Inference Test for Two Proportions

Permutation
Monte Carlo Simulation
Random Number Generation
Distributions
Generating Parallel Worlds
Author

Arvind V.

Published

November 10, 2022

Modified

October 1, 2025

Abstract
Inference Test for Two Proportions

1 Setting up R packages

library(tidyverse)
library(mosaic)
library(vcd)
library(visStatistics) # One package to test them all

### Dataset from Chihara and Hesterberg's book (Second Edition)
library(resampledata)

Plot Fonts and Theme

Show the Code
library(systemfonts)
library(showtext)
library(ggrepel)
library(marquee)
## Clean the slate
systemfonts::clear_local_fonts()
systemfonts::clear_registry()
##
showtext_opts(dpi = 96) # set DPI for showtext
sysfonts::font_add(
  family = "Alegreya",
  regular = "../../../../../../fonts/Alegreya-Regular.ttf",
  bold = "../../../../../../fonts/Alegreya-Bold.ttf",
  italic = "../../../../../../fonts/Alegreya-Italic.ttf",
  bolditalic = "../../../../../../fonts/Alegreya-BoldItalic.ttf"
)

sysfonts::font_add(
  family = "Roboto Condensed",
  regular = "../../../../../../fonts/RobotoCondensed-Regular.ttf",
  bold = "../../../../../../fonts/RobotoCondensed-Bold.ttf",
  italic = "../../../../../../fonts/RobotoCondensed-Italic.ttf",
  bolditalic = "../../../../../../fonts/RobotoCondensed-BoldItalic.ttf"
)
showtext_auto(enable = TRUE) # enable showtext
##
theme_custom <- function() {
  theme_bw(base_size = 10) +

    theme_sub_axis(
      title = element_text(
        family = "Roboto Condensed",
        size = 8
      ),
      text = element_text(
        family = "Roboto Condensed",
        size = 6
      )
    ) +

    theme_sub_legend(
      text = element_text(
        family = "Roboto Condensed",
        size = 6
      ),
      title = element_text(
        family = "Alegreya",
        size = 8
      )
    ) +

    theme_sub_plot(
      title = element_text(
        family = "Alegreya",
        size = 14, face = "bold"
      ),
      title.position = "plot",
      subtitle = element_text(
        family = "Alegreya",
        size = 10
      ),
      caption = element_text(
        family = "Alegreya",
        size = 6
      ),
      caption.position = "plot"
    )
}

## Use available fonts in ggplot text geoms too!
ggplot2::update_geom_defaults(geom = "text", new = list(
  family = "Roboto Condensed",
  face = "plain",
  size = 3.5,
  color = "#2b2b2b"
))
ggplot2::update_geom_defaults(geom = "label", new = list(
  family = "Roboto Condensed",
  face = "plain",
  size = 3.5,
  color = "#2b2b2b"
))

ggplot2::update_geom_defaults(geom = "marquee", new = list(
  family = "Roboto Condensed",
  face = "plain",
  size = 3.5,
  color = "#2b2b2b"
))
ggplot2::update_geom_defaults(geom = "text_repel", new = list(
  family = "Roboto Condensed",
  face = "plain",
  size = 3.5,
  color = "#2b2b2b"
))
ggplot2::update_geom_defaults(geom = "label_repel", new = list(
  family = "Roboto Condensed",
  face = "plain",
  size = 3.5,
  color = "#2b2b2b"
))

## Set the theme
ggplot2::theme_set(new = theme_custom())

## tinytable options
options("tinytable_tt_digits" = 2)
options("tinytable_format_num_fmt" = "significant_cell")
options(tinytable_html_mathjax = TRUE)


## Set defaults for flextable
flextable::set_flextable_defaults(font.family = "Roboto Condensed")

2 Introduction

Many experiments gather qualitative data across different segments of a population, for example, opinion about a topic among people who belong to different income groups, or who live in different parts of a city. This should remind us of the Likert Plots that we plotted earlier. In this case the two variables, dependent and independent, are both Qualitative, and we can calculate counts and proportions.

How does one Qual variable affect the other? How do counts/proportions of the dependent variable vary with the levels of the independent variable? This is our task for this module.

Here is a quick example of the kind of data we might look at here, taken from the British Medical Journal:

Figure 1: Breast Feeding

Clearly, we can see differences in counts/proportions of women who breast-fed their babies for three months or more, based on whether they were “printers wives” or “farmers’ wives”!

Is there a doctor in the House?

2.1 The CLT for Two Proportions

We first need to establish some model assumptions prior to making our analysis. As before, we wish to see if the CLT applies here, and if so, in what form. The difference between two proportions \(\hat{p_1}-\hat{p_2}\) can be modeled using a normal distribution when:

  • Independence (extended): The data are independent within and between the two groups. Generally this is satisfied if the data come from two independent random samples or if the data come from a randomized experiment.
  • Success-failure condition: The success-failure condition holds for both groups, where we check successes and failures in each group separately. That is, we should have at least 10 successes and 10 failures in each of the two groups.

