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title Inferential Analysis of Tooth Growth Data
author Justin Z
date March 7, 2019
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html_document pdf_document
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Introduction

This document is for the Statistical Inference course from Johns Hopkins University within the Data Science Specialization on Coursera. This is part two of a two part assignment. The instructions say:

Now in the second portion of the project, we're going to analyze the ToothGrowth data in the R datasets package.

  1. Load the ToothGrowth data and perform some basic exploratory data analyses
  2. Provide a basic summary of the data.
  3. Use confidence intervals and/or hypothesis tests to compare tooth growth by supp and dose. (Only use the techniques from class, even if there's other approaches worth considering)
  4. State your conclusions and the assumptions needed for your conclusions.

The dplyr and ggplot2 packages are required to execute this code. The input for this document is the ToothGrowth dataset, and the output is this report which is generated from a markdown file using knitr.

Part 1: Load the data and perform some basic EDA

The first step is to get the data loaded and explore it:

# Load required packages
library(dplyr)
library(ggplot2)


# Load the ToothGrowth data and perform some basic exploratory data analysis
tooth.data <- as_tibble(ToothGrowth)  # Load data as a tibble
str(tooth.data)  # Display structure
## Classes 'tbl_df', 'tbl' and 'data.frame':	60 obs. of  3 variables:
##  $ len : num  4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
##  $ supp: Factor w/ 2 levels "OJ","VC": 2 2 2 2 2 2 2 2 2 2 ...
##  $ dose: num  0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 ...

head(tooth.data)  # View first 6 rows
## # A tibble: 6 x 3
##     len supp   dose
##   <dbl> <fct> <dbl>
## 1   4.2 VC      0.5
## 2  11.5 VC      0.5
## 3   7.3 VC      0.5
## 4   5.8 VC      0.5
## 5   6.4 VC      0.5
## 6  10   VC      0.5

It is recommended to check the dataset documentation with ?ToothGrowth for additional detail. The first variable, len, is the length of the cells responsible for tooth growth in guinea pigs. In the documentation the units of the len variable are not provided. Based on its values it seems possible that the measurements are in millimeters. The analysis will proceed on this assumption. The supp variable indicates the supplement type, either OJ for orange juice, or VC for ascorbic acid. The last variable dose indicates the dosage level of vitamin C that was received.

# Prepare data for plot
tooth.data <- mutate(tooth.data, dose = as.factor(dose))  # Make "dose" a factor

# Plot the data
print(qplot(x = dose, y = len, data = tooth.data, facets = ~ supp, color = dose,
            main = "The Effect of Vitamin C on Tooth Growth in Guinea Pigs",
            xlab = "Dose of Vitamin C [mg/day]", ylab = "Tooth Length [mm]"))

In the plot above higher dose appears to correlate with longer tooth length, and it is difficult to say if there is a correlation to supplement type.

Part 2: Provide a basic summary of the data

Next a basic summary of the data is provided

summary(tooth.data)
##       len        supp     dose   
##  Min.   : 4.20   OJ:30   0.5:20  
##  1st Qu.:13.07   VC:30   1  :20  
##  Median :19.25           2  :20  
##  Mean   :18.81                   
##  3rd Qu.:25.27                   
##  Max.   :33.90

The summary shows that the len variable ranges from 4.20 to 33.90 with a mean of 18.81. There are 60 total observations - 30 each with the two supplement types: orange juice and ascorbic acid. There are 20 observations each of the 3 dosage levels. Basically there are six groups and each of them has 10 observations.

Part 3: Compare tooth growth by supp and dose

Next, hypothesis tests will be used to compare tooth growth by supplement type and dosage amount. However, before proceeding the assumptions required for t-tests should be reviewed:

  1. Means of populations follow a normal distribution
  2. Variances of populations are equal
  3. Data are sampled independently from the populations

These assumptions will be checked before continuing; first check if data follow a normal distribution:

# Make plots to check normality
par(mfrow = c(1, 2))  # Setup plot space
hist(tooth.data$len, main = "Histogram of Tooth Length",
     xlab = "Tooth Length [mm]")  # Plot histogram

# Check a Q-Q Plot also
qqnorm(tooth.data$len)  # Plot data
qqline(tooth.data$len)  # Add the normal line

The histogram doesn't look very normal, but it's reasonably symmetric. A Q-Q plot is also shown, and while it isn't perfect, the t-test is robust to the normality assumption. The analysis will continue assuming the data are normal enough.

