Independent
Means and Medians

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 independent means (\mu_1-\mu_2)

R Setup

library(tidyverse)
library(ssstats)

• Recall our Zootopia data,

zootopia <- read_csv("https://raw.githubusercontent.com/samanthaseals/SDSI/refs/heads/main/files/data/lectures/1-zootopia.csv")

Confidence Interval for \mu_1-\mu_2

• (1–\alpha)100% CI for a population mean, \mu_1-\mu_2

(\bar{y}_1 - \bar{y}_2) \pm t_{\alpha/2} \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2}}

• where
  • (\bar{x}_1-\bar{x}_2) is the point estimate for \mu_1-\mu_2
    • \bar{x}_i is the sample mean for group i
  • t_{\alpha/2} has \min(n_1-1, n_2-1) degrees of freedom
  • s_i^2 is the sample variance for group i
  • n_i is the sample size of group i

Hypothesis Test for \mu_1-\mu_2

Hypotheses

• \mu_i is the population mean for group i,

• \mu_0 is the hypothesized difference between the groups

Hypothesis Test for \mu_1-\mu_2

Test Statistic:

t_0 = \frac{(\bar{x}_1-\bar{x}_2)-\mu_0}{{\sqrt{\frac{s_1^2}{n_1} + \frac{s^2_2}{n_2}}}} \sim t_{\text{df}}

• where
  • \bar{x}_i is the mean for group i
  • \mu_0 is the hypothesized difference
  • s_i^2 is the sample variance for group i
  • n_i is the sample size of group i
    
  • \text{df} = \text{min}(n_1-1, n_2-1)

Statistical Inference on Two Means (R)

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

• Generic syntax:

1dataset_name |> independent_means(
2  outcome = outcome_variable,
3  grouping = grouping_variable,
4  alternative = "two" | "less" | "greater",
5  mu = hypothesized_difference,
6  alpha = alpha_level
  ) 

Example 1

• Okonkwo pings you on your first morning back from vacation: “Patrol runs early shifts while Desk runs later ones and I keep hearing officers gripe about who’s more rested. Tell me if there’s actually a difference in how much sleep our two groups are getting.”

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

  • Is there a difference in average nightly sleep between Patrol and Desk officers?

• First, we need to describe the data.

  • We start with point and interval estimates.

• Then, we will perform a hypothesis test to answer the question.

Example 1

• Point and interval estimates:

zootopia |> independent_means(outcome = sleep_hours,
                              grouping = department)

Independent means (Welch's t-test)

x̅[Desk] = 6.2667
x̅[Patrol] = 7.1444
Point estimate: x̅[Desk] − x̅[Patrol] = -0.8777

95% CI for μ[Desk] − μ[Patrol]: (-1.9269, 0.1713)

Hypotheses:
H₀: μ[Desk] − μ[Patrol] = 0
H₁: μ[Desk] − μ[Patrol] ≠ 0
Test statistic: t(21.52) = -1.737
p-value: 0.097
Conclusion: Fail to reject the null hypothesis (p = 0.097 ≥ α = 0.050)

Example 1

• Then, we will perform a hypothesis test to answer the question.

  • Is there a difference in average nightly sleep between Patrol and Desk officers?

• How do we translate this into hypotheses?

  • H_0: \mu_{\text{Patrol}} = \mu_{\text{Desk}}
  • H_1: \mu_{\text{Patrol}} \ne \mu_{\text{Desk}}

• For the independent_means() function, we will set

  • mu = 0
  • alternative = "two".

Example 1

• Is there a difference in average nightly sleep between Patrol and Desk officers?

zootopia |> independent_means(outcome = sleep_hours,
                              grouping = department,
                              mu = 0,
                              alternative = "two")

Independent means (Welch's t-test)

x̅[Desk] = 6.2667
x̅[Patrol] = 7.1444
Point estimate: x̅[Desk] − x̅[Patrol] = -0.8777

95% CI for μ[Desk] − μ[Patrol]: (-1.9269, 0.1713)

Hypotheses:
H₀: μ[Desk] − μ[Patrol] = 0
H₁: μ[Desk] − μ[Patrol] ≠ 0
Test statistic: t(21.52) = -1.737
p-value: 0.097
Conclusion: Fail to reject the null hypothesis (p = 0.097 ≥ α = 0.050)

Example 1

• Hypotheses

  • H_0: \mu_{\text{Desk}} = \mu_{\text{Patrol}}
  • H_1: \mu_{\text{Desk}} \ne \mu_{\text{Patrol}}

• Test Statistic and p-Value

  • t_0 = -1.737
  • p = 0.097

• 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 average sleep differs between Patrol and Desk officers.

