First a simulation excercise, Second a inferential analysis excercise.
1.Simulation Excercises:
Tasks:
i.Sample mean and compare it to the theoretical mean
ii.How variable the sample is compared to the theoretical variance
iii.Show that the distribution is approximately normal
1.1 Investigate the exponential distribution in R and compare it with the Central Limit Theorem.
lambda <- 0.2 # sets the value for lambda in the instructions
n<- 1000 # number of simulatins
hist(rexp(1:n, rate = 1/lambda))# histogram of 1000 simulations
mean(rexp(1:n, rate = 1/lambda)) # mean of 1000 of abve.
## [1] 0.1989961
(sd(rexp(1:n, rate = 1/lambda)))^2 # variance of above
## [1] 0.0429375
1.2 Investigate the distribution of averages of 40 exponentials.
mns = NULL
for (i in 1 : 1000) mns = c(mns, mean(rexp(40, rate = 1/lambda))) # 1000 sims of 40 means
hist(mns) # histogram of above
mean(mns) #mean of above
## [1] 0.2001591
(sd(mns))^2 #variance of above
## [1] 0.0009468359
# Explains understanding of the differences of the variances: Variance of 1.2 is narrower than 1.1 because of the CLT.
2. Inferential Analysis
To analyze the ToothGrowth data in the R datasets.
Tasks:
i. Load the ToothGrowth data
ii. Provide a basic summary of the data.
data("ToothGrowth")
x <- ToothGrowth$len #vector of len
y<-mean(x) #mean len
(sd(x))^2 # variance of len
## [1] 58.51202
iii. Hypothesis tests to compare tooth growth by supp and dose
iv. Conclusions and the assumptions
y + c(-1, 1) * qnorm(0.975) * sd(x)/sqrt(length(x))# confidence interval of len
## [1] 16.87783 20.74884
t.test(len ~ supp, paired = FALSE, var.equal = TRUE, data = ToothGrowth)#Test Ho: Supps have no effect on len
##
## Two Sample t-test
##
## data: len by supp
## t = 1.9153, df = 58, p-value = 0.06039
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.1670064 7.5670064
## sample estimates:
## mean in group OJ mean in group VC
## 20.66333 16.96333
ToothGrowt12 <- subset(ToothGrowth, dose %in% c(1, 2))# new ssubset with doses 1 & 2
t.test(len ~ dose, paired = FALSE, var.equal = TRUE, data = ToothGrowt12)#Test Ho: Doses 1&2 have same effect on len
##
## Two Sample t-test
##
## data: len by dose
## t = -4.9005, df = 38, p-value = 1.811e-05
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -8.994387 -3.735613
## sample estimates:
## mean in group 1 mean in group 2
## 19.735 26.100
# Assumptions: paired = FALSE, var.equal = TRUE
# Conclusions: Because it's not significant at .05, we can't reject Ho: Supps have no effect on len. Because it's significant at .05 we reject Ho: Doses 1&2 have same effect on len