Synopsis - Basic Inferential Data Analysis

This prject analyses ToothGrowth data in the R datasets package. It analyses ToothGrowth with regards to type of supplement and amount of dosage and concludes the effect of both on Tooth Growth.

Data Analysis

Load the ToothGrowth data and perform some basic exploratory data analyses
Provide a basic summary of the data.

data("ToothGrowth")
summary(ToothGrowth) #displaying a summary of Tooth Growth data

##       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

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

Plots

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

We will do the hypothesis test in two parts. A> First we will check the effect of supplement type on tooth growth. B> Second we will check the effect of dosage on tooth growth.

A> Effect of supplement type on Tooth Growth

H0: The supplement type (VC or OJ) has no effect on Tooth Growth (length)

pval <- round(t.test(len~supp,data=ToothGrowth)$p.value, 2)
confint <- round(t.test(len~supp,data=ToothGrowth)$conf.int,2)

The p-value for the t-test is 0.06 and the confidence interval is (-0.17, 7.57). The p-value is extremely close to the alpha value of 0.05 so we cannot concusively reject the null hypothesis. The confidence intervl has 0 in it so the test is not significant. Hence, we conclude that supplement type (VC or OJ) has no effect on Tooth Growth.

B> Effect of Dosage on Tooth Growth

H0: Dosage has no effect on Tooth Growth (Different Doses will have the same effect on Tooth Growth)

#Comparing the toothgrowth for dosages 0.5 and 1
pval_0.5_1 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 0.5],ToothGrowth$len[ToothGrowth$dose == 1], paired=FALSE)$p.value, 2)
confint_0.5_1 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 0.5],ToothGrowth$len[ToothGrowth$dose == 1], paired=FALSE)$conf.int, 2)
mean_0.5_1 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 0.5],ToothGrowth$len[ToothGrowth$dose == 1], paired=FALSE)$estimate, 2) 
names(mean_0.5_1) <- c("Mean of 0.5", "Mean of 1")

The p-value of the comparision for dosages 0.5 and 1 is 0 and the confidence interval is (-11.98, -6.28). The p-value is close to 0 so we can reject the null hypothesis. The confidence interval doesn’t contain 0 so our test statistic is significant. So Dosage has an effect on Tooth Growth. We can also see the means for the dosages 0.5 and 1 as 10.61, 19.73 respectively.

Similarly we can run the same analysis for dosages 0.5&2 and 1&2. The results are shown below.

Comparing the toothgrowth for dosages 0.5 and 2

#Comparing the toothgrowth for dosages 0.5 and 2
pval_0.5_2 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 0.5],ToothGrowth$len[ToothGrowth$dose == 2], paired=FALSE)$p.value, 2)
confint_0.5_2 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 0.5],ToothGrowth$len[ToothGrowth$dose == 2], paired=FALSE)$conf.int, 2)
mean_0.5_2 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 0.5],ToothGrowth$len[ToothGrowth$dose == 2], paired=FALSE)$estimate, 2) 
names(mean_0.5_2) <- c("Mean of 0.5", "Mean of 2")

The p-value of the comparision for dosages 0.5 and 2 is 0 and the confidence interval is (-18.16, -12.83). The p-value is close to 0 so we can reject the null hypothesis. The confidence interval doesn’t contain 0 so our test statistic is significant.So Dosage has an effect on Tooth Growth. We can also see the means for the dosages 0.5 and 1 as 10.61, 26.1 respectively.

#Comparing the toothgrowth for dosages 0.5 and 1
pval_1_2 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 1],ToothGrowth$len[ToothGrowth$dose == 2], paired=FALSE)$p.value, 2)
confint_1_2 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 1],ToothGrowth$len[ToothGrowth$dose == 2], paired=FALSE)$conf.int, 2)
mean_1_2 <- round(t.test(ToothGrowth$len[ToothGrowth$dose == 1],ToothGrowth$len[ToothGrowth$dose == 2], paired=FALSE)$estimate, 2) 
names(mean_1_2) <- c("Mean of 1", "Mean of 2")

The p-value of the comparision for dosages 0.5 and 1 is 0 and the confidence interval is (-9, -3.73). The p-value is close to 0 so we can reject the null hypothesis. The confidence interval doesn’t contain 0 so our test statistic is significant.So Dosage has an effect on Tooth Growth. We can also see the means for the dosages 1 and 2 as 19.73, 26.1 respectively.

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

The conclusions we draw by analysing the Tooth Growth data are the following: 1. The type of supplement i.e. Orange Juice or Vitamin C has no effect on Tooth Growth 2. We see that as the dosage is increased from 0.5 to 1 to 2 our mean estimates increase significantly. Based on the t-tests we can therefore conclude that Dosage has an effect on Tooth Growth. 3. Increase in Dosage is directly proportional to Tooth Growth.

Assumptions

Sample is representative of the entire population.
For the t-tests, the variances are assumed to be different for the two groups being compared.

Proj 1 - Statistical Inference - TOoth Growth data

Anupam Acharya

26 October 2018