When these conditions are satisfied, the standard error of \(\hat{p_1}-\hat{p_2}\) is well-approximated by:

\[ SE(\hat{p_1}-\hat{p_2}) = \sqrt{\frac{\hat{p_1}*(1-\hat{p_1})}{n_1}} + \sqrt{\frac{\hat{p_2}*(1-\hat{p_2})}{n_2}} \tag{1}\]

where \(\hat{p_1}\) and \(\hat{p_2}\) represent the sample proportions, and \(n_1\) and \(n_2\) represent the sample sizes.

We can represent the Confidence Intervals as:

\[ \begin{eqnarray} CI(p_1 - p_2) &=& (\hat{p_1} - \hat{p_2}) \pm 1.96 * SE(\hat{p_1}-\hat{p_2})\\ &=& (\hat{p_1} - \hat{p_2}) \pm 1.96 * \left(\sqrt{\frac{\hat{p_1}*(1-\hat{p_1})}{n_1}} + \sqrt{\frac{\hat{p_2}*(1-\hat{p_2})}{n_2}}\right) \end{eqnarray} \tag{2}\]

3 Case Study-1: GSS2002 dataset

We saw how we could perform inference for a single proportion. We can extend this idea to multiple proportions too.

Let us try a dataset with Qualitative / Categorical data. This is the General Social Survey GSS dataset from the resampledata package, and we have people with different levels of Education stating their opinion on the Death Penalty. We want to know if these two Categorical variables have a correlation, i.e. can the opinions in favour of the Death Penalty be explained by the Education level?

Since data is Categorical ( both variables ), we need to take counts in a table, and then implement a chi-square test. In the test, we will permute the Education variable to see if we can see how significant its effect size is.

data(GSS2002, package = "resampledata")
glimpse(GSS2002)
Rows: 2,765
Columns: 21
$ ID            <int> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 1…
$ Region        <fct> South Central, South Central, South Central, South Centr…
$ Gender        <fct> Female, Male, Female, Female, Male, Male, Female, Female…
$ Race          <fct> White, White, White, White, White, White, White, White, …
$ Education     <fct> HS, Bachelors, HS, Left HS, Left HS, HS, Bachelors, HS, …
$ Marital       <fct> Divorced, Married, Separated, Divorced, Divorced, Divorc…
$ Religion      <fct> Inter-nondenominational, Protestant, Protestant, Protest…
$ Happy         <fct> Pretty happy, Pretty happy, NA, NA, NA, Pretty happy, NA…
$ Income        <fct> 30000-34999, 75000-89999, 35000-39999, 50000-59999, 4000…
$ PolParty      <fct> "Strong Rep", "Not Str Rep", "Strong Rep", "Ind, Near De…
$ Politics      <fct> Conservative, Conservative, NA, NA, NA, Conservative, NA…
$ Marijuana     <fct> NA, Not legal, NA, NA, NA, NA, NA, NA, Legal, NA, NA, NA…
$ DeathPenalty  <fct> Favor, Favor, NA, NA, NA, Favor, NA, NA, Favor, NA, NA, …
$ OwnGun        <fct> No, Yes, NA, NA, NA, Yes, NA, NA, Yes, NA, NA, NA, NA, N…
$ GunLaw        <fct> Favor, Oppose, NA, NA, NA, Oppose, NA, NA, Oppose, NA, N…
$ SpendMilitary <fct> Too little, About right, NA, About right, NA, Too little…
$ SpendEduc     <fct> Too little, Too little, NA, Too little, NA, Too little, …
$ SpendEnv      <fct> About right, About right, NA, Too little, NA, Too little…
$ SpendSci      <fct> About right, About right, NA, Too little, NA, Too little…
$ Pres00        <fct> Bush, Bush, Bush, NA, NA, Bush, Bush, Bush, Bush, NA, NA…
$ Postlife      <fct> Yes, Yes, NA, NA, NA, Yes, NA, NA, Yes, NA, NA, NA, NA, …


Note how all variables are Categorical !! Education has five levels, and of course DeathPenalty has three:

GSS2002 %>% count(Education)
GSS2002 %>% count(DeathPenalty)

4 Data Munging

Let us drop NA entries in Education and Death Penalty and also check the levels of the two factors and set them up as we need:

