Vitamin C Dosis

Since the pigs were treated with different doses of Vitamin C, the first question we need to address is whether the amount of vitamin C administered to each pig does any difference in their tooth growth.

The different doses were:

## [1] 0.5 1.0 2.0

So, since there is not a control group, lets test the hypothesis that the doses matter: is the growth associated with the 2.0 mg doses greater than the growth associated with 0.5 doses?

\[H_0: \mu_{0.5 mg} = \mu_{2.0 mg} \leftrightarrow H_\alpha: \mu_{2.0 mg} > \mu_{0.5 mg} \]

Lets split the 2 groups and test the hypothesis through a one-sided T-test

g05 <- subset(dados$len, dados$dose == 0.5)
g2 <- subset(dados$len, dados$dose == 2)

t.test(g2, g05, alternative = "greater", paired = FALSE, var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  g2 and g05
## t = 11.799, df = 36.883, p-value = 0.00000000000002199
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  13.27926      Inf
## sample estimates:
## mean of x mean of y 
##    26.100    10.605

Notice the very small p-value.

Just for comparison, lets do this test by hand through its confidence interval:

n2 <- length(g2)
n05 <- length(g05)
sd2 <- sd(g2)
sd05 <- sd(g05)
se1 <- sqrt(((n2-1)*sd2^2 + (n05-1)*sd05^2) / (n2 + n05 -2))
se2 <- sqrt(1/n2 + 1/n05)
se <- se1*se2
ci <- mean(g2) - mean(g05) + c(-1,1)*qt(.95,df = (n2+n05 -2))*se
ci <- round(ci,2)
ci

## [1] 13.28 17.71

Notice that 0 isn’t in the confidence interval, so \(\mu_{0.5 mg} - \mu_{2.0 mg} \neq 0\), ever! The true difference of means will always be between 13.28, 17.71 mm.

We can infer by this 95% test that \(H_0\) must be rejected and therefore, a larger dosis results in larger tooth growth!

Vitamin C Delivery Method

What is the best delivery method?

The delivery methods are:

## [1] VC OJ
## Levels: OJ VC

OJ means Orange Juice while VC means ascorbic acid.

Tooth Growth by delivery method

Lets test the null hypothesis that the deliver methods yields the same results:

\[H_0: \mu_{juice} = \mu_{acid} \leftrightarrow H_\alpha: \mu_{juice} \neq \mu_{acid}\]

The first thing to do is to separate the data between the two delivery methods and then run a two-sided T-test on the groups.

juicers <- subset(dados$len, dados$supp=="OJ")
trippin <- subset(dados$len, dados$supp=="VC")

t.test(juicers, trippin, paired = FALSE, var.equal = FALSE, alternative = "two.sided")

## 
##  Welch Two Sample t-test
## 
## data:  juicers and trippin
## 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:
##  -0.1710156  7.5710156
## sample estimates:
## mean of x mean of y 
##  20.66333  16.96333

Interesting enough, the T-test yields a confidence interval and a p-value that led us to fail to reject \(H_0\).

Therefore, the two delivery methods produces the same results