OVERVIEW

This project has two parts:

Part 1: Simulation Exercise Instructions

In this project you will investigate the exponential distribution in R and compare it with the Central Limit Theorem. The exponential distribution can be simulated in R with rexp(n, lambda) where lambda is the rate parameter. The mean of exponential distribution is 1/lambda and the standard deviation is also 1/lambda. Set lambda = 0.2 for all of the simulations. You will investigate the distribution of averages of 40 exponentials. Note that you will need to do a thousand simulations.

Part 2: Basic Inferential Data Analysis

In the second portion of the project, we’re going to analyze the ToothGrowth data in the R datasets package.

Part 1 - SIMULATION

Part 1: Simulation Exercise Instructions

1) Illustrate via simulation and associated explanatory text the properties of the distribution of the mean of 40 exponentials. You should Show the sample mean and compare it to the theoretical mean of the distribution.

Setup work environment

## Warning: package 'knitr' was built under R version 3.3.3

## Warning: package 'ggplot2' was built under R version 3.3.3

## Warning: package 'dplyr' was built under R version 3.3.3

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

Make a dataframe of a 1000 rows of 40 observations (n=40) - lamda=.2 and find mean values for each row

a1<- rexp(40000,.2)
m1<- matrix(a1,1000)
d1<-tbl_df(m1)
mean1<-apply(d1,1,mean)

Look at theortical and actual: mean, standard deviation and variance. Note that the theortical mean is 1/lamda or 5. The actual mean is computed below. Note that the difference between the theortical mean and the actual mean is very small.

mean_actual<-mean(mean1)
mean_actual

## [1] 5.03249

Let’s graph the mean as a density histogram for the 1000 rows of observations

meandf<-data.frame(mean1)
g<-ggplot(meandf, aes(x=mean1))
plotmean<-g + geom_histogram(binwidth = .1,fill="yellow",color="black")+labs(x="Mean Values", y="Frequency",title="Simulated distribution of the mean for 1000 rows of 40 exponentials")
plotmean

2) Show how variable the sample is (via variance) and compare it to the theoretical variance of the distribution.

The theortical SD is 1/lamda*1/sqrt(n)=.7905694. The actual SD is computed below. Note that the difference between the theortical SD and the actual SD is very small. The theortical variance is SD^2 = .7905694^2 = .625. The actual variance is also computed below. Note that the difference between the theortical variance and the actual variance is also small.

sd_actual<-sd(mean1)
sd_actual

## [1] 0.7802573

var_actual<-var(mean1)
var_actual

## [1] 0.6088015

2) Show how variable the sample is (via variance) and compare it to the theoretical variance of the distribution and 3) Show that the distribution is approximately normal. Let’s compare simulated means with a stat_fuction line using theortical mean of 5 and theortical sd of .625.

g2<-ggplot(meandf, aes(x=mean1) )

plot2<- g2 +geom_histogram(binwidth=.1,fill="yellow",color="black",aes(y=..density..))+labs(x="Mean Values", y="Density",title="Compare distribution of the theortical mean (RedLine) vrs actual mean (GreenLine)") + stat_function(fun = dnorm, args = list(mean=5,sd=.7776059),color="red",size=1) + stat_density(geom="line",color="green",size=1) 
plot2

Note that the theortical and actual are similar

Part 2: Basic Inferential Data Analysis of “ToothGrowth data”

First load data

library(tidyr)

## Warning: package 'tidyr' was built under R version 3.3.3

data("ToothGrowth")
dtooth<-ToothGrowth
dftooth<-tbl_df(dtooth)

Overview/summary of “ToothGrowth”

summary(dftooth)

##       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

dftooth

## # A tibble: 60 × 3
##      len   supp  dose
##    <dbl> <fctr> <dbl>
## 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
## 7   11.2     VC   0.5
## 8   11.2     VC   0.5
## 9    5.2     VC   0.5
## 10   7.0     VC   0.5
## # ... with 50 more rows

Find mean of supply mode:

df2<-group_by(dtooth,supp)
df3<-df3<-summarize(df2,length=mean(len))
df3

## # A tibble: 2 × 2
##     supp   length
##   <fctr>    <dbl>
## 1     OJ 20.66333
## 2     VC 16.96333

Plot 3 - Tooth size vrs kind of supplement used

g3<-ggplot(dftooth, aes(x=supp,y=len))
plot3<-g3+geom_boxplot(aes(fill=supp))+labs(x="Type of Supplement", y="Length of Tooth", title="Length of Tooth vrs supplement type")
plot3

Plot 4 - Tooth size vrs Vitamin Dose

dftooth$dose<-as.factor(dftooth$dose)
g4<-ggplot(dftooth, aes(x=dose,y=len))
plot4<-g4+geom_boxplot(aes(fill=dose))+labs(x="Dose-milligrams", y="Length of Tooth", title="Length of Tooth vrs Dose Strength")
plot4

Use confidence intervals to compare tooth growth by supp.

t.test(dftooth$len[dftooth$supp=="OJ"], dftooth$len[dftooth$supp=="VC"], paired = FALSE, var.equal = FALSE)$conf.int

## [1] -0.1710156  7.5710156
## attr(,"conf.level")
## [1] 0.95

The above confience interval contains 0 therefore we can conclude that the supplement type has no impact on tooth growth

Use confidence intervals to compare tooth growth by dose.

t.test(dftooth$len[dftooth$dose==2], dftooth$len[dftooth$dose==1], paired = FALSE, var.equal = FALSE)$conf.int

## [1] 3.733519 8.996481
## attr(,"conf.level")
## [1] 0.95

The above confience interval does not contain 0 therefore we can conclude that the supplement type does impact on tooth growth - the higher dose has higher growth rate.

Statistical Inference Project

GaryFH

April 25, 2017

OVERVIEW

This project has two parts:

Part 1: Simulation Exercise Instructions

Part 2: Basic Inferential Data Analysis

In the second portion of the project, we’re going to analyze the ToothGrowth data in the R datasets package.

Part 1 - SIMULATION

Part 1: Simulation Exercise Instructions

1) Illustrate via simulation and associated explanatory text the properties of the distribution of the mean of 40 exponentials. You should Show the sample mean and compare it to the theoretical mean of the distribution.

Setup work environment

Make a dataframe of a 1000 rows of 40 observations (n=40) - lamda=.2 and find mean values for each row

Look at theortical and actual: mean, standard deviation and variance. Note that the theortical mean is 1/lamda or 5. The actual mean is computed below. Note that the difference between the theortical mean and the actual mean is very small.

Let’s graph the mean as a density histogram for the 1000 rows of observations

2) Show how variable the sample is (via variance) and compare it to the theoretical variance of the distribution.

2) Show how variable the sample is (via variance) and compare it to the theoretical variance of the distribution and 3) Show that the distribution is approximately normal. Let’s compare simulated means with a stat_fuction line using theortical mean of 5 and theortical sd of .625.

Note that the theortical and actual are similar

Part 2: Basic Inferential Data Analysis of “ToothGrowth data”

First load data

Overview/summary of “ToothGrowth”

Find mean of supply mode:

Plot 3 - Tooth size vrs kind of supplement used

Plot 4 - Tooth size vrs Vitamin Dose

Use confidence intervals to compare tooth growth by supp.

The above confience interval contains 0 therefore we can conclude that the supplement type has no impact on tooth growth

Use confidence intervals to compare tooth growth by dose.

The above confience interval does not contain 0 therefore we can conclude that the supplement type does impact on tooth growth - the higher dose has higher growth rate.