GSS2002_modified <- GSS2002 %>%
  dplyr::select(Education, DeathPenalty) %>%
  tidyr::drop_na(., c(Education, DeathPenalty)) %>%
  # Re-level the factors
  mutate(
    Education = factor(Education,
      levels = c("Left HS", "HS", "Jr Col", "Bachelors", "Graduate"),
      labels = c("Left HS", "HS", "Jr Col", "Bachelors", "Graduate"),
      # ordered = TRUE # Nope can't use this. See below
    ),
    DeathPenalty = factor(DeathPenalty,
      levels = c("Favor", "Oppose"),
      labels = c("Favor", "Oppose")
      # ordered = TRUE # Nope can't use this. See below
    )
  )
glimpse(GSS2002_modified)
Rows: 1,307
Columns: 2
$ Education    <fct> HS, Bachelors, HS, HS, HS, HS, HS, Jr Col, HS, Bachelors,…
$ DeathPenalty <fct> Favor, Favor, Favor, Favor, Favor, Favor, Favor, Favor, O…

We can now set up a Contingency Table.

gss_vcd_table <- vcd::structable(Education ~ DeathPenalty, # cols ~ rows
  data = GSS2002_modified
)
gss_vcd_table %>% addmargins()
            Education
DeathPenalty Left HS  HS Jr Col Bachelors Graduate  Sum
      Favor      117 511     71       135       64  898
      Oppose      72 200     16        71       50  409
      Sum        189 711     87       206      114 1307
Table 1: Contingency Table of Education vs Death Penalty

4.1 Contingency Table Plots

The Contingency Table can be plotted using a mosaic plot using two packages, as we have seen. Let us do a quick recap:

Show the Code
# Already computed
# gss_vcd_table <- vcd::structable(Education ~ DeathPenalty, # cols ~ rows
#                              data = GSS2002_modified)

gss_vcd_table %>%
  vcd::mosaic(
    gp = shading_hsv, direction = "h",
    main = "GSS2002 Education vs Death Penalty Mosaic Chart",
    legend = TRUE,
    labeling = labeling_border(
      varnames = c("F", "F"), # Remove variable name labels
      rot_labels = c(90, 0, 0, 0), # t,r,b,l?
      just_labels = c(
        "left", # Top Side. How?
        "left", # Right side
        "left", # Bottom side
        "right"
      )
    )
  ) # Left side. How?

NoteBusiness Insights

We see that:

  • the proportion of people in favour of the Death Penalty decreases with increasing levels of Education.
  • there are some imbalances in the counts of people with different levels of Education (vertical divisions are not straight), which we will need to account for in our analysis.

As discussed in the Descriptive Analysis on Proportions, visStatistics is a recent package that allows a very wide variety of statistical charts to be created auto-magically based on the variables chosen. Let us plot a mosaic chart directly with this package: with one function visstats(), we obtain mosaic and bar charts, as well as a statistical analysis of the data. We will discuss this analysis shortly.

chisq_visstat <- visstat(
  x = GSS2002_modified$DeathPenalty,
  y = GSS2002_modified$Education
)

chisq_visstat %>% summary()
Summary of visstat object

--- Named components ---
 [1] "statistic"    "parameter"    "p.value"      "method"       "data.name"   
 [6] "observed"     "expected"     "residuals"    "stdres"       "mosaic_stats"

--- Contents ---

$statistic:
X-squared 
 23.45093 

$parameter:
df 
 4 

$p.value:
[1] 0.0001028891

$method:
[1] "Pearson's Chi-squared test"

$data.name:
[1] "counts"

$observed:
        Education
groups    HS Bachelors Left HS Graduate Jr Col
  Favor  511       135     117       64     71
  Oppose 200        71      72       50     16

$expected:
        Education
groups         HS Bachelors   Left HS Graduate   Jr Col
  Favor  488.5065 141.53634 129.85616 78.32594 59.77506
  Oppose 222.4935  64.46366  59.14384 35.67406 27.22494

$residuals:
        Education
groups           HS  Bachelors    Left HS   Graduate     Jr Col
  Favor   1.0177047 -0.5494154 -1.1281841 -1.6187145  1.4518580
  Oppose -1.5079895  0.8140992  1.6716928  2.3985389 -2.1512983

$stdres:
        Education
groups          HS Bachelors   Left HS  Graduate    Jr Col
  Favor   2.694093 -1.070092 -2.180585 -3.028754  2.686322
  Oppose -2.694093  1.070092  2.180585  3.028754 -2.686322

$mosaic_stats:
             Education  HS Bachelors Left HS Graduate Jr Col
DeathPenalty                                                
Favor                  511       135     117       64     71
Oppose                 200        71      72       50     16

