Two-Sample Proportions

Introduction

• In the last module, we reviewed statistical inference for one sample.

• In this module, we will focus on tests for two samples.

  • Two-sample independent means (\mu_1-\mu_2)
  • Two-sample dependent means (\mu_d)
  • Two-sample proportions (\pi_1-\pi_2)
  • Two-sample variances (\sigma^2_1/\sigma^2_2)

• In this lecture: two-sample proportions (\pi_1-\pi_2)

R Setup

library(tidyverse)
library(ssstats)

• Both examples today use a new dataset for two ZPD precincts, Meadowbrook and Coldharbor, each with 60 officers who just went through recertification.

zootopia <- read_csv("https://raw.githubusercontent.com/samanthaseals/SDSI/refs/heads/main/files/data/lectures/3-zootopia-props.csv")

Confidence Interval for \pi_1-\pi_2

• (1–\alpha)100% CI for a difference between population proportions, \pi_1-\pi_2

(\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}}

Hypothesis Test for \pi_1-\pi_2

• Hypotheses:

• \pi_i is the population proportion for group i

• \pi_0 is the hypothesized value of the difference in proportions (often 0, for “no change”).

Hypothesis Test for \pi_1-\pi_2

• Test Statistic

z_0 = \frac{\left( \hat{p}_1 - \hat{p}_2 \right)- d_0}{\sqrt{\hat{p}\left(1-\hat{p}\right)\left( \frac{1}{n_1}+\frac{1}{n_2} \right)}}

• where
  • \hat{p}_i is the sample proportion for group i,
  • n_i is the sample size for group i, and

Hypothesis Test for \pi_1-\pi_2

• Test Statistic

z_0 = \frac{\left( \hat{p}_1 - \hat{p}_2 \right)- d_0}{\sqrt{\hat{p}\left(1-\hat{p}\right)\left( \frac{1}{n_1}+\frac{1}{n_2} \right)}}

• where
  • \hat{p} is the pooled proportion, given by

\hat{p} = \frac{x_1+x_2}{n_1+n_2}

Statistical Inference on Two Proportions (R)

• We will use the independent_proportions() function from library(ssstats) to perform statistical inference.

• Generic syntax:

1dataset_name |> independent_proportions(outcome = binary_variable,
2                                        grouping = grouping_variable,
3                                        event = "Name of Event",
4                                        p = hypothesized_value,
5                                        alternative = "two" | "less" | "greater",
6                                        alpha = specified_alpha)

Example 1

• Meadowbrook and Coldharbor just finished the same taser recertification course, same instructors, same exam. Deputy Chief Okonkwo wants to verify that Meadowbrook and Coldharbor recertification rates look the same.

• We will use the zootopia dataset to answer the following question:

  • Does the taser recertification pass rate differ between Meadowbrook and Coldharbor?

• We’ll work with taser_recert and precinct.

Example 1

• Point and interval estimates:

zootopia |> independent_proportions(outcome = taser_recert,
                                    grouping = precinct,
                                    event = "Pass")

Independent proportions

p̂[Coldharbor] = 0.65
p̂[Meadowbrook] = 0.7
Point estimate: p̂[Coldharbor] − p̂[Meadowbrook] = -0.05

95% CI for π[Coldharbor] − π[Meadowbrook]: (-0.2174, 0.1174)

Hypotheses:
H₀: π[Coldharbor] − π[Meadowbrook] = 0
H₁: π[Coldharbor] − π[Meadowbrook] ≠ 0
Test statistic: z = -0.58
p-value: 0.559
Conclusion: Fail to reject the null hypothesis (p = 0.559 ≥ α = 0.050)

Example 1

• Does the taser recertification pass rate differ between Meadowbrook and Coldharbor?

• How do we translate this into hypotheses?

  • No direction or hypothesized value given.

• Hypotheses:

  • H_0: \pi_{\text{MB}} = \pi_{\text{CH}}
  • H_1: \pi_{\text{MB}} \ne \pi_{\text{CH}}

• For the independent_proportion() function, we will set

  • p = 0
  • alternative = "two".

