Recap

So far, we've talked about performing interval estimation and hypothesis testing for means using

• simulation-based methods, such as bootstrap or direct simulation, and
• the Central Limit Theorem

In all cases so far, we've only compared one sample against a hypothesized value.

Two-sample inference for means

Suppose we have two (representative) samples, and wanted to either

• estimate the difference in means in the two populations

  • confidence interval for μ1−μ2

• Test the hypotheses

H0:μ1=μ2Ha:μ1≠μ2,

where μ1 and μ2 are the population means in groups 1 and 2.

Today's data

Some voice analysis terminology

• Jitter: frequency variation from cycle to cycle
• Shimmer: amplitude variation of the sound wave

Jitter and shimmer are affected by lack of control of vocal cord vibration, and pathological differences from average values may be indicative of Parkinson's Disease (PD).

(from Teixeira, Oliveira, and Lopes, 2013)

Question of interest

Is there a difference in average voice jitter between patients with Parkinson's disease (PD) and those who don't have Parkinson's disease (control group)?

parkinsons.csv contains repeated voice recordings from a number of patients, some with PD and some serving as non-PD controls (Erdogdu B et al.). For now, assume that all samples were taken independently from each other (this is not actually the case, but we'll make this assumption).

Jitter is given in milliseconds (ms), and shimmer is given in decibels (dB).

Bootstrap estimation

Let's construct the bootstrap distribution for the difference in means.

set.seed(2020)
parkinsons <- read_csv("data/parkinsons.csv")
library(infer)
boot_diffs <- parkinsons %>% 
  specify(jitter ~ status) %>%
  generate(reps = 1000, type = "bootstrap") %>% 
  calculate(stat = "diff in means",
            order = c("Healthy", "PD"))

Bootstrap estimation

Let's construct the bootstrap distribution for the difference in means.

CI for difference in means

Let's construct the bootstrap distribution for the difference in means.

boot_diffs %>% 
  summarize(lower = quantile(stat, 0.025),
            upper = quantile(stat, 0.975))

## # A tibble: 1 x 2
##      lower    upper
##      <dbl>    <dbl>
## 1 -0.00413 -0.00220

CI for difference in means

## # A tibble: 1 x 2
##      lower    upper
##      <dbl>    <dbl>
## 1 -0.00413 -0.00220

Interpretation: We are 95% confident that the mean voice jitter for people without Parkinson's disease is about 0.002 to 0.004 ms less than the mean voice jitter for those with Parkinson's disease.

Hypothesis testing

Let μP be the mean voice jitter among PD patients, and μH be the mean voice jitter among healthy patients. Let's test

H0:μP=μHHa:μP≠μH

If the two means are truly equal (i.e., if H0 is true), then the difference, μH−μP, should be zero.

Hypothesis testing

Let's construct the simulated null distribution for the difference in means, μH−μP. If the two means are truly equal (i.e., if H0 is true), then this difference should be zero.

null_dist <- parkinsons %>%
  specify(jitter ~ status) %>%
  hypothesize(null = "independence") %>%
  generate(reps = 1000, type = "permute") %>%
  calculate(stat = "diff in means", 
            order = c("Healthy", "PD"))

Hypothesis testing

Hypothesis testing

obs_diff <- parkinsons %>% 
  specify(jitter ~ status) %>% 
  calculate(stat = "diff in means", order = c("Healthy", "PD")) %>% 
  pull()
obs_diff

## [1] -0.00312321

null_dist %>% 
  filter(abs(stat) >= abs(obs_diff)) %>%
  summarise(p_val = n() / nrow(null_dist))

## # A tibble: 1 x 1
##   p_val
##   <dbl>
## 1     0

Conclusion

The p-value is very small, so we reject H0. The data provide sufficient evidence that there is a difference in the mean voice jitter between patients who have Parkinson's disease and those who don't have the disease.

Difference in means using CLT

CLT-based inference for a difference in means relies on the two-sample t-test for independent samples. Like the t-test we've seen before, the test statistic takes on the following form:

T=(ˉX1−ˉX2)−(μ1−μ2)^SEdiff

The test statistic depends on whether we can assume that the two groups have the same underlying variability in their observations.

The exact form of the test statistic under the null hypothesis, including the degrees of freedom, are a complicated fraction that no one calculates by hand. Let's let R handle this!

CLT: Difference in means

parkinsons %>%
  t_test(jitter ~ status, 
         mu = 0,
         order = c("Healthy", "PD"), 
         alternative = "two-sided", 
         conf_int = TRUE, conf_level = 0.95)

## # A tibble: 1 x 6
##   statistic  t_df      p_value alternative lower_ci upper_ci
##       <dbl> <dbl>        <dbl> <chr>          <dbl>    <dbl>
## 1     -5.96  187. 0.0000000124 two.sided   -0.00416 -0.00209

CLT: Difference in means

## # A tibble: 1 x 6
##   statistic  t_df      p_value alternative lower_ci upper_ci
##       <dbl> <dbl>        <dbl> <chr>          <dbl>    <dbl>
## 1     -5.96  187. 0.0000000124 two.sided   -0.00416 -0.00209