--- Attributes ---
$plot_paths
character(0)
(a) Dodged Bar Plot of Death Penalty by Education
(b) Mosaic Plot of Death Penalty by Education
Figure 2: visStatistics output for Chi-Square Test 
chisq_visstat$observed
        Education
groups    HS Bachelors Left HS Graduate Jr Col
  Favor  511       135     117       64     71
  Oppose 200        71      72       50     16
chisq_visstat$expected
        Education
groups         HS Bachelors   Left HS Graduate   Jr Col
  Favor  488.5065 141.53634 129.85616 78.32594 59.77506
  Oppose 222.4935  64.46366  59.14384 35.67406 27.22494
chisq_visstat$stdres
        Education
groups          HS Bachelors   Left HS  Graduate    Jr Col
  Favor   2.694093 -1.070092 -2.180585 -3.028754  2.686322
  Oppose -2.694093  1.070092  2.180585  3.028754 -2.686322
Table 2: visStatistics Tables output for Chi-Square Test

This is a very comprehensive output, with just one line of code. We obtain:

  • The original Contingency Table
  • the Expected Contingency Table ( if the two variables were independent )
  • the contribution each cell makes to the overall Chi-Square statistic i.e. each cell’s Pearson Residual
  • the test statistic, degrees of freedom and p-value
  • a mosaic plot
  • a dodged bar plot

We will use all of these results to obtain a clear understanding of the test.

NoteOrdering of levels between vcd and visstat

The ordering of the levels (Left HS, HS, Jr Col, Bachelors, Graduate) is different between the two packages. This is because vcd orders the levels in the order they appear in the data, while visStatistics orders them in alphabetical strange order. If we attempt to munge the data into ordinal factors ( ordered = TRUE), visStatistics does not accept it at all, and wants plain factors.

We will therefore continue to use the outputs from vcd.

5 Hypotheses Definition

What would our Hypotheses be relating to the proportions of votes for or against the DeathPenalty?

\(H_0: \text{Education does not affect votes for Death Penalty}\\\)

\(H_a: \text{Education affects votes for Death Penalty}\\\)

6 Inference for Two Proportions

We are now ready to perform our statistical inference. We will use the standard Pearson chi-square test, and develop and intuition for it. We will then do a permutation test to have an alternative method to complete the same task.

Let us now perform the base chisq test: We need a contingency table and then the xchisq test: We will calculate the observed-chi-squared value, and compare it with the critical value. ( The xchisq function is from the mosaic package, and is a wrapper around the base chisq.test function. It provides more verbose output, which is easier to work with. )

contingency_table <- vcd::structable(Education ~ DeathPenalty,
  data = GSS2002_modified
)
contingency_table
             Education Left HS  HS Jr Col Bachelors Graduate
DeathPenalty                                                
Favor                      117 511     71       135       64
Oppose                      72 200     16        71       50
xq_test_object <- xchisq.test(contingency_table)

    Pearson's Chi-squared test

data:  x
X-squared = 23.451, df = 4, p-value = 0.0001029

  117      511       71      135       64   
(129.86) (488.51) ( 59.78) (141.54) ( 78.33)
 [1.27]   [1.04]   [2.11]   [0.30]   [2.62] 
<-1.13>  < 1.02>  < 1.45>  <-0.55>  <-1.62> 
         
   72      200       16       71       50   
( 59.14) (222.49) ( 27.22) ( 64.46) ( 35.67)
 [2.79]   [2.27]   [4.63]   [0.66]   [5.75] 
< 1.67>  <-1.51>  <-2.15>  < 0.81>  < 2.40> 
         
key:
    observed
    (expected)
    [contribution to X-squared]
    <Pearson residual>
xq_test_object %>%
  broom::tidy() %>%
  select(statistic) %>%
  as.numeric() -> X_squared_observed
X_squared_observed
[1] 23.45093

We have the CT and the quite verbose test results, which we will examine shortly.

Let us also evaluate the critical value for the Chi-Square distribution, with alpha = 0.05 and df = (nrows-1)*(ncols-1) = (5-1)*(2-1) = 4:

# Determine the Chi-Square critical value
X_squared_critical <- qchisq(
  p = .05,
  df = (5 - 1) * (2 - 1), # (nrows-1) * (ncols-1)
  lower.tail = FALSE
)
X_squared_critical
[1] 9.487729

We see that our observed \(X^2_{obs} = 23.45\); the critical value \(X^2_{crit} = 9.48\), which is much smaller! The p-value is \(0.0001029\), very low as we would expect, indicating that the NULL Hypothesis should be rejected in favour of the alternate hypothesis, that proportions of DeathPenalty are affected by Education.

