Statistical Inference Coursera Assignment Part 2

1.Load the ToothGrowth data and perform some basic exploratory data analyses

#load the plotting library
library(ggplot2)
# Load the data ToothGrowth
data(ToothGrowth)
# Look at the structure 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 ...

# Look at the first 5 rows of the data
head(ToothGrowth, 5)

##    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

2. Provide a basic summary of the data.

# Look at the summary of the data
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

# Compare means of the different delivery methods
tapply(ToothGrowth$len,ToothGrowth$supp, mean)

##       OJ       VC 
## 20.66333 16.96333

# Make a plot to look at data graphically
ggplot(ToothGrowth, aes(factor(dose), len, fill = factor(dose))) +
      geom_boxplot() +
      # facet_grid(.~supp)+
      facet_grid(.~supp, labeller = as_labeller(
            c("OJ" = "Orange juice", 
              "VC" = "Ascorbic Acid"))) +
      labs(title = "Tooth growth of 60 guinea pigs 
           by dosage and\nby delivery method of vitamin C",
           x = "Dose in milligrams/day", 
           y = "Tooth Lengh") +
      scale_fill_discrete(name = "Dosage of\nvitamin C\nin mg/day")

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

# Comparison by delivery method for the same dosage
t05 <- t.test(len ~ supp, 
       data = rbind(ToothGrowth[(ToothGrowth$dose == 0.5) & 
                                      (ToothGrowth$supp == "OJ"),],
                    ToothGrowth[(ToothGrowth$dose == 0.5) & 
                                      (ToothGrowth$supp == "VC"),]), 
       var.equal = FALSE)

t1 <- t.test(len ~ supp, 
       data = rbind(ToothGrowth[(ToothGrowth$dose == 1) & 
                                      (ToothGrowth$supp == "OJ"),],
                    ToothGrowth[(ToothGrowth$dose == 1) & 
                                      (ToothGrowth$supp == "VC"),]), 
       var.equal = FALSE)

t2 <- t.test(len ~ supp, 
       data = rbind(ToothGrowth[(ToothGrowth$dose == 2) & 
                                      (ToothGrowth$supp == "OJ"),],
                    ToothGrowth[(ToothGrowth$dose == 2) & 
                                      (ToothGrowth$supp == "VC"),]), 
       var.equal = FALSE)

# Make summary of the conducted t.tests, which compare the delivery methods by dosage
# take p-values and CI
summaryBYsupp <- data.frame(
      "p-value" = c(t05$p.value, t1$p.value, t2$p.value),
      "Conf.Low" = c(t05$conf.int[1],t1$conf.int[1], t2$conf.int[1]),
      "Conf.High" = c(t05$conf.int[2],t1$conf.int[2], t2$conf.int[2]),
      row.names = c("Dosage .05","Dosage 1","Dosage 2"))
# Show the data table 
summaryBYsupp

##                p.value  Conf.Low Conf.High
## Dosage .05 0.006358607  1.719057  8.780943
## Dosage 1   0.001038376  2.802148  9.057852
## Dosage 2   0.963851589 -3.798070  3.638070

4. State your conclusions and the assumptions needed for your conclusions.

With 95% confidence we reject the null hypothesis, stating that there is no difference in the tooth growth by the delivery method for .5 and 1 milligrams/day. We observe p-values less than the treshold of .05 and the confidence levels don’t include 0. So, for dosage of .5 milligrams/day and 1 milligrams/day does matter the delivery method. With 95% confidence we fail to reject the null hypothesis, stating that there is no difference in the tooth growth by the delivery method for 2 milligrams/day. We observe p-values more than the treshold of .05 and the confidence levels include 0. So, for dosage of 2 milligrams/day the delivery method doesn’t matter.