The second check is of the variance, which does appear to be reasonably equal in this case. However, the t.test() function is capable of compensating for unequal variance, so that method will be used to remain conservative.

Lastly, there is no reason to believe the data were not sampled independently. Additionally, there is no way to test for this assumption, so analysis will continue assuming that all conditions for the two-sided t-test are met.

The first hypothesis test investigates the effect of the supplement type. The hypotheses are stated below:

  • H_0: mu_oj = mu_vc (There is no significant difference in the means)
  • H_a: mu_oj != mu_vc (There is a significant difference in the means)

These will be evaluted using the t.test() function:

# The t.test function will be used to evaluate the hypotheses
t.test(x = filter(tooth.data, supp == "VC")$len,
       y = filter(tooth.data, supp == "OJ")$len)
## 
## 	Welch Two Sample t-test
## 
## data:  filter(tooth.data, supp == "VC")$len and filter(tooth.data, supp == "OJ")$len
## t = -1.9153, df = 55.309, p-value = 0.06063
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -7.5710156  0.1710156
## sample estimates:
## mean of x mean of y 
##  16.96333  20.66333

The p-value of this test is 0.06063, so the test is not significant at the 0.95 confidence level. There does not appear to be a significant difference between the supplement types.

Next, the effect of the dose will be explored, but the instructions specify that only methods taught in the course up to this point can be used. Because of this t-tests will be used again even though other methods could be superior.

The next test will compare the 0.5 dose level to the 1.0 dose level, and the hypotheses are stated below:

  • H_0: mu_05 = mu_10 (There is no significant difference in the means)
  • H_a: mu_05 != mu_10 (There is a significant difference in the means)

These are evaluated below:

# The t.test function will be used to evaluate the hypotheses
t.test(x = filter(tooth.data, dose == "0.5")$len,
       y = filter(tooth.data, dose == "1")$len)
## 
## 	Welch Two Sample t-test
## 
## data:  filter(tooth.data, dose == "0.5")$len and filter(tooth.data, dose == "1")$len
## t = -6.4766, df = 37.986, p-value = 1.268e-07
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -11.983781  -6.276219
## sample estimates:
## mean of x mean of y 
##    10.605    19.735

The p-value of this test is 1.268e-07 which is extreemly small and is significant at the 0.95 confidence level. One more test will be checked below.

The last test will compare the 1.0 dose level to the 2.0 dose level, and the hypotheses are stated below:

  • H_0: mu_10 = mu_20 (There is no significant difference in the means)
  • H_a: mu_10 != mu_20 (There is a significant difference in the means)

These are evaluated below:

# The t.test function will be used to evaluate the hypotheses
t.test(x = filter(tooth.data, dose == "1")$len,
       y = filter(tooth.data, dose == "2")$len)
## 
## 	Welch Two Sample t-test
## 
## data:  filter(tooth.data, dose == "1")$len and filter(tooth.data, dose == "2")$len
## t = -4.9005, df = 37.101, p-value = 1.906e-05
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -8.996481 -3.733519
## sample estimates:
## mean of x mean of y 
##    19.735    26.100

The p-value of this test is 1.906e-05 which is extreemly small and is significant at the 0.95 confidence level. Between these two tests it does appear that the dose level is significantly correlated with tooth length. A correction could be applied to compensate for the fact that two tests were conducted, but the p-values are low enough that the results would still be significant.

Part 4: State your conclusions and assumptions

The conclusions of this analysis are that tooth growth in guinea pigs does have a statistically significant correlation with the dosage level of vitamin C. Another conclusion is that tooth growth was not significantly correlated with the types of vitamin C used in the experiment.

The analysis only required two assumptions to reach these conclusions:

  1. Means of populations follow a normal distribution
  2. Data are sampled independently from the populations

While the t-test is robust to the normality assumption, plots were created above which showed that the data are reasonably normal. While the variance appears to be homogenous in this case, it was compensated for in the t-test which assumed unequal variance. The assumption for independent sampling seems reasonable as well, so the conclusions reached here are based on reasonable assumptions.