As always, we analyze our plots, and plot our analysis. Here is a plot for this test:

Show the Code
ggplot2::theme_set(new = theme_custom())

mosaic::xqchisq(
  p = 0.95, df = 4,
  return = c("plot"), verbose = F,
  system = "gg"
) %>%
  gf_labs(
    x = "F value", title = "Critical and Observed Chi-Square Values",
    x = ""
  ) %>%
  gf_vline(
    xintercept = X_squared_observed,
    color = "red", linewidth = 1
  ) %>%
  gf_vline(
    xintercept = X_squared_critical,
    color = "dodgerblue",
    linewidth = 1
  ) %>%
  gf_annotate(
    x = X_squared_observed - 2.5, y = 0.15,
    geom = "label", label = "Observed\n Chi-Square",
    fill = "red", alpha = 0.3
  ) %>%
  gf_annotate("curve", X_squared_observed - 2.5,
    y = 0.135, yend = 0.10,
    xend = X_squared_observed - 0.5,
    linewidth = 0.5, curvature = 0.3,
    arrow = arrow(length = unit(0.25, "cm"))
  ) %>%
  gf_annotate(
    x = X_squared_critical + 2.5, y = 0.15,
    geom = "label", label = "Critical\n Chi-Square",
    fill = "lightblue"
  ) %>%
  gf_annotate("curve", X_squared_critical + 2.5,
    y = 0.135, yend = 0.10,
    xend = X_squared_critical + 0.5,
    linewidth = 0.5, curvature = -0.3,
    arrow = arrow(length = unit(0.25, "cm"))
  ) %>%
  gf_refine(
    scale_y_continuous(expand = c(0, 0)),
    scale_x_continuous(expand = c(0, 0))
  )
Figure 3: Results of Chi-Square Test

Let us now dig into that cryptic-looking table above!

Let us once again look at the output of the xchisq.test() function:

xchisq.test(contingency_table)

    Pearson's Chi-squared test

data:  x
X-squared = 23.451, df = 4, p-value = 0.0001029

  117      511       71      135       64   
(129.86) (488.51) ( 59.78) (141.54) ( 78.33)
 [1.27]   [1.04]   [2.11]   [0.30]   [2.62] 
<-1.13>  < 1.02>  < 1.45>  <-0.55>  <-1.62> 
         
   72      200       16       71       50   
( 59.14) (222.49) ( 27.22) ( 64.46) ( 35.67)
 [2.79]   [2.27]   [4.63]   [0.66]   [5.75] 
< 1.67>  <-1.51>  <-2.15>  < 0.81>  < 2.40> 
         
key:
    observed
    (expected)
    [contribution to X-squared]
    <Pearson residual>

As per the key below, this is actually four 2-row tables interleaved together into one. The first row is the Observed counts, the second row is the Expected counts, if the two variables were independent. The third row is the Pearson Residuals, which we will explain shortly. The last row is the Contribution of each cell to the overall Chi-Square statistic.

Let us plot these tables and develop first a visual intuition, and then a mathematical one. If we pull apart the four tables, here are the mosaic plots for the actual and expected Contingency Tables, along with the association plot showing the differences, as we did when plotting Proportions:

(a) Actual
(b) Expected
(c) Tile-Wise Differences Contribute to Chi-Square Statistic
Figure 4: Mosaic Plots of Actual, Expected and Differences in Contingency Tables

In Figure 4 (a), we have the Observed Contingency Table as a mosaic, with its imbalances. The horizontal slice-line provides the overall proportion Favour::Oppose. The vertical slice-lines tell us that this proportion is not the same across levels of Education.

What if we straighten out the vertical slice-lines?

Figure 4 (b) does exactly this: we have a fictitious mosaic plot of an Expected Contingency Table, with straight vertical divisions, as if Education had no effect on the Favour::Oppose proportion. (NULL hypothesis!!)

The third plot, Figure 4 (c), is a tile-plot of the tile-wise differences between the two tables, which Contribute to the overall X-statistic. The blue tiles indicate that the Actual count is higher than the Expected count, and the red tiles indicate that the Actual count is lower than the Expected count. The intensity of the colour indicates the magnitude of the difference.

When we scale these differences by the standard deviation of each cell, we get the Pearson Residuals, which we will explain next.

Let us perform these computations manually, to see how this works. Here are all the four tables, side by side: (Are you tired of these yet?)

Show the Code
# xq_test_object <- xchisq.test(contingency_table)
# Already computed