Example 2

• Okonkwo’s back with another one: “Word around the precinct is that Patrol officers have more access to the public, so it’s naturally easier to run their side hustles. Let’s see what the data says – do Patrol officers earn more than Desk officers?”

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

  • Do patrol officers earn more than desk officers?

• First, we need to describe the data.

  • We start with point and interval estimates.

• Then, we will perform a hypothesis test to answer the question.

Example 2

• Point and interval estimates:

zootopia |> independent_means(outcome = hustle_earnings,
                              grouping = department)

Independent means (Welch's t-test)

x̅[Desk] = 31.6483
x̅[Patrol] = 40.0883
Point estimate: x̅[Desk] − x̅[Patrol] = -8.44

95% CI for μ[Desk] − μ[Patrol]: (-27.8211, 10.9411)

Hypotheses:
H₀: μ[Desk] − μ[Patrol] = 0
H₁: μ[Desk] − μ[Patrol] ≠ 0
Test statistic: t(17.45) = -0.917
p-value: 0.372
Conclusion: Fail to reject the null hypothesis (p = 0.372 ≥ α = 0.050)

Example 2

• Then, we will perform a hypothesis test to answer the question.

  • Do patrol officers earn more than desk officers?

• How do we translate this into hypotheses?

  • H_0: \mu_{\text{Desk}} \ge \mu_{\text{Patrol}}
  • H_1: \mu_{\text{Desk}} < \mu_{\text{Patrol}}

• For the independent_means() function, we will set

  • mu = 0
  • alternative = "greater".

Example 2

• Do patrol officers earn more than desk officers?

zootopia |> independent_means(outcome = hustle_earnings,
                              grouping = department,
                              mu = 0,
                              alternative = "less")

Independent means (Welch's t-test)

x̅[Desk] = 31.6483
x̅[Patrol] = 40.0883
Point estimate: x̅[Desk] − x̅[Patrol] = -8.44

95% CI for μ[Desk] − μ[Patrol]: (-27.8211, 10.9411)

Hypotheses:
H₀: μ[Desk] − μ[Patrol] ≥ 0
H₁: μ[Desk] − μ[Patrol] < 0
Test statistic: t(17.45) = -0.917
p-value: 0.186
Conclusion: Fail to reject the null hypothesis (p = 0.186 ≥ α = 0.050)

Example 2

• Hypotheses

  • H_0: \mu_{\text{Desk}} \ge \mu_{\text{Patrol}}
  • H_1: \mu_{\text{Desk}} < \mu_{\text{Patrol}}

• Test Statistic and p-Value

  • t_0 = -0.917
  • p = 0.186

• 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 patrol officers earn more in side hustles than desk officers.

Assumption on Two-Sample t

• We assume that the data has an approximate mound-shaped and symmetric distribution.

y_{11}, y_{12}, \ldots, y_{1n_1} \sim N(\mu_1, \sigma_1^2) y_{21}, y_{22}, \ldots, y_{2n_2} \sim N(\mu_2, \sigma_2^2)

• We will use a quantile-quantile plot (Q-Q plot) to check the assumption of normality.

  • We will also eyeball a histogram to further check the assumption of normality.

Checking Normality (R)

• We will use the independent_qq() function from library(ssstats) to assess normality.

dataset_name |> independent_qq(continuous = continuous_variable,
                               grouping = grouping_variable)

• This will provide the the Q-Q plots and the histograms for the two independent groups under consideration.

A Note About Homoskedasticity

• Recall the assumptions,

y_{11}, y_{12}, \ldots, y_{1n_1} \sim N(\mu_1, \sigma_1^2) y_{21}, y_{22}, \ldots, y_{2n_2} \sim N(\mu_2, \sigma_2^2)

• Note that we are not assuming \sigma_1^2 = \sigma_2^2.

  • The independent_means() function assumes \sigma_1^2 \ne \sigma_2^2.

Example 1

• Let’s check the normality assumption for the sleep data.

zootopia |> independent_qq(outcome = sleep_hours,
                           grouping = department)

Example 2

• Let’s check the normality assumption for the side hustle earnings data.

zootopia |> independent_qq(outcome = hustle_earnings,
                           grouping = department)

Confidence Interval for M_1-M_2

• The confidence interval is built on every possible pairwise difference between the two groups.

