## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests (i.e. \$chi^{2}\$ tests)? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests (i.e. \$chi^{2}\$ tests)? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests (i.e. \$chi^{2}\$ tests), either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests (i.e. \$chi^{2}\$ tests), either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!

## #StackBounty: #hypothesis-testing #chi-squared #contingency-tables #pooling Pooled variance for post hoc tests for contingency table \$…

### Bounty: 100

The 2-by-2 contingency table test provides an alternate hypothesis testing approach to the \$z\$ test for proportion difference. Conveniently enough, a 2-by-k contingency table test provides an omnibus test for proportion difference, akin to the one-way ANOVA’s test for mean difference:

One-way ANOVA \$H_{0}: mu_{1} = mu_{2} = cdots = mu_{k}\$

2-by-\$k\$ contingency table test \$H_{0}: p_{1} = p_{2} = cdots = p_{k}\$

If we reject the null hypothesis in the 2-by-\$k\$ contingency table test, we could proceed to conduct post hoc pairwise comparisons using the z test for proportion difference between groups \$i\$ and \$j\$, where the test statistic is given by:

\$\$z = frac{hat{p}{i}-hat{p}{j}}{sqrt{hat{p}left(1-hat{p}right)left[frac{1}{n_{i}}+frac{1}{n_{j}}right]}}\$\$

And where (I think) \$hat{p}\$ creates the pooled estimate assuming the contingency table’s null hypothesis is true (i.e. \$hat{p}\$ is the total number of events divided by the total sample size across all \$k\$ groups).

Question 1: How to incorporate this pooled estimate into post hoc pairwise 2-by-2 contingency table tests? (bonus points if you can speak to continuity corrections)

Contingency table tests are also useful for posing questions about evidence of association between two categorical variables where both have more than 2 categories. For an \$l\$-by-\$k\$ contingency table test, where \$l>2\$ and \$k>2\$, if we reject the null hypothesis and wish to proceed to conduct post hoc subgroup tests:

Question 2a: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc 2-by-2 table tests? (bonus for continuity corrections)

Question 2b: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE disjoint (i.e. they do not overlap on the \$l\$-by-\$k\$ contingency table).

Question 2c: How do we incorporate the pooled variance under the \$l\$-by-\$k\$ contingency table test’s null hypothesis for post hoc \$m\$-by-\$n\$ contingency table tests, either \$2<mle l\$ OR \$2<nle k\$, or \$2<mle l\$ AND \$2<nle k\$, and these tests ARE NOT disjoint (i.e. they do overlap on the \$l\$-by-\$k\$ contingency table).

Get this bounty!!!