xq_test_object$observed %>%
  addmargins() %>%
  as_tibble(rownames = "DeathPenalty") %>% # Convert to tibble; ensure row names!
  tt(caption = "Observed") %>%
  style_tt(color = "grey20") %>%
  style_tt(i = 1, j = 2, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 2, j = 5, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 3, j = 1:7, color = "black", bold = TRUE, background = "palegreen") %>%
  style_tt(j = 7, i = 1:3, color = "black", bold = TRUE, background = "palegreen")
xq_test_object$expected %>%
  addmargins() %>%
  as_tibble(rownames = "DeathPenalty") %>% # Convert to tibble; ensure row names!
  tt(digits = 2, caption = "Expected") %>%
  style_tt(color = "grey20") %>%
  style_tt(i = 1, j = 2, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 2, j = 5, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 3, j = 1:7, color = "black", bold = TRUE, background = "palegreen") %>%
  style_tt(j = 7, i = 1:3, color = "black", bold = TRUE, background = "palegreen")
xq_test_object$residuals %>%
  addmargins() %>%
  as_tibble(rownames = "DeathPenalty") %>% # Convert to tibble; ensure row names!
  tt(digits = 2, caption = "Pearson Residuals") %>%
  style_tt(color = "grey20") %>%
  style_tt(i = 1, j = 2, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 2, j = 5, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 3, j = 1:7, color = "black", bold = TRUE, background = "palegreen") %>%
  style_tt(j = 7, i = 1:3, color = "black", bold = TRUE, background = "palegreen")
xq_test_object$contribution %>%
  as.matrix() %>%
  as_tibble(rownames = "DeathPenalty") %>% # Convert to tibble; ensure row names!
  tt(digits = 2, caption = "Contribution to Chi-Square Statistic") %>%
  style_tt(color = "grey20") %>%
  style_tt(i = 1, j = 2, color = "black", bold = TRUE, background = "yellow") %>%
  style_tt(i = 2, j = 5, color = "black", bold = TRUE, background = "yellow")
Observed
DeathPenalty Left HS HS Jr Col Bachelors Graduate Sum
Favor 117 511 71 135 64 898
Oppose 72 200 16 71 50 409
Sum 189 711 87 206 114 1307
Expected
DeathPenalty Left HS HS Jr Col Bachelors Graduate Sum
Favor 130 489 60 142 78 898
Oppose 59 222 27 64 36 409
Sum 189 711 87 206 114 1307
Pearson Residuals
DeathPenalty Left HS HS Jr Col Bachelors Graduate Sum
Favor -1.1 1 1.5 -0.55 -1.6 -0.83
Oppose 1.7 -1.5 -2.2 0.81 2.4 1.2
Sum 0.54 -0.49 -0.7 0.26 0.78 0.4
Contribution to Chi-Square Statistic
DeathPenalty Left HS HS Jr Col Bachelors Graduate
Favor 1.3 1 2.1 0.3 2.6
Oppose 2.8 2.3 4.6 0.66 5.8
Table 3: All Contingency Tables
A. Expected Counts

How would we calculate the Expected table? The numbers that we might expect (under independence) in each cell there is the probability of an entry landing in that square times the total number of entries:

\[ \begin{align} \text{Expected Value[1,1]} &= p_{row_1} * p_{col_1} * Total~Scores\\\ &= \Large{\frac{\sum_{r_{1}}}{\sum_{r_{all}c_{all}}} * \frac{\sum_{c_{1}}}{\sum_{r_{all}c_{all}}} * \sum_{r_{all}c_{all}}} \\ &= \frac{898}{1307} * \frac{189}{1307} * 1307\\\ &= 130 \end{align} \]

Proceeding in this way for all the 15 entries in the Contingency Table, we get the “Expected” Contingency Table. Remember expect is another way of saying mean !!

B. Pearson Residuals

Now, the Pearson Residual in each cell is equivalent to the z-score of that cell. Recall the z-score idea: we subtract the mean and divide by the std. deviation to get the z-score.

  1. To get a z-score, we need the value, the mean, and sd.

  2. Now, the mean is the Expected value, as stated, and the value is the Actual value.

  3. What of the sd? In the Contingency Table, we have counts which are usually modeled as an (integer-valued) Poisson distribution, for which mean and variance are identical. And \(sd = \sqrt{variance}\). Thus we get the z-scores aka Pearson Residuals as:

    \[ r_{i,j} = \frac{(Actual - Expected)}{\sqrt{\displaystyle Expected}} \tag{3}\]

  4. The sum of all the squared Pearson residuals is the chi-square statistic, χ2, upon which the inferential analysis follows. We square them because we want to measure the magnitude of the difference, not the direction ( + or - ):

    \[ χ2 = \sum_{i=1}^R\sum_{j=1}^C{r_{i,j}^2} \tag{4}\]

    where R and C are number of rows and columns in the Contingency Table, the levels in the two Qual variables. So for cell-location [1,1], its contribution to χ2 would be: \((117-130)^2/130 = -1.13\). Do try to compute all of these and the \(X^2\) statistic by hand !!

