Question

7. This problem will us R. The following code will simulate flipping a coin 100 times. n 100 space <-c("H","T") p c0.5,0.5) observationample (space, size-n, prob-p, replace-TRUE) р.hat <-sum (observation.. "Н")/n р.hat (a) Run the code 5 times, and report the proportion of times a head showed up for each time you ran the code. Label them .aP.P (b) Now change n to 10000, n-10000. Run the code 5 times, and report the proportion of times a head showed up for each time you ran the code. Label them p, pa. Pa. P/P (c) Noting that the true proportion should be 0.5, what do you notice as n increses (d) Now, wet the weed to your computer to 55. Run the code 5 times report the proportion of times a head showed up for each time you ran the code set.seed (55) n<-100 space-H, T p c(0.5,0.5) observationample(apace, size-, prob-p, р.hat <-sun(observation.. "Н")/n p.hat replace-TRUE) (e) Is your answer the same or different than in part (a)? What do you notice when you set the weed?

Answer 1

a)

Code:

for(i in 1:5)

{

n<- 100

space < - c("H","T")

p<- c(0.5,0.5)

observation <- sample(space,size= n, prob = p, replace = TRUE)

p.hat<- sum(observation =="H")/n

cat("p.hat = ", p.hat,"\n")

}

Output:

> for(i in 1:5){

+ n<- 100

+ space <- c("H","T")

+ p<- c(0.5,0.5)

+ observation <- sample(space,size= n, prob = p, replace = TRUE)

+ p.hat<- sum(observation =="H")/n

+ cat("p.hat = ", p.hat,"\n")

+ }

p.hat = 0.48

p.hat = 0.45

p.hat = 0.5

p.hat = 0.54

p.hat = 0.46

b)            Code:

                                for(i in 1:5)

{

n<- 1000

space <- c("H","T")

p<- c(0.5,0.5)

observation <- sample(space,size= n, prob = p, replace = TRUE)

p.hat<- sum(observation =="H")/n

cat("p.hat = ", p.hat,"\n")

}

Output:

> for(i in 1:5)

+ {

+ n<- 1000

+ space <- c("H","T")

+ p<- c(0.5,0.5)

+ observation <- sample(space,size= n, prob = p, replace = TRUE)

+ p.hat<- sum(observation =="H")/n

+ cat("p.hat = ", p.hat,"\n")

+ }

p.hat = 0.519

p.hat = 0.508

p.hat = 0.478

p.hat = 0.498

p.hat = 0.477

c) We can see Output from (b) is closer to 0.5 than output from (a). So we noticed that the as n increases proportion of heads gets more closer to 0.5

d) Code:

for(i in 1:5)

{

set.seed(55)

n<- 100

space <- c("H","T")

p<- c(0.5,0.5)

observation <- sample(space,size= n, prob = p, replace = TRUE)

p.hat<- sum(observation =="H")/n

cat("p.hat = ", p.hat,"\n")

}

Output:

> for(i in 1:5)

+ {

+ set.seed(55)

+ n<- 100

+ space <- c("H","T")

+ p<- c(0.5,0.5)

+ observation <- sample(space,size= n, prob = p, replace = TRUE)

+ p.hat<- sum(observation =="H")/n

+ cat("p.hat = ", p.hat,"\n")

+ }

p.hat = 0.53

p.hat = 0.53

p.hat = 0.53

p.hat = 0.53

p.hat = 0.53

e) Answer in part (d) is different from answer in part (a)

When I set seed then I noticed that estimated proportion of head is constant for all 5 times run.

So can conclude that random sample of size n = 100 is same for all 5 times run.