Tooth Growth Analysis

In this project, we are going to explore the ToothGrowth R dataset and do inferential data analysis aiming to answer the following questions:

1 - Is there a relation between tooth length and supplement used?

2 - Does changing the supplement dosage interfere with the tooth length?

The project is set in literate programming, allowing full research reproducibility.

Tools Used

• R language compiler
• R base graphic devices
• Tidyverse library packages
• RMarkdown library package
• Knitr library package

Files

• CODEBOOK: step-by-step book explaining the code processing.
• Figures: the plotted images
• Infsim.Rmd: the script to compile the project from source

CODEBOOK

Setup

knitr::opts_chunk$set(echo = TRUE, cache = TRUE, warning = FALSE)

Provide a basic summary of the data

Loading required packages and data set

require(tidyverse)

## Loading required package: tidyverse

## ── Attaching packages ───────────────────────────────────────────────────────────────────────────────── tidyverse 1.3.0 ──

## ✓ ggplot2 3.3.0     ✓ purrr   0.3.4
## ✓ tibble  3.0.1     ✓ dplyr   0.8.5
## ✓ tidyr   1.0.2     ✓ stringr 1.4.0
## ✓ readr   1.3.1     ✓ forcats 0.5.0

## ── Conflicts ──────────────────────────────────────────────────────────────────────────────────── tidyverse_conflicts() ──
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()

data <- ToothGrowth

Exploratory Data Analysis

Summary of the data

means <- data %>% 
    group_by(supp, dose) %>% 
    summarize(tooth.length = mean(len))

means

## # A tibble: 6 x 3
## # Groups:   supp [2]
##   supp   dose tooth.length
##   <fct> <dbl>        <dbl>
## 1 OJ      0.5        13.2 
## 2 OJ      1          22.7 
## 3 OJ      2          26.1 
## 4 VC      0.5         7.98
## 5 VC      1          16.8 
## 6 VC      2          26.1

# Classes
nclass = round(sqrt(length(data$len)))
hist(data$len, col = "slateblue1", main = "Histogram of Tooth Length", xlab = "Tooth Length", nclass = nclass)

# OJ - orange juice
# VC - vitamin C

Inferential Analysis

Use confidence intervals and/or hypothesis tests to compare tooth growth by supp and dose

Hypotesis 1 - Relation between tooth length and the supplement used

# Separate samples for analysis
supp = data$supp

len_by_OJ = data$len[supp == "OJ"] #sample_b
len_by_VC = data$len[supp == "VC"] #sample_a

delta = abs(mean(len_by_OJ) - mean(len_by_VC))
pooledsd = (sd(len_by_OJ) + sd(len_by_VC))/2
t1_error = 0.05

# Statistical Power

power.t.test(n = length(len_by_VC), sig.level = t1_error, delta = delta, sd = pooledsd, alternative = "one.sided", type = "two.sample")$power

## [1] 0.6024765

The statistical power for the t-test is too low!! (below .80)

The outcome of the t-test can result in a false positive!

# The Null Hypothesis
# H0: µa >= µb
# The Alternative Hypothesis
# Ha: µa < µb

# T-test: non-paired, variance based on approx of df, µa < H0
t.test(len_by_VC, len_by_OJ, conf.level = .95, var.equal = TRUE, paired = FALSE, alternative = "less")$p.value

## [1] 0.03019669

The p-value shows strong evidence to reject the null hypothesis (pval < 0.05), indicating high significance for the alternative hypothesis between the tooth length and the supplement used.

Hypothesis Conclusion

We can imply that the p-value for this observation is too high in relation to the threshold, and the statistical power of the analysis is too low, meaning that the alternative hypothesis may be a false positive.

Hypothesis 2 - Comparing tooth growth by dose

# Separate samples for analysis
half = data$len[data$dose == .5] # sample_a
one = data$len[data$dose == 1] # sample_b
two = data$len[data$dose == 2] # sample_c

length(half) == length(one)

## [1] TRUE

delta = abs(mean(half) - mean(one))
pooledsd = (sd(half) + sd(one))/2
t1_error = 0.05

# Statistical Power

power.t.test(n = length(half), sig.level = t1_error, delta = delta, sd = pooledsd, alternative = "one.sided", type = "two.sample")$power

## [1] 0.9999988

Very high Statistical Power, meaning the outcome of the t-test is very accurate!!

# The Null Hypothesis
# H0: µa >= µb
# The Alternative Hypothesis
# Ha: µa < µb

# T-test, non-paired, var by approx of df, µa < µb
t.test(half, one, alternative = "less", paired = FALSE, var.equal = TRUE, conf.level = .95)$p.value

## [1] 6.331485e-08

The p-value shows strong evidence to reject the null hypothesis (pval < 0.05), indicating proof for the alternative hypothesis between lower supplement doses and tooth length.

length(two) == length(one)

## [1] TRUE

delta = abs(mean(two) - mean(one))
pooledsd = (sd(two) + sd(one))/2
t1_error = 0.05

# Statistical Power

power.t.test(n = length(two), sig.level = t1_error, delta = delta, sd = pooledsd, alternative = "one.sided", type = "two.sample")$power

## [1] 0.9992657

Very high Statistical Power, meaning the outcome of the t-test is very accurate!!

# The Null Hypothesis
# H0: µc <= µb
# The Alternative Hypothesis
# Ha: µc > µb

# T-test, non-paired, var by approx of df, µc > µb
t.test(two, one, alternative = "greater", paired = FALSE, var.equal = TRUE, conf.level = .95)$p.value

## [1] 9.054143e-06

The p-value shows strong evidence to reject the null hypothesis (pval < 0.05), indicating proof for the alternative hypothesis between higher supplement doses and tooth length.

Hypothesis Conclusion

We can confirm with a 95% confidence interval that, for the true population distribution, the supplement dosage level interfere with tooth growth.

We can visualize this statement in the following plot:

ggplot(data, aes(dose, len, group = supp, fill = supp)) +
    geom_bar(stat = "identity", position = "dodge", color = "white") +
    labs(
        x = "Dose",
        y = "Tooth Length",
        fill = "Supplement"
    ) +
    coord_flip() +
    theme_minimal()