Statistical Inference - ToothGrowth Analysis

Overview

ToothGrowth dataset contains information about impact on length of tooth on intake of Orange Juice(OJ) and Vitamin C(VC) on Guinea Pigs.
Three doses of both the supplements are given i.e. 0.5, 1 & 2.
I will analyse the impact of these suppliments on the length of Tooth.

Exploratory Data Analysis

A plot is made with two different supplements. Also a line connecting mean value is drawn for each supplement.
A table is created containing mean and median for each supplement corresponding to each dose.
From EDA(in figure), I found that as the dose increases the length of tooth increases.

Summary of Tooth Length
dose	supp	meanLen	medLen
0.5	OJ	13.23	12.25
0.5	VC	7.98	7.15
1.0	OJ	22.70	23.45
1.0	VC	16.77	16.50
2.0	OJ	26.06	25.95
2.0	VC	26.14	25.95

Also in table for dose of 2 of each supplement mean is approximately equal.

For Code - refer Appendix

Confidence Interval Test and Hypothesis test.

I have done T-Confidence Interval Test(paired) for the data.For all the Hypothesis testing p-value >= 0.95.First, data for Orange Juice and Vitamin C is compared irrespective of dose.Assumption(Null Hypothesis) - Orange Juice and Vitamin C have same impact on tooth growth i.e. difference in means of both supplements is 0.

tVC <- ToothGrowth[1:30,]
tOJ <- ToothGrowth[31:60,]
tIntervalOJVC <- t.test(tOJ$len,tVC$len,paired = TRUE)

In the above test we found that 95% confidence interval for mean of differences is 1.4086586, 5.9913414.
It states that with .95 probability the mean of differences will lie in this interval and actual mean is 3.7.
P-value is 0.0025498 which is less than 0.95 so we reject the null hypothesis.
In this case(difference of Orange Juice & Vitamin C) the result is positive. It signifies that Orange Juice has large effect on length of Tooth than Vitamin C.

Second, at a dosage level comparison of two supplements is done.Assumption(Null Hypothesis) - Orange Juice and Vitamin C have same impact on tooth growth i.e. difference in means of both supplements for each dose is 0.For dose = 0.5

t05 <- filter(ToothGrowth, dose == 0.5)
tInterval05 <- t.test(filter(t05, supp == 'OJ')$len, filter(t05, supp == 'VC')$len, paired = TRUE)

In the above test we found that 95% confidence interval for mean of differences is 1.2634583, 9.2365417.
It states that with .95 probability the mean of differences will lie in this interval and actual mean is 5.25.
P-value is 0.015472 which is less than 0.95 so we reject the null hypothesis.
In this case(difference of Orange Juice & Vitamin C) the result is positive. It signifies that Orange Juice has large effect on length of Tooth than Vitamin C for 0.5 dose.

For dose = 1

t1 <- filter(ToothGrowth, dose == 1)
tInterval1 <- t.test(filter(t1, supp == 'OJ')$len, filter(t1, supp == 'VC')$len, paired = TRUE)

In the above test we found that 95% confidence interval for mean of differences is 1.9519109, 9.9080891.
It states that with .95 probability the mean of differences will lie in this interval and actual mean is 5.93.
P-value is 0.0082292 which is less than 0.95 so we reject the null hypothesis.
In this case(difference of Orange Juice & Vitamin C) the result is positive. It signifies that Orange Juice has large effect on length of Tooth than Vitamin C for 1 dose.

For dose = 2

t2 <- filter(ToothGrowth, dose == 2)
tInterval2 <- t.test(filter(t2, supp == 'OJ')$len, filter(t2, supp == 'VC')$len, paired = TRUE)

In the above test we found that 95% confidence interval for mean of differences is -4.3289765, 4.1689765.
It states that with .95 probability the mean of differences will lie in this interval and actual mean is -0.08.
P-value is 0.9669567 which is greater than 0.95 so we accept the null hypothesis i.e both the supplements have almost similar impact at this dose level.
In this case(difference of Orange Juice & Vitamin C) the result is negative. It signifies that Orange Juice has less effect on length of Tooth than Vitamin C for 2 dose.

Third, For a particular supplement, effect of increasing the dose.Assumption(Null Hypothesis) - Effect of increasing the dose from 0.5 to 1 is same as 1 to 2 for each supplement.For Vitamin C

tVC105 <- filter(tVC,dose==1)$len - filter(tVC,dose==0.5)$len
tVC21 <- filter(tVC,dose==2)$len - filter(tVC,dose==1)$len
tIntervalVC <- t.test(tVC21,tVC105,paired = TRUE)

For the above test, difference of length at 1, 0.5 and that at 2, 1 are taken and then they are compared with T confidence intervals.
For Vitamin C, the mean is 0.58, which is very less and confidence interval is -5.3223375, 6.4823375
P-value is 0.8290479 which is less than 0.95 so we reject the null hypothesis.
Above results shows that increasing Vitamin C dose from 1 to 2 has more impact than from 0.5 to 1.

For Orange Juice

tOJ105 <- filter(tOJ,dose==1)$len - filter(tOJ,dose==0.5)$len
tOJ21 <- filter(tOJ,dose==2)$len - filter(tOJ,dose==1)$len
tIntervalOJ <- t.test(tOJ21,tOJ105,paired = TRUE)

For the above test, difference of length at 1, 0.5 and that at 2, 1 are taken and then they are compared with T confidence intervals.
For Orange Juice, the mean is -6.11, which is negative and confidence interval is -14.3884822, 2.1684822
P-value is 0.1293383 which is less than 0.95 so we reject the null hypothesis.
Above results shows that increasing Orange Juice dose from 1 to 2 has very less impact than from 0.5 to 1.

Appendix

Code of Exploratory Data Analysis

For plotting the graph

library(ggplot2)
data("ToothGrowth")
g <- ggplot(ToothGrowth,aes(x = dose,y = len, color = supp))
g <- g + geom_point()
g <- g + stat_summary(aes(group=1),geom = "line",fun.y = mean, col="black", size = 1)
g + facet_grid(. ~ supp)

For creating the table

library(xtable); library(dplyr); library(knitr)
tGroup <- group_by(ToothGrowth,dose,supp)
tSummary <- kable(summarize(tGroup,meanLen = mean(len), medLen = median(len)),
format = "pandoc", caption = "Summary of Tooth Length")
tSummary