Basic statistical inference

Basic inference/Basic Inference.Rmd

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+---
+title: "Basic Inference"
+author: "Robinson Montes Gómez"
+date: "11/16/2020"
+output: pdf_document
+---
+
+```{r setup, include=FALSE}
+knitr::opts_chunk$set(echo = TRUE)
+```
+## PART 2: BASIC INFERENTIAL DATA ANALYSIS
+
+In this second part, we will analyze the ToothGrowth data. This data set has 60 observations on 3 variables; the first column is tooth length, the second is a factor (VC: ascorbic acid, a form of vitamin C and OJ: orange juice) and the last one is dose, that is the dose in milligrams/day received for each pig. The first step is to load the data and do some basic summary analysis
+
+```{r}
+data_pig <- ToothGrowth
+summary(data_pig)
+```
+
+Next, filter the data set by supp variable and them do the summary analysis for each factor
+
+```{r}
+pigs_VC <- filter(data_pig, data_pig$supp == 'VC')
+pigs_OJ <- filter(data_pig, data_pig$supp == 'OJ')
+summary(pigs_VC)
+```
+
+```{r}
+summary(pigs_OJ)
+```
+It's look like the pigs with the orange juice (OJ) treatment has highest length in his toot, let's prove calculating an interval for each mean with a confidence of 90%
+
+```{r}
+LS <- mean(pigs_VC[,1])-qt(0.025, length(pigs_VC[,1])-1)*sd(pigs_VC[,1])/(length(pigs_VC[,1])^0.5)
+LI <- mean(pigs_VC[,1])+qt(0.025, length(pigs_VC[,1])-1)*sd(pigs_VC[,1])/(length(pigs_VC[,1])^0.5)
+paste('(', round(LI, 4), ',', round(LS, 4), ')')
+```
+With a 95% of confidence, the conclusion is that the average length of the pigs' tooth is between 13.88 and 20.05.
+
+```{r}
+LS <- mean(pigs_OJ[,1])-qt(0.025, length(pigs_OJ[,1])-1)*sd(pigs_OJ[,1])/(length(pigs_OJ[,1])^0.5)
+LI <- mean(pigs_OJ[,1])+qt(0.025, length(pigs_OJ[,1])-1)*sd(pigs_OJ[,1])/(length(pigs_OJ[,1])^0.5)
+paste('(', round(LI, 4), ',', round(LS, 4), ')')
+```
+With a 95% of confidence, the conclusion is that the average length of the pigs' tooth is between 18.20 and 20.13.
+
+For the interval of difference of means, first let's calculate the standard deviation of the two samples (OJ, VC), to know if the samples have the sample variance
+
+```{r}
+sd_VC <- sd(pigs_VC[,1])
+sd_OJ <- sd(pigs_OJ[,1])
+paste('Sd VC = ', sd(pigs_VC[,1]), ', ', 'Sd OJ = ', sd(pigs_OJ[,1]))
+```
+Both samples looks like have different standard deviation, we have to compute the freedom degrees with the next formula:
+
+```{r}
+sd_VC <- sd(pigs_VC[,1])
+sd_OJ <- sd(pigs_OJ[,1])
+n_VC <- length(pigs_VC[,1])
+n_OJ <- length(pigs_OJ[,1])
+v <- (sd_VC/n_VC+sd_OJ/n_OJ)^2/((sd_VC/n_VC)^2/(n_VC-1)+(sd_OJ/n_OJ)^2/(n_OJ-1))
+round(v, 0)
+
+```
+Continue with the confidence interval of the difference of means
+
+```{r}
+LS <- mean(pigs_OJ[,1])-mean(pigs_VC[,1])-qt(0.025,v)*(sd_OJ/n_OJ+sd_VC/n_VC)^0.5
+LI <- mean(pigs_OJ[,1])-mean(pigs_VC[,1])+qt(0.025,v)*(sd_OJ/n_OJ+sd_VC/n_VC)^0.5
+paste('(', round(LI, 4), ',', round(LS, 4), ')')
+```
+
+As we see the values of the interval has the same sings, now we can conclude, with a confidence of 95% that there is difference in the length of pigs' tooth mean.