  5. How did this \(X^2\) distribution come from? Here is a lovely, brief explanation from this StackOverflow Post:

  • In a Contingency Table the Null Hypothesis states that the variables in the rows and the variable in the columns are independent.
  • The (each) cell counts \(E_{ij}\) are assumed to be Poisson distributed with mean = \(E_{ij}\) and as they are Poisson, their variance is also \(E_{ij}\).
  • Asymptotically (when cell counts are large) the Poisson distribution approaches the normal distribution, with mean = \(E_{ij}\) and standard deviation with \(\sqrt{E_{ij}}\) so, asymptotically \(\large{\frac{(X_{ij} - E_{ij})}{\sqrt{E_{ij}}}}\) is approximately standard normal \(N(0,1)\).
  • If you square standard normal variables and sum these squares then the result is a chi-square random variable so \(\sum_{i,j}\left(\frac{(X_{ij}-E_{ij})}{\sqrt{E_{ij}}}\right)^2\) has a (asymptotically) a chi-square distribution.
  • Asymptotics must hold and that is why most textbooks state that the result of the test is valid when all expected cell counts \(E_{ij}\) are larger than 5, but that is just a rule of thumb that makes the approximation ‘’good enough’’.

Hence, after all this calculation, we have the \(X^2\) statistic, which we can compare with the critical value, and make our inference.

7 Permutation Test for Education

We will now perform the permutation test for the difference between proportions. We will first get an intuitive idea of the permutation, and then perform it using both mosaic and infer.

We saw from the diagram created by Allen Downey that there is only one test! We will now use this philosophy to develop a technique that allows us to mechanize several Statistical Models in that way, with nearly identical code. We will first look visually at a permutation exercise. We will create dummy data that contains the following case study:

A set of identical resumes was sent to male and female evaluators. The candidates in the resumes were of both genders. We wish to see if there was difference in the way resumes were evaluated, by male and female evaluators. (We use just one male and one female evaluator here, to keep things simple!)

         M 
-0.3333333

So, we have a solid disparity in percentage of selection between the two evaluators! Now we pretend that there is no difference between the selections made by either set of evaluators. So we can just:

  • Pool up all the evaluations
  • Arbitrarily re-assign a given candidate(selected or rejected) to either of the two sets of evaluators, by permutation.

How would that pooled shuffled set of evaluations look like?

 

As can be seen, the ratio is different!

We can now check out our Hypothesis that there is no bias. We can shuffle the data many many times, calculating the ratio each time, and plot the distribution of the differences in selection ratio and see how that artificially created distribution compares with the originally observed figure from Mother Nature.

ggplot2::theme_set(new = theme_custom())

null_dist <- do(4999) * diff(mean(
  candidate_selected ~ shuffle(evaluator),
  data = data
))
# null_dist %>% names()
null_dist %>%
  gf_histogram(~M,
    fill = ~ (M <= obs_difference),
    bins = 25, show.legend = FALSE,
    xlab = "Bias Proportion",
    ylab = "How Often?",
    title = "Permutation Test on Difference between Groups",
    subtitle = ""
  ) %>%
  gf_vline(xintercept = ~obs_difference, color = "red") %>%
  gf_label(500 ~ obs_difference,
    label = "Observed\n Bias",
    show.legend = FALSE
  )
mean(~ M <= obs_difference, data = null_dist)

 

[1] 0.00220044

We see that the artificial data can hardly ever (\(p = 0.0022\)) mimic what the real world experiment is showing. Hence we had good reason to reject our NULL Hypothesis that there is no bias.

We should now repeat the test with permutations on Education:

ggplot2::theme_set(new = theme_custom())

set.seed(42)
null_chisq <- do(4999) *
  chisq.test(mosaic::tally(DeathPenalty ~ shuffle(Education),
    data = GSS2002_modified
  ))

head(null_chisq)
gf_histogram(~X.squared, data = null_chisq) %>%
  gf_vline(
    xintercept = X_squared_observed,
    color = "red"
  ) %>%
  gf_labs(
    title = "Permutation Test on Chi-Square Statistic",
    x = "Chi-Square Statistic",
    y = "How Often?"
  ) %>%
  gf_annotate(
    geom = "label", y = 500, x = X_squared_observed,
    label = "Observed\n Chi-Square", fill = "moccasin"
  ) %>%
  gf_refine(
    scale_x_continuous(
      breaks = c(0, 5, 10, 15, 20, round(X_squared_observed, 2)),
      expand = expansion(mult = c(0, .1))
    ),
    scale_y_continuous(expand = c(0, 0))
  )

prop1(~ X.squared >= X_squared_observed, data = null_chisq)
prop_TRUE 
    2e-04

The p-value is well below our threshold of \(0.05\), so we would conclude that Education has a significant effect on DeathPenalty opinion!

8 Inference for Proportions Case Study-2: TBD dataset

To be Written Up. Yes, but when, Arvind?

9 Conclusion

In our basic \(X^2\) test, we calculate the test statistic of \(X^2\) and look up a theoretical null distribution for that statistic, and see how unlikely our observed value is.

Why would a permutation test be a good idea here? With a permutation test, there are no assumptions of the null distribution: this is computed based on real data. We note in passing that, in this case, since the number of cases in each cell of the Contingency Table are fairly high ( >= 5) the resulting NULL distribution is of the \(X^2\) variety.

