Question 1

Please explain CLT in your own words. You can, and should read your textbook and/or online references to understand what is CLT, its uses, et ctera. Furthermore, if you find any useful resource, include it in your post so that the rest of the class can have a look at it to. EG - Aspect 3 in the section Overview of four aspects. (1:16) in the YouTube video substantiates my claim more formally that xxx…

It is the theory that when you take the mean of all the samples of a large enough population, with a finite level of variance, it will roughly equal the mean of the population.

Click here: This site is a good resource

Question 2

Pick up any distribution apart from normal, uniform or poisson. You can Wikipedia about the distribution and/or read how to implement the distribution in R (what parameters are required to generate the distribution). Apply the CLT on the sample mean of this chosen distribution in R (adapt our class R code, or you can find alternative code on the web too). Does central limit theorem hold as expected? Elaborate.
Step 0
rm(list = ls()) 
gc()            
##          used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 515625 27.6    1138545 60.9         NA   669337 35.8
## Vcells 946738  7.3    8388608 64.0      16384  1839792 14.1
cat("\f")

##### Step 1

?rbinom # simulates a series of Bernoulli trials and return the results               

mybinom <- rbinom(1000, # observations
                  1000, # trials/observation
                  0.05  # prob of success
                  )
mybinom[1:16]
##  [1] 58 66 33 47 48 57 54 59 62 50 57 41 56 50 57 53
mu <- mean(mybinom)    
mu
## [1] 50.101
sigma <- sd(mybinom)   
sigma
## [1] 6.878817
library("psych")
describe(mybinom)     
##    vars    n mean   sd median trimmed  mad min max range skew kurtosis   se
## X1    1 1000 50.1 6.88     50   49.97 7.41  28  71    43 0.17    -0.01 0.22
hist(x    = mybinom, 
     main = "Histogram of Binomial Distribution (n=1000)",
     xlab = ""
     )

Step 2
?matrix     # creates a matrix from the given set of values.
?rep        # replicates the values in x lenth times

z <- matrix(data = rep(x     = 0, 
                       times = 1000
                       ), 
            nrow = 1000, 
            ncol = 1)

z[1:16]
##  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
describe(z)
##    vars    n mean sd median trimmed mad min max range skew kurtosis se
## X1    1 1000    0  0      0       0   0   0   0     0  NaN      NaN  0
Step 3
?sample # makes a sample of the specified size from the elements of x using either with or without replacement.
# sample(x, size, replace = FALSE, prob = NULL)

for (i in 1:1000){ 
    z[i,] <- mean(sample( x        = mybinom, 
                          size     = 100,    
                          replace  = TRUE    
                        )
                  )
  }
z[1:16]
##  [1] 50.43 48.47 50.34 49.99 50.82 50.50 51.23 50.12 49.16 49.45 50.96 49.43
## [13] 49.13 50.65 49.51 50.81
describe(z)
##    vars    n mean   sd median trimmed  mad   min   max range  skew kurtosis
## X1    1 1000 50.1 0.69  50.12    50.1 0.73 47.92 52.43  4.51 -0.04    -0.09
##      se
## X1 0.02
hist(z)

Step 4
z <- matrix(data = rep(x     = 0, 
                       times = 4000
                       ), 
            nrow = 1000, 
            ncol = 4)

n <- c(2, 6, 30, 100) 

for (j in 1:4){        
 
for (i in 1:1000){  # indexes the rows of matrix
  
    z[i,j] <- mean(sample( x       = mybinom,  # compute mean and assign
                           size    = n[j], 
                           replace = TRUE
                         )
                         )
    }
}

Step 5

colnames(z) <- c("Sample size=2", "Sample size=6", "Sample size=30", "Sample size=100") 
summary(z)  
##  Sample size=2   Sample size=6   Sample size=30  Sample size=100
##  Min.   :37.50   Min.   :40.17   Min.   :46.83   Min.   :47.91  
##  1st Qu.:47.00   1st Qu.:48.33   1st Qu.:49.27   1st Qu.:49.62  
##  Median :50.00   Median :50.17   Median :50.10   Median :50.09  
##  Mean   :50.19   Mean   :50.23   Mean   :50.10   Mean   :50.08  
##  3rd Qu.:53.00   3rd Qu.:52.33   3rd Qu.:50.97   3rd Qu.:50.56  
##  Max.   :66.00   Max.   :59.00   Max.   :54.93   Max.   :52.10
par(mfrow=c(3,2))
length(mybinom)
## [1] 1000
hist(x    = mybinom, 
     main = "Histogram of a Binomial Distribution, N=1000",
     xlab = ""
     )

?hist

for (k in 1:4){
    hist(x    = z[,k],     
         main = "Histogram of Binomial Distribution", 
         xlim = c(1, 100),
         xlab = paste0("Sample Size ", n[k], " (Column ", k, " from matrix)") 
         )
}

Question 3

Alternatively, note you can apply the CLT not only on the sample mean statistic, but also on any other sample statistics like say the median, 25th percentile or even the 80th percentile. Thus, instead of part 2 above, you can choose to apply the CLT on uniform or poisson distribution for any sample statistic except the mean (as we have seen that in class) and submit your code. This may be marginally harder than part 2 though, but you can try to submit both.

Not attempted