Project - Statistical Inference - part 1

Sergio Vicente Simioni

May, 18, 2015

1. Synopsis
2. Data Processing
3. Results
4. Summary

Synopsis

This project 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. It will be investigated the distribution of averages of 40 exponentials.

Data processing

Expo_Distr <- NULL
lambda<- 0.2
z<- 40
d<- 1000
for ( i in 1:d) 
Expo_Distr <- c(Expo_Distr, mean(rexp(z,lambda)))

Results

1. Sample mean size 40 compared with the theoretical mean of the distribution.

theorectical_mean<- mean(rexp(d,lambda))
sample_mean<- mean(Expo_Distr)

print(theorectical_mean)

## [1] 4.957975

print(sample_mean)

## [1] 5.013667

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

theorectical_variance <- var(rexp(d,lambda))
sample_variance<- var(Expo_Distr)

print(theorectical_variance)

## [1] 22.69022

print(sample_variance)

## [1] 0.6174074

3. Show that the distribution is approximately normal.

z<- Expo_Distr
k<- rexp(1000, 0.2)

par(mfrow=c(1,2))

b<- hist(rexp(k, lambda), breaks=10, col="red", main= " Exponential Distribution", 
 xlab=expression(paste( ,mu, " ~ 5" ,"     ",  sigma^2, " ~ 25")))
abline(v = theorectical_mean, col = "blue", lwd = 2)

a<- hist(z, prob=T, breaks=10, col="red", main= " Sample Mean Distribution", 
         xlab=expression(paste( ,hat(x), " ~ 5" ,"        ",  hat(s)^2, " ~ 25/40")))
abline(v = sample_mean, col = "blue", lwd = 2)
lines(density(z, bw=.5),col = "blue", lwd = 1)

Summary

The exponential distribution graphic on the left was built with 1000 values and lambda = 0.2, its mean and variance are approximately 1/lambda and 1/(lambda^2) respectively ( ~ 5 and ~ 25). At the right was built a new distribution utilizing the mean of the samples from the exponential distribution, it is amazing to observe that the new distribution is normal, which mean is also ( ~ 5 ) and its variance has a proportionality with the population variance which is inversely proportional to the sample size, for this case the variance is (1/(lambda^2))/(size of samples) which represents 25/40 ~ 0.6.

May/19/15 SVS