Basic Inferential Data Analysis

Doris Chen
2023-01-26

Overview

This is the Statistical Inference Course Project Part 2. In this project, we are going to do some basic inferential data analysis with the ToothGrowth data in R datasets package.

Loading the packages and the data

library(tidyverse)

## ── Attaching packages ─────────────────────────────────────── tidyverse 1.3.2 ──
## ✔ ggplot2 3.4.1      ✔ purrr   1.0.1 
## ✔ tibble  3.1.8      ✔ dplyr   1.0.10
## ✔ tidyr   1.3.0      ✔ stringr 1.5.0 
## ✔ readr   2.1.3      ✔ forcats 0.5.2 
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag()    masks stats::lag()

library(datasets)
data("ToothGrowth")

Looking at the data

First, we check the structure, first few rows and summary of the data.

str(ToothGrowth)

## '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(ToothGrowth)

##    len supp dose
## 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 10.0   VC  0.5

summary(ToothGrowth)

##       len        supp         dose      
##  Min.   : 4.20   OJ:30   Min.   :0.500  
##  1st Qu.:13.07   VC:30   1st Qu.:0.500  
##  Median :19.25           Median :1.000  
##  Mean   :18.81           Mean   :1.167  
##  3rd Qu.:25.27           3rd Qu.:2.000  
##  Max.   :33.90           Max.   :2.000

Then, we check the count of subjects taking different dose of different supplements.

ToothGrowth %>% 
        ggplot(aes(x=dose))+
        geom_histogram(aes(fill=supp), binwidth = 0.1)+
        facet_wrap(.~ supp)

As can be seen from the plot, the count of subjects taking different doses of different supplements are the same (10 for each).

Last, we check the tooth growth of subjects taking different dose of different supplements.

ToothGrowth %>% 
        ggplot(aes(x=dose, y=len))+
        geom_boxplot(aes(group=dose, fill=supp))+
        facet_wrap(.~ supp)

The plot shows that, as the dose increases, the tooth growth increases too.

Basic Summary:

It’s a data frame with 60 observations on 3 variables:

The variable “len” is numeric, showing the length of tooth growth, ranging from 4.2 to 33.9;
The variable “supp” is factor, with two levels – OJ and VC – each has 30 observations; * The variable “dose” is numeric, with the values, 0.5, 1 and 2;
The number of subjects taking different dose (0.5, 1.0 or 2.0) of different supplements (OJ / VC) are the same (10 for each);
The boxplot shows that as the dose increases, the tooth growth increases too.

Comparing tooth growth by supp and dose

1. Comparing tooth growth by supp

We will do hypothesis test to compare tooth growth with same doses of different supplements.

First, we subset six groups of data by supp and dose for the pairwise comparison.

oj_0.5 <- ToothGrowth[which(ToothGrowth$supp=="OJ" & ToothGrowth$dose==0.5),1]
oj_1 <- ToothGrowth[which(ToothGrowth$supp=="OJ" & ToothGrowth$dose==1),1]
oj_2 <- ToothGrowth[which(ToothGrowth$supp=="OJ" & ToothGrowth$dose==2),1]
vc_0.5 <- ToothGrowth[which(ToothGrowth$supp=="VC" & ToothGrowth$dose==0.5),1]
vc_1 <- ToothGrowth[which(ToothGrowth$supp=="VC" & ToothGrowth$dose==1),1]
vc_2 <- ToothGrowth[which(ToothGrowth$supp=="VC" & ToothGrowth$dose==2),1]

Then we proceed the following hypothesis test:

(1) 0.5 dose of OJ versus 0.5 dose of VC

H₀: Mean_{oj_0.5} = Mean_{vc_0.5} (The average tooth growth between two groups– one taking 0.5 dose of OJ, the other taking 0.5 dose of VC – are equal.)
H₁: Mean_{oj_0.5} != Mean_{vc_0.5} (The mean tooth growth between two groups– one taking 0.5 dose of OJ, the other taking 0.5 dose of VC – are not equal.)
We assume their variance are not equal.

t.test(oj_0.5, vc_0.5, paired = FALSE, var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  oj_0.5 and vc_0.5
## t = 3.1697, df = 14.969, p-value = 0.006359
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  1.719057 8.780943
## sample estimates:
## mean of x mean of y 
##     13.23      7.98

The p-value=0.006, which is less than 0.05, we reject H₀. The 95% confidence interval is 1.72 to 8.78.

