probability of rejecting $\text{H}_{0}$, assuming it is true, is $p = P(G_{\text{df}=1}\ge G)$, where $\text{df} = \text{No. rows}-1 \times \text{No. columns} - 1$, and the $p$ value is based on the $\chi^{2}$ distribution.

  3. Reject $H_{0}$ if $p\le \alpha$.

  4. If you rejected $\text{H}_0$, conclude that you found evidence of a difference in probability of positives by group/found that probability of positive is associated with group. If you did not reject $\text{H}_0$, conclude that you failed to find such evidence.

  1. The test statistic $G = 2\sum_{ij}O_{ij}\ln\left(\frac{O_{ij}}{E_{ij}}\right)$, where $i,j$ index the cells across rows and columns.

  2. The probability of rejecting $\text{H}_{0}$, assuming it is true, is $p = P(G_{\text{df}=1}\ge G)$, where $\text{df} = \text{No. rows}-1 \times \text{No. columns} - 1$, and the $p$ value is based on the $\chi^{2}$ distribution with $\text{df}$ degrees of freedom.

  3. Reject $H_{0}$ if $p\le \alpha$.

  4. If you rejected $\text{H}_0$, conclude that you found evidence of a difference in probability of positives by group/found that probability of positive is associated with group. If you did not reject $\text{H}_0$, conclude that you failed to find such evidence.

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Alexis
Alexis
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Contingency table tests (e.g., the $G$ test, the $\chi^{2}$ test) can often test for evidence of difference/evidence of association (and also equivalence, though those are more sophisticated) in the probability distributions of categories of rows by categories of columns in tables of counts (i.e. contingency tables).

A simple example is the 2×2 table (the 2×2 being the observed counts, $O_{ij}$):

Group A Group B Marginal Total
Count of Positive Count of Positive No. Positive in A + No. Positive in B
Count of Negative Count of Negative No. Negative in A + No. Negative in B
Total Count in A Total Count in B Total

$\boldsymbol{G}$ Test:

  1. $\text{H}_{0}\text{: }P(\text{Positive}|\text{Group = A}) = P(\text{Positive}|\text{Group = B})$, $\text{H}_{\text{A}}\text{: }P(\text{Positive}|\text{Group = A}) \ne P(\text{Positive}|\text{Group = B})$

This null hypothesis implies that the best estimate of $P(\text{Positive})$ combines counts from both groups, or (No. Positive in A + No. Positive in B) ÷ Total. When multiplied by the total counts in each group, these create the expected counts, $E_{ij}$ for the top row in a table of expected counts. For example, the expected count of positives for Group A = Total Count in A × (No. Positive in A + No. Positive in B) ÷ Total and (No. Negative in A + No. Negative in B) ÷ Total gives $P(\text{Negative})$ under the null, which does similar for the bottom row in the table of expected counts:

Group A Group B Marginal Total
Expected Count of Positive Expected Count of Positive No. Positive in A + No. Positive in B
Expected Count of Negative Expected Count of Negative No. Negative in A + No. Negative in B
Total count in A Total count in B Total

  1. The test statistic $G = 2\sum{ij}O_{ij}\ln\left(\frac{O_{ij}}{E_{ij}}\right)$.

  2. The probability of rejecting $\text{H}_{0}$, assuming it is true, is $p = P(G_{\text{df}=1}\ge G)$, where $\text{df} = \text{No. rows}-1 \times \text{No. columns} - 1$, and the $p$ value is based on the $\chi^{2}$ distribution.

  3. Reject $H_{0}$ if $p\le \alpha$.

  4. If you rejected $\text{H}_0$, conclude that you found evidence of a difference in probability of positives by group/found that probability of positive is associated with group. If you did not reject $\text{H}_0$, conclude that you failed to find such evidence.

This test extends to contingency tables of arbitrary number of rows and columns, although the null becomes "No association between row variable and column variable," with the alternative "Association between row variable and column variable."

$\boldsymbol{\chi^2}$ test

As above, but the test statistic is $\chi^{2} = \sum_{ij} \frac{\left(O_{ij}-E_{ij}\right)^{2}}{E_{ij}}$, and the probability of rejecting is $p = P(X^{2}_{\text{df}} \ge \chi^{2})$.

There are a few nuances (e.g., continuity corrections for 2×2 tables, corresponding equivalence tests and their hypotheses), but that is the meat of it. You will want to make your rows correspond to a single main category or sub-category (rather than use categories and subcategories), and will need to organize your columns by group (A, B, C, etc.).