  • We examine the n_1 \times n_2 differences between the observations.

    • i.e., We take each observation in Group 1 and subtract each observation in Group 2.

• The median of that pool of differences is the Hodges–Lehmann estimate.

  • It estimates the typical shift between the two groups.

• The CI is then found by

  • ranking that entire pool of n_1 \times n_2 differences from smallest to largest,

  • then we pick two values from that ranked list as the lower and upper bounds.

Hypothesis Test for M_1-M_2

Hypotheses

• M_i is the population median for group i, and

• M_0 is the hypothesized difference between the groups

Hypothesis Test for M_1-M_2

Test Statistic

T_0 = \sum R_{\text{1}} - \frac{n_1(n_1+1)}{2}

• where
  • \sum R_1 is the sum of the ranks for the first group
  • n_1 is the sample size of the first group

Statistical Inference on Two Medians (R)

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

• Generic syntax:

1dataset_name |> independent_medians(
2  outcome = outcome_variable,
3  grouping = grouping_variable,
4  alternative = "two" | "less" | "greater",
5  m = hypothesized_difference,
6  alpha = alpha_level
  ) 

Example 1

• Then, we will perform a hypothesis test to answer the question.

  • Is there a difference in average (median) nightly sleep between Patrol and Desk officers?

• How do we translate this into hypotheses?

  • H_0: M_{\text{Patrol}} = M_{\text{Desk}}
  • H_1: M_{\text{Patrol}} \ne M_{\text{Desk}}

• For the independent_medians() function, we will set

  • m = 0
  • alternative = "two".

Example 1

zootopia |> independent_medians(outcome = sleep_hours,
                                grouping = department)

Independent medians

M[Desk] = 6.45
M[Patrol] = 7.35
Hodges-Lehmann estimate of median difference: -0.7

95% CI for M[Desk] − M[Patrol]: (-1.7, 0.1)

Hypotheses:
H₀: M[Desk] − M[Patrol] = 0
H₁: M[Desk] − M[Patrol] ≠ 0
Test statistic: W = 68
p-value: 0.092
Conclusion: Fail to reject the null hypothesis (p = 0.092 ≥ α = 0.050)

Example 1

• Hypotheses

  • H_0: M_{\text{Desk}} = M_{\text{Patrol}}
  • H_1: M_{\text{Desk}} \ne M_{\text{Patrol}}

• Test Statistic and p-Value

  • T_0 = 68
  • p = 0.092

• 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 median sleep differs between Patrol and Desk officers.

Example 2

• Then, we will perform a hypothesis test to answer the question.

  • Do patrol officers earn more than desk officers?

• How do we translate this into hypotheses?

  • H_0: M_{\text{Desk}} \ge M_{\text{Patrol}}
  • H_1: M_{\text{Desk}} < M_{\text{Patrol}}

• For the independent_means() function, we will set

  • mu = 0
  • alternative = "greater".

Example 2

zootopia |> independent_medians(outcome = hustle_earnings,
                              grouping = department, alternative = "less")

Independent medians

M[Desk] = 22.265
M[Patrol] = 34.17
Hodges-Lehmann estimate of median difference: -11.695

95% CI for M[Desk] − M[Patrol]: (-23.54, 0.7)

Hypotheses:
H₀: M[Desk] − M[Patrol] ≥ 0
H₁: M[Desk] − M[Patrol] < 0
Test statistic: W = 63
p-value: 0.029
Conclusion: Reject the null hypothesis (p = 0.029 < α = 0.050)

Example 2

• Hypotheses

  • H_0: M_{\text{Desk}} \ge M_{\text{Patrol}}
  • H_1: M_{\text{Desk}} < M_{\text{Patrol}}

• Test Statistic and p-Value

  • T_0 = 63
  • p = 0.029

• Rejection Region

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

• Conclusion and Interpretation

  • Reject H_0. There is sufficient evidence to suggest that median side hustle earnings are greater among those on patrol.

Wrap Up

• Using a t-test to examine the difference between two independent means (\mu_1-\mu_2) requires normality.

  • We check normality with a quantile-quantile plot (independent_qq()).

• The analysis you choose should match the needed method:

  • If we meet normality (independent_means()):

    • Interval estimate: CI for \mu_1-\mu_2
    • One-sample t-test for \mu_1-\mu_2

  • If we do not meet normality (independent_medians()):

    • Interval estimate: CI for M_1-M_2
    • One-sample sign test for M_1-M_2