(2) 1 dose of OJ versus 1 dose of VC

H₀: Mean_{oj_1} = Mean_{vc_1}
H₁: Mean_{oj_1} != Mean_{vc_1}
We assume their variance are not equal.

t.test(oj_1, vc_1, paired = FALSE, var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  oj_1 and vc_1
## t = 4.0328, df = 15.358, p-value = 0.001038
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  2.802148 9.057852
## sample estimates:
## mean of x mean of y 
##     22.70     16.77

The p-value=0.001, which is much less than 0.05, we reject H₀. The 95% confidence interval is 2.8 to 9.06.

(3) 2 doses of OJ versus 2 doses of VC

H₀: Mean_{oj_2} = Mean_{vc_2}
H₁: Mean_{oj_2} != Mean_{vc_2}
We assume their variance are not equal.

t.test(oj_2, vc_2, paired = FALSE, var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  oj_2 and vc_2
## t = -0.046136, df = 14.04, p-value = 0.9639
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -3.79807  3.63807
## sample estimates:
## mean of x mean of y 
##     26.06     26.14

The p-value=0.96, which is greater than 0.05, we fail to reject H₀. The 95% confidence interval is -3.8 to 3.64. Also, the mean value of the two groups are very close, with the number of 26.06 and 26.14 respectively.

2. Comparing tooth growth by dose

Now we want to analysis the efficacy of dosage independent from the supplement.

First, we subset three groups with different doses of VC/OJ.

d_0.5 <- ToothGrowth[which(ToothGrowth$dose==0.5), 1]
d_1 <- ToothGrowth[which(ToothGrowth$dose==1), 1]
d_2 <- ToothGrowth[which(ToothGrowth$dose==2), 1]

Then we proceed the following hypothesis test:

(1) 0.5 dose versus 1 dose

H₀: Mean_{d_0.5} = Mean_{d_1}
H₁: Mean_{d_0.5} != Mean_{d_1}
We assume their variance are not equal.

t.test(d_0.5, d_1, paired=FALSE, var.equal=FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  d_0.5 and d_1
## 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=1.268e-07, which is much smaller than 0.05, we reject the H₀. The 95% confidence interval is -11.98 to -6.28.

(2) 0.5 dose versus 2 doses

H₀: Mean_{d_0.5} = Mean_{d_2}
H₁: Mean_{d_0.5} != Mean_{d_2}
We assume that their variance are not equal.

t.test(d_0.5, d_2, paired=FALSE, var.equal=FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  d_0.5 and d_2
## t = -11.799, df = 36.883, p-value = 4.398e-14
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -18.15617 -12.83383
## sample estimates:
## mean of x mean of y 
##    10.605    26.100

The p-value=4.398e-14, which is much smaller than 0.05, we reject the H₀. The 95% confidence interval is -18.16 to -12.83.

(3) 1 dose versus 2 dose

H₀: Mean_{d_1} = Mean_{d_2}
H₁: Mean_{d_1} != Mean_{d_2}
We assume that their variance are not equal.

t.test(d_1, d_2, paired=FALSE, var.equal=FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  d_1 and d_2
## 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=1.906e-05, which is much smaller than 0.05, we reject the H₀. The 95% confidence interval is -9 to -3.73.

Conclusions

Based on our assumptions that the groups are independent and the variance are equal, we draw the conclusions from the test as following:

1. Comparing by supplements

There are significantly difference between the groups taking different supplements with 0.5 or 1 dose. With 95% confidence, OJ works better in tooth growth than VC in this case;
However, there is no significantly difference between the groups taking different supplements when the dose goes up to 2. The 95% confidence interval is -3.8 to 3.64.

2. Comparing by dose

The average tooth growth of groups taking 0.5, 1 and 2 doses of VC/OJ are 10.6, 19.7, and 26.1 respectively. The hypothesis test showed that there are significantly difference between these groups;
We can say with 95% confidence that taking more dose of VC/OJ brings more tooth growth.