Statistical Inference Course Project 1
Date: 22/08/2020
Overview
In this project I 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. I will set lambda = 0.2 for all of the simulations. I will investigate the distribution of averages of 40 exponentials. Note that I will need to do a thousand simulations.
Simulations
# load neccesary libraries
library(ggplot2)
# set constants
lambda <- 0.2 # lambda for rexp
n <- 40 # number of exponetials
numberOfSimulations <- 1000 # number of tests
# set the seed to create reproducability
set.seed(11081979)
# run the test resulting in n x numberOfSimulations matrix
exponentialDistributions <- matrix(data=rexp(n * numberOfSimulations, lambda), nrow=numberOfSimulations)
exponentialDistributionMeans <- data.frame(means=apply(exponentialDistributions, 1, mean))
Sample Mean versus Theoretical Mean
The expected mean \(\mu\) of a exponential distribution of rate \(\lambda\) is
\(\mu= \frac{1}{\lambda}\)
mu <- 1/lambda
mu## [1] 5Let \(\bar X\) be the average sample mean of 1000 simulations of 40 randomly sampled exponential distributions.
meanOfMeans <- mean(exponentialDistributionMeans$means)
meanOfMeans## [1] 5.027126As you can see the expected mean and the avarage sample mean are very close
Sample Variance versus Theoretical Variance
The expected standard deviation \(\sigma\) of a exponential distribution of rate \(\lambda\) is
\(\sigma = \frac{1/\lambda}{\sqrt{n}}\)
The e
sd <- 1/lambda/sqrt(n)
sd## [1] 0.7905694The variance \(Var\) of standard deviation \(\sigma\) is
\(Var = \sigma^2\)
Var <- sd^2
Var## [1] 0.625Let \(Var_x\) be the variance of the average sample mean of 1000 simulations of 40 randomly sampled exponential distribution, and \(\sigma_x\) the corresponding standard deviation.
sd_x <- sd(exponentialDistributionMeans$means)
sd_x## [1] 0.8020334Var_x <- var(exponentialDistributionMeans$means)
Var_x## [1] 0.6432577As you can see the standard deviations are very close Since variance is the square of the standard deviations, minor differnces will we enhanced, but are still pretty close.
Distribution
Comparing the population means & standard deviation with a normal distribution of the expected values. Added lines for the calculated and expected means
As you can see from the graph, the calculated distribution of means of random sampled exponantial distributions, overlaps quite nice with the normal distribution with the expected values based on the given lamba
Statistical Inference Course Project 2
Overview
Load the ToothGrowth data and perform some basic exploratory data analyses - Provide a basic summary of the data. - Use confidence intervals and/or hypothesis tests to compare tooth growth by supp and dose. (Only use the techniques from class, even if thereโs other approaches worth considering) - State your conclusions and the assumptions needed for your conclusions.
Load Data
# load neccesary libraries
library(ggplot2)
library(datasets)
library(gridExtra)
library(GGally)
# The Effect of Vitamin C on Tooth Growth in Guinea Pigs
data(ToothGrowth)
toothGrowth <- ToothGrowth
toothGrowth$dose <- as.factor(toothGrowth$dose) # convert to factorBasic Summary of the data
str(toothGrowth)## 'data.frame': 60 obs. of 3 variables:
## $ len : num 4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
## $ supp: Factor w/ 2 levels "OJ","VC": 2 2 2 2 2 2 2 2 2 2 ...
## $ dose: Factor w/ 3 levels "0.5","1","2": 1 1 1 1 1 1 1 1 1 1 ...summary(toothGrowth)## len supp dose
## Min. : 4.20 OJ:30 0.5:20
## 1st Qu.:13.07 VC:30 1 :20
## Median :19.25 2 :20
## Mean :18.81
## 3rd Qu.:25.27
## Max. :33.90head(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 0.5table(toothGrowth$supp, toothGrowth$dose)##
## 0.5 1 2
## OJ 10 10 10
## VC 10 10 10
Do some analysis based on Analysis of Variance (ANOVA)
anova.out <- aov(len ~ supp * dose, data=toothGrowth)
summary(anova.out)## Df Sum Sq Mean Sq F value Pr(>F)
## supp 1 205.4 205.4 15.572 0.000231 ***
## dose 2 2426.4 1213.2 92.000 < 2e-16 ***
## supp:dose 2 108.3 54.2 4.107 0.021860 *
## Residuals 54 712.1 13.2
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The results show there is a notable interaction between the length (len) and dosage (dose) (F(1,54)=15.572;p<0.01) Also a very clear effect on length(len) by supplement type (supp) (F(2,54)=92;p<0.01). Last but not least there is a minor interaction between the combination of supplement type (supp) and dosage (dose) compared to the length (len) (F(2,54)=4.107;p<0.05).
TukeyHSD(anova.out)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = len ~ supp * dose, data = toothGrowth)
##
## $supp
## diff lwr upr p adj
## VC-OJ -3.7 -5.579828 -1.820172 0.0002312
##
## $dose
## diff lwr upr p adj
## 1-0.5 9.130 6.362488 11.897512 0.0e+00
## 2-0.5 15.495 12.727488 18.262512 0.0e+00
## 2-1 6.365 3.597488 9.132512 2.7e-06
##
## $`supp:dose`
## diff lwr upr p adj
## VC:0.5-OJ:0.5 -5.25 -10.048124 -0.4518762 0.0242521
## OJ:1-OJ:0.5 9.47 4.671876 14.2681238 0.0000046
## VC:1-OJ:0.5 3.54 -1.258124 8.3381238 0.2640208
## OJ:2-OJ:0.5 12.83 8.031876 17.6281238 0.0000000
## VC:2-OJ:0.5 12.91 8.111876 17.7081238 0.0000000
## OJ:1-VC:0.5 14.72 9.921876 19.5181238 0.0000000
## VC:1-VC:0.5 8.79 3.991876 13.5881238 0.0000210
## OJ:2-VC:0.5 18.08 13.281876 22.8781238 0.0000000
## VC:2-VC:0.5 18.16 13.361876 22.9581238 0.0000000
## VC:1-OJ:1 -5.93 -10.728124 -1.1318762 0.0073930
## OJ:2-OJ:1 3.36 -1.438124 8.1581238 0.3187361
## VC:2-OJ:1 3.44 -1.358124 8.2381238 0.2936430
## OJ:2-VC:1 9.29 4.491876 14.0881238 0.0000069
## VC:2-VC:1 9.37 4.571876 14.1681238 0.0000058
## VC:2-OJ:2 0.08 -4.718124 4.8781238 1.0000000The Tukey HSD analysis shows that there are significant differences between each of the groups in supp and dose Only the interactions between VC:0.5-OJ:0.5; VC:1-OJ:0.5; OJ:2-OJ:1; VC:2-OJ:1 and VC:2-OJ:2 are not significant
confint(anova.out)## 2.5 % 97.5 %
## (Intercept) 10.9276907 15.532309
## suppVC -8.5059571 -1.994043
## dose1 6.2140429 12.725957
## dose2 9.5740429 16.085957
## suppVC:dose1 -5.2846186 3.924619
## suppVC:dose2 0.7253814 9.934619print(model.tables(anova.out,"means"),digits=3)## Tables of means
## Grand mean
##
## 18.81333
##
## supp
## supp
## OJ VC
## 20.66 16.96
##
## dose
## dose
## 0.5 1 2
## 10.60 19.73 26.10
##
## supp:dose
## dose
## supp 0.5 1 2
## OJ 13.23 22.70 26.06
## VC 7.98 16.77 26.14Conclusions
There are clear indications that both the supplement as the dosage have clear indipendent effects on the length of teeth guinea pigs. More those means on avarage longer teeth. Supplement type has a clear influence too, but OJ has a greater avarage teethgrowth in combination with dosages 0.5 and 1 then for the VC supplement, while teeth length for the VC supplement vs the OJ in combiantion with dosage 2 has no significant effect (almost same mean & same confidence interval)
The fact remains however that these assumpionts are based on the facts:
- that the guinea pigs are repesentative for the population of guinea pigs,
- that dosage and supplement were randomly assigned and
- that the distribution of the means is normal.