10 Wait, But Why?

The \(X^2\) test is a very powerful tool for testing the independence of two categorical variables. It is widely used in various fields, including social sciences, biology, and market research. The test is based on the comparison of observed frequencies in a contingency table with expected frequencies under the assumption of independence.

When performing research for a design project, peasants, you are likely to have only several Qual variables, and you will want to see if there is a relationship between them. The \(X^2\) test is a great way to do that. It allows you to test the independence of two categorical variables, which can help you understand the relationships between different factors in your design project.

11 Your Turn

  1. DaytonSurvey: Substance Abuse among highschoolers. Part of the vcdExtra package. ( Install it, peasants!). dataset: https://friendly.r-universe.dev/vcdExtra/doc/manual.html#DaytonSurvey

    • Is there a relationship between Gender and Substance use?
    • Is there a relationship between Race and Substance use?

  2. Titanic dataset: https://www.kaggle.com/c/titanic/data

    • Is there a relationship between Sex and Survived?
    • Is there a relationship between Pclass and Survived?
    • Is there a relationship between Embarked and Survived?

    Can you prove that Jack would have survived, had Rose been a nice-R person?

  3. Gilby dataset: https://friendly.r-universe.dev/vcdExtra/doc/manual.html#Gilby

    • Is there a relationship between Clothing and Dullness?

  4. Can you show that people who wear ethnic are more likely to be considered dull? Find a dataset to prove or disprove this hypothesis!

12 References

  1. OpenIntro Modern Statistics: Chapter 17
  2. Chapter 8: The Chi-Square Test, from Statistics at Square One. The British Medical Journal. https://www.bmj.com/about-bmj/resources-readers/publications/statistics-square-one/8-chi-squared-tests. Very readable and easy to grasp. Especially if you like watching Grey’s Anatomy and House.
  3. Exploring the underlying theory of the chi-square test through simulation - part 1 https://www.rdatagen.net/post/a-little-intuition-and-simulation-behind-the-chi-square-test-of-independence/
  4. Exploring the underlying theory of the chi-square test through simulation - part 2 https://www.rdatagen.net/post/a-little-intuition-and-simulation-behind-the-chi-square-test-of-independence-part-2/
  5. An Online \(\Xi^2\)-test calculator. https://www.statology.org/chi-square-test-of-independence-calculator/
  6. https://saylordotorg.github.io/text_introductory-statistics/s13-04-comparison-of-two-population-p.html
R Package Citations
Package Version Citation
ggmosaic 0.3.3 Jeppson, Hofmann, and Cook (2021)
resampledata 0.3.2 Chihara and Hesterberg (2018)
scales 1.4.0 Wickham, Pedersen, and Seidel (2025)
vcd 1.4.13 Meyer, Zeileis, and Hornik (2006); Zeileis, Meyer, and Hornik (2007); Meyer et al. (2024)
Chihara, Laura M., and Tim C. Hesterberg. 2018. Mathematical Statistics with Resampling and r. John Wiley & Sons Hoboken NJ. https://github.com/lchihara/MathStatsResamplingR?tab=readme-ov-file.
Jeppson, Haley, Heike Hofmann, and Di Cook. 2021. ggmosaic: Mosaic Plots in the “ggplot2” Framework. https://doi.org/10.32614/CRAN.package.ggmosaic.
Meyer, David, Achim Zeileis, and Kurt Hornik. 2006. “The Strucplot Framework: Visualizing Multi-Way Contingency Tables with Vcd.” Journal of Statistical Software 17 (3): 1–48. https://doi.org/10.18637/jss.v017.i03.
Meyer, David, Achim Zeileis, Kurt Hornik, and Michael Friendly. 2024. vcd: Visualizing Categorical Data. https://doi.org/10.32614/CRAN.package.vcd.
Wickham, Hadley, Thomas Lin Pedersen, and Dana Seidel. 2025. scales: Scale Functions for Visualization. https://doi.org/10.32614/CRAN.package.scales.
Zeileis, Achim, David Meyer, and Kurt Hornik. 2007. “Residual-Based Shadings for Visualizing (Conditional) Independence.” Journal of Computational and Graphical Statistics 16 (3): 507–25. https://doi.org/10.1198/106186007X237856.
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Citation

BibTeX citation:
@online{v.2022,
  author = {V., Arvind},
  title = {Inference {Test} for {Two} {Proportions}},
  date = {2022-11-10},
  url = {https://madhatterguide.netlify.app/content/courses/Analytics/20-Inference/Modules/190-TwoProp/},
  langid = {en},
  abstract = {Inference Test for Two Proportions}
}
For attribution, please cite this work as:
V., Arvind. 2022. “Inference Test for Two Proportions.” November 10, 2022. https://madhatterguide.netlify.app/content/courses/Analytics/20-Inference/Modules/190-TwoProp/.