Example 1

• Does the taser recertification pass rate differ between Meadowbrook and Coldharbor?

zootopia |> independent_proportions(outcome = taser_recert,
                                    grouping = precinct,
                                    event = "Pass")

Independent proportions

p̂[Coldharbor] = 0.65
p̂[Meadowbrook] = 0.7
Point estimate: p̂[Coldharbor] − p̂[Meadowbrook] = -0.05

95% CI for π[Coldharbor] − π[Meadowbrook]: (-0.2174, 0.1174)

Hypotheses:
H₀: π[Coldharbor] − π[Meadowbrook] = 0
H₁: π[Coldharbor] − π[Meadowbrook] ≠ 0
Test statistic: z = -0.58
p-value: 0.559
Conclusion: Fail to reject the null hypothesis (p = 0.559 ≥ α = 0.050)

Example 1

• Hypotheses

  • H_0: \pi_{\text{MB}} = \pi_{\text{CH}}
  • H_1: \pi_{\text{MB}} \ne \pi_{\text{CH}}

• Test Statistic and p-Value

  • t_0 = -0.58
  • p = 0.559

• Rejection Region

  • Reject H_0 if p < \alpha; \alpha = 0.05

• Conclusion and Interpretation

  • Fail to reject H_0. There is not sufficient evidence to suggest that the recertification rate is different between the two precincts.

Example 2

• Coldharbor got a new firearms range last year… the one Meadowbrook’s been asking for forever. If it’s actually helping, Coldharbor’s requalification numbers should be better. Let’s see if the budget was worth it.

• We will use the zootopia dataset to answer the following question:

  • Does Coldharbor have a higher firearms requalification pass rate than Meadowbrook?

• We’ll work with firearms_requal and precinct.

Example 2

• Point and interval estimates:

zootopia |> independent_proportions(outcome = firearms_requal,
                                    grouping = precinct,
                                    event = "Pass")

Independent proportions

p̂[Coldharbor] = 0.8333
p̂[Meadowbrook] = 0.6667
Point estimate: p̂[Coldharbor] − p̂[Meadowbrook] = 0.1667

95% CI for π[Coldharbor] − π[Meadowbrook]: (0.0146, 0.3187)

Hypotheses:
H₀: π[Coldharbor] − π[Meadowbrook] = 0
H₁: π[Coldharbor] − π[Meadowbrook] ≠ 0
Test statistic: z = 2.11
p-value: 0.035
Conclusion: Reject the null hypothesis (p = 0.035 < α = 0.050)

Example 2

• Does Coldharbor have a higher firearms requalification pass rate than Meadowbrook?

• How do we translate this into hypotheses?

  • Direction given: “higher”

• Hypotheses:

  • H_0: \pi_{\text{CH}} \le \pi_{\text{MB}}
  • H_1: \pi_{\text{CH}} > \pi_{\text{MB}}

• For the independent_proportion() function, we will set

  • p = 0
  • alternative = "greater".

Example 2

• Does Coldharbor have a higher firearms requalification pass rate than Meadowbrook?

zootopia |> independent_proportions(outcome = firearms_requal,
                                    grouping = precinct,
                                    event = "Pass",
                                    alternative = "greater")

Independent proportions

p̂[Coldharbor] = 0.8333
p̂[Meadowbrook] = 0.6667
Point estimate: p̂[Coldharbor] − p̂[Meadowbrook] = 0.1667

95% CI for π[Coldharbor] − π[Meadowbrook]: (0.0146, 0.3187)

Hypotheses:
H₀: π[Coldharbor] − π[Meadowbrook] ≤ 0
H₁: π[Coldharbor] − π[Meadowbrook] > 0
Test statistic: z = 2.11
p-value: 0.018
Conclusion: Reject the null hypothesis (p = 0.018 < α = 0.050)

Example 2

• Hypotheses

  • H_0: \pi_{\text{CH}} \le \pi_{\text{MB}}
  • H_1: \pi_{\text{CH}} > \pi_{\text{MB}}

• Test Statistic and p-Value

  • z_0 = 2.11
  • p = 0.018

• Rejection Region

  • Reject H_0 if p < \alpha; \alpha = 0.05

• Conclusion and Interpretation

  • Reject H_0. There is sufficient evidence to suggest that the requalification rate is higher in Coldharbor as compared to Meadowbrook.

Wrap Up

• Independent (two-sample) proportion data comes from two separate, unrelated groups.

  • We work with the difference in sample proportions, \hat{p}_1 - \hat{p}_2.

